UBC Wiki - User contributions [en]

Electrostatics

2010-04-15T21:52:29Z

RaymondGoerke: /* Examples */

The electric field is a fundamental property of the universe. It is a vector field defined over all of space, which means that at every
point in space it has a value that is a vector with a direction and magnitude. Roughly speaking, this value represents the force per
unit charge that a point charge would experience if placed at that point.

The study of electrostatics is the study of the electric field in situations where charges are no longer permitted to move, or are so
small that their motions do not effect the electric field. Eventually you will learn about magnetostatics, which is the study of
constant, steady flows of charge in closed loops or long curves, and then if you study physics at a higher level you will learn about
electrodynamics, which deals with the generalized interactions between moving charges. For now though, it is useful to learn about the
simpler field of electrostatics.

The foundational principle of electrostatics which was derived for experiment from which we can derive all of electrostatics is the
definition of the electric field of a point particle of charge <math>q</math> at a point a distance <math>r</math> from the particle
in the direction <math>\hat{r}</math>

:<math>E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}</math>

We will see how this equation can be used to derive many more interesting properties of electrostatics.

==The Principle of Superposition==

The most important idea in electrostatics is the principle of superposition. This principle states that if you know the
electric field due to a specific configuration of charge and the electric field of some other configuration of charge, then the electric
field due to putting both of those configurations on top of each other is simply the sum of the electric fields of each individual
configuration.

The simplest posible case where we can use the principle of superposition is to calculate the electrostatic
field from a collection of point charges.

====Example====
[[File:efield2.jpg|right]]
'''Q''':A point charge <math>+q</math> is located a distance 1 from the origin along the positive x-axis.
Another point charge <math>-q</math> is located a distance 1 from the origin along the negative x-axis.
What is the electric field at a distance 1 from the origin along the y-axis (point A in the diagram)?

'''A''':We can calculate the electric field from each individual charge using Coulomb's Law. Then using
the principle of superposition, we know that the total electric field is the sum of the field from each
charge. [[File:efield1.jpg|right]]

Since point A is a total distance <math>\sqrt{2}</math> from each charge we use Coulomb's Law:

<math>E_+ = \frac{kq}{(\sqrt{2})^2} = \frac{kq}{2}</math>

<math>E_- = \frac{k(-q)}{(\sqrt{2})^2} = \frac{-kq}{2}</math>

Where <math>E_+</math> and <math>E_-</math> are the contributions from the positive and negative point
charges respectively.

When we add the contributions together we must remember that they are vectors. The value for the electric field
at contribution at A from the positive charge is a vector pointing along the direction between the positive charge
and A. Since the value is positive, the vector points towards A. The same is true for the contribution from the
negative charge, except that it is negative, and thus points away from A. As depicted in the diagram, the vertical
components of the vectors cancel each other and the horizontal components add. The result is an electric field
vector pointing to the left with magnitude (with <math>\sin(\pi /4) = \cos(\pi /4) = 1/\sqrt{2}</math>):

<math> E = \frac{kq}{2}\frac{1}{\sqrt{2}} + \frac{kq}{2}\frac{1}{\sqrt{2}} = \frac{kq}{\sqrt{2}}</math>

==Gauss's Law==

In principle, we can use the principle of superposition to calculate the electrostatic field of any charge
distribution, even solid objects with complicated shapes and some (not necessarily constant) charge density
(amount of charge per unit volume). The way this is acheived is by splitting up the object into very small
pieces which act alomst like a point charge and then adding their contributions. If we take the limit as the
pieces get smaller and smaller, we will get closer and closer to the real value of the electrostatic field.
Those who have taken second-term first year calculus may recognize this process as integration. Those who
have taken second year calculus will recognize this more specifically as a volume integral.

As you may expect this process is in general very difficult, and Physics 102 you will only need to worry
about point charges like in the previous examples. But there exists a special technique for calculating
the electric field of more complicated charge distributions more easily in situations with lots of
symmetry. This technique uses Gauss's Law.

Gauss's Law is a very general law of electrostatics and is in fact one of the four famous Maxwell Equations,
which when taken together desribe the entirely of classical electromagnetism. It is possible to derive
Guass's Law from Coulomb's Law, but this requires a complicated calculation which requires integral
calculus. It is however very easy to derive Coulomb's Law from Guass's Law, and as a result Guass's Law is
usually considered to be more fundamental, even though Coulombs's Law was discovered through experiment.

Guass's Law states that the electric flux through any closed surface is proportional to the net charge
enclosed by the surface. More specifically:

<math>\oint \mathbf{E} \cdot d\mathbf{a} = 4\pi k Q_{enc}</math> [[File:eflux2.gif|right]]

Where this denotes a surface integral. A surface integral consists of splitting a surface
into many small, approximately flat and square pieces. For each piece, we define a vector
perpendicular to the surface and pointing outwards with magnitude equal to the area of the
square. Then taking the dot product with the electric field effectively multiples the
component of the electric field perpendicular to the surface by the area of the small piece.
Taking the sum of all these products and taking the limit as the squares get very small
gives us a surface integral, which is in some sense a measure of how much the electric field
points out of the surface, as it ignores the component that is points along the surface.

One useful way to think about a surface integral is to imagine that the electric field is actually
representing the flow of water, which the magnitude and direction of the electric field represent the
speed and direction of water flowing through that point. In this case the electric flux through a
surface is how much water flows out of the surface per unit time. A positive electric flux means
water is flowing out, a negative electric flux means water is flowing in, and a electric flux of zero
means that the total flow in equals the total flow out.

Continuing with the water flow analogy, we can consider the implication of Gauss's Law, which states
that the electric flux through a closed surface is proportional to the enclosed charge. This is equivalent
to saying (in out water analogy) that if there is a net flow of water out of a closed surface, then
there must be some source (such as a faucet constantly pouring out water), and that if there is more
outward flow then there must either be more faucets or they must be pouring out water faster. Similarly,
if there is a net flow of water into a surface, then there must be a sink constantly sucking up water,
and that if there is more inward flow then there must be more sinks or they must be sucking up water
faster. In terms of the electric field positive charges play the role of faucets and negative charges
play the role of sinks. If there is a positive electric flux through a closed surface, then there
must be a net positive charge inside, and similarly a negative flux through a closed surface means
there must be a net negative charge inside.

The fascinating property of Gauss's Law is that it does not depend on where inside the surface the
charges are locate, or even how the surface is shaped. If two completely different closed surfaces
contain the same net charge, the electric flux through them will be the same. If the charges are
distributed in completely different configurations, so long as the total net charge is the same,they
will have the same electric flux. We can use this property to easily calculate the electric field
due to charge distributions in which is it easy to calculate the electric flux. We will see some
examples.

====Examples====

'''Q''': Derive Coulombs Law using Gauss's Law. [[File:spherefluxpic3.jpg|right]]

'''A''': Suppose there is a point charge <math>q</math>. Imagine a surface surrounding the point with
radius <math>r</math>. We must calculate the electric flux through this surface from from the point
charge. First we note that the electric field on the surface of the sphere must always be perpendicular
to the sphere. We can say this not because we secretly already know Coulomb's Law and thus know the
electric field must point radially out from the point charge, but by appealing to symmetry. It is worth
labouring a bit on the concept of symmetry because it is an important tool in physics: Suppose that
the electric field at the surface were not perpendicular to the surface. Then it must be pointing in
some particular direction, say slightly to the left. But then we could simply walk around to the other
side of the sphere and it would look like the field was pointing slightly to the right. In both views
however, the physical situation was identical since the point charge still is at the center
of the sphere. If the charge were not in the center we could not make this argument, since when you moved
to the other side, the electric field would change direction, but the charge would have flipped sides as
well. It is a general principle of physics that the Laws of Physics should be true no matter
what angle you look at the situation. Thus we can conclude that the above situation cannot be true if the
charge is at the center and that the electric field must be perpendicular to the surface everywhere.

Once we've done that we can also conclude by symmetry that the electric field must have the same
magnitude at every point. (Suppose this were not the case, and that the electric field was stronger on
one side. Then if you looked at it from the other side, we would have the same geometric situation but
the electric field would now be stronger on the right). With these assumptions, we can simplify Gauss's Law:

<math>\oint\mathbf{E}\cdot d\mathbf{a} = E_r\oint d\mathbf{a} = E_r A</math>

We can take the electric field out of the integral because we know that it only has a radial component
and is the same magnitude everywhere on the surface. By using the formula for the surface area of a
sphere and applying Gauss's Law, we can conclude:

<math>E_r 4 \pi r^2 = 4 \pi k q~</math>

<math>E_r = \frac{kq}{r^2}</math>

Which is Coulomb's Law!

'''Q''': Suppose there is an infinitely long straight wire carrying uniform
charge per unit length <math>\lambda</math> (see diagram). What is the electric field of this charge
distribution?
[[File:gauss-wire.gif|right]]

'''A''': We can use Gauss's Law because this problem has so-called 'cylindrical' symmetry. Imagine
a cylindrical surface along some finite length <math>l</math> of the wire with the wire right on top of
the cylinder axis. We can argue by symmetry that the electric field must be pointing directly away from
the wire and perpendicular to the outer surface of the cylinder at all points. Try to imagine how, if this
were not the case, you could look at the cylinder from a different angle or shift the cylinder along the
wire to find a contradiction in direction of the field as we did above. Note that if the wire were not
infinitely long, this would not and is not the case, since the field at the ends could be different than
the field at the center. In practice, this result will hold as a good approximation of the field near
the center of very long wires.

In order to apply Gauss's Law, we must close our surface. To do so we can simply add flat caps to the ends
of the cylinder. If the caps are flat, then we know that they will not contribute to the electric flux
since we determined that the electric field will always be pointing radially away from the wire and thus
parallel to these caps (recall from the definition of surface integral that it ignores parallel components).

Now that we have a closed surface, we apply Gauss's Law:

<math>\oint \mathbf{E} \cdot d \mathbf{a} = E_r A = E_r 2\pi r l = 4 \pi k Q_{enc} = 4 \pi k (\lambda l)</math>

Where we have used the formula for the surface area of a cylinder (ignoring the caps as we discussed) and
the total charge enclosed is simply the length of the cylinder times the linear charge density. Now we solve
for <math>E_r</math> to find:

<math>E_r = \frac{2k\lambda}{r}</math>

Thus this is the magnitude of the field at a distance <math>r</math> from the wire, with the direction
pointing radially away from the wire.

'''Q''': Find the electric field of an infinite charged plate with uniform surface charge density
<math>\sigma</math>.
[[File:plate.gif|right]]

'''A''': By symmetry, we know that the electric field will be perpendicular to the plate, and that it
must be symmetric with respect to reflection about the plate (ie: the electric field should act the same
above the plate as below the plate). We can imagine a cylindrical surface with two faces of the
cylinder perpendiculat to the charged plate and at equal distances from it. Since the surface integral ignores parallel components of the electric field, we can ignore the contributions to the surface integral from the sides of the cylinder, since they are parallel to the electric field. Also, since the remaining faces of the box are at equal distance from the charged plate, the electric field must have the same magnitude on each face (by symmetry). Furthermore, the field at each face will either both point in or both point out of the box (depending on the sign of sigma), again by symmetry. Thus Gauss's Law becomes:

<math>\oint \mathbf{E} d\cdot\mathbf{a} = E_r (2A)= 4\pi k Q_{enc} = 4\pi k \sigma A</math>

Where <math>A</math> is the area of a face. Thus we solve for <math>E_r</math>:

<math>E_r = 2\pi k \sigma~</math>

Electrostatics

2010-03-21T23:55:24Z

RaymondGoerke: /* Examples */

File:Plate.gif

2010-03-21T23:51:43Z

RaymondGoerke:

User:RaymondGoerke

2010-03-21T23:15:44Z

RaymondGoerke: /* Raymond Liam Goerke */

== Raymond Liam Goerke ==
email: rgoerke@physics.ubc.ca

----
[[File:profile.jpg|thumb|left]]

I'm a third year physics undergrad attempting to complete the Combined Honours in Physics and Mathematics degree. This year I am also an AMS physics tutor tutoring online on Sundays and in IKB on Thursdays. I'm interesting in becoming a theoretical researcher in some area of fundamental theoretical physics. Last summer I worked with Dr. Thomas Mattison at UBC on his project relating to nanometer-vibration control; this coming summer I will be working with Sonia Bacca at Triumf on a
project in nuclear physics, likely related to halo neuclei.

I have been working since September on the physicsHelp wiki as part of my responsibilites as an AMS tutor. Last term I cleaned up and expanded the Phys 101 section. This term I'm working on putting together a Phys 102 section, which was pretty much non-existant previously.

Please help contribute to this wiki so that one day it can become a useful and comprehensive supplement to the regular course materials in first year physics.

[[physicsHelp]]

Electrostatics

2010-03-15T02:29:50Z

RaymondGoerke: /* Examples */

File:Gauss-wire.gif

2010-03-15T02:07:33Z

RaymondGoerke:

Electrostatics

2010-03-15T02:05:39Z

RaymondGoerke: /* Example */

Electrostatics

2010-03-15T02:05:03Z

RaymondGoerke: /* Examples */

The electric field is a fundamental property of the universe. It is a vector field defined over all of space, which means that at every
point in space it has a value that is a vector with a direction and magnitude. Roughly speaking, this value represents the force per
unit charge that a point charge would experience if placed at that point.

The study of electrostatics is the study of the electric field in situations where charges are no longer permitted to move, or are so
small that their motions do not effect the electric field. Eventually you will learn about magnetostatics, which is the study of
constant, steady flows of charge in closed loops or long curves, and then if you study physics at a higher level you will learn about
electrodynamics, which deals with the generalized interactions between moving charges. For now though, it is useful to learn about the
simpler field of electrostatics.

The foundational principle of electrostatics which was derived for experiment from which we can derive all of electrostatics is the
definition of the electric field of a point particle of charge <math>q</math> at a point a distance <math>r</math> from the particle
in the direction <math>\hat{r}</math>

:<math>E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}</math>

We will see how this equation can be used to derive many more interesting properties of electrostatics.

==The Principle of Superposition==

The most important idea in electrostatics is the principle of superposition. This principle states that if you know the
electric field due to a specific configuration of charge and the electric field of some other configuration of charge, then the electric
field due to putting both of those configurations on top of each other is simply the sum of the electric fields of each individual
configuration.

The simplest posible case where we can use the principle of superposition is to calculate the electrostatic
field from a collection of point charges.

====Example====
[[File:efield2.jpg|right]]
'''Q''':A point charge <math>+q</math> is located a distance 1 from the origin along the positive x-axis.
Another point charge <math>-q</math> is located a distance 1 from the origin along the negative x-axis.
What is the electric field at a distance 1 from the origin along the y-axis (point A in the diagram)?

'''A''':We can calculate the electric field from each individual charge using Coulomb's Law. Then using
the principle of superposition, we know that the total electric field is the sum of the field from each
charge. [[File:efield1.jpg|right]]

Since point A is a total distance <math>\sqrt{2}</math> from each charge we use Coulomb's Law:

<math>E_+ = \frac{kq}{(\sqrt{2})^2} = \frac{kq}{2}</math>

<math>E_- = \frac{k(-q)}{(\sqrt{2})^2} = \frac{-kq}{2}</math>

Where <math>E_+</math> and <math>E_-</math> are the contributions from the positive and negative point
charges respectively.

When we add the contributions together we must remember that they are vectors. The value for the electric field
at contribution at A from the positive charge is a vector pointing along the direction between the positive charge
and A. Since the value is positive, the vector points towards A. The same is true for the contribution from the
negative charge, except that it is negative, and thus points away from A. As depicted in the diagram, the vertical
components of the vectors cancel each other and the horizontal components add. The result is an electric field
vector pointing to the left with magnitude (with <math>\sin(\pi /4) = \cos(\pi /4) = 1/\sqrt{2}~</math>):

<math> E = \frac{kq}{2}\frac{1}{\sqrt{2}} + \frac{kq}{2}\frac{1}{\sqrt{2}} = \frac{kq}{\sqrt{2}}</math>

==Gauss's Law==

In principle, we can use the principle of superposition to calculate the electrostatic field of any charge
distribution, even solid objects with complicated shapes and some (not necessarily constant) charge density
(amount of charge per unit volume). The way this is acheived is by splitting up the object into very small
pieces which act alomst like a point charge and then adding their contributions. If we take the limit as the
pieces get smaller and smaller, we will get closer and closer to the real value of the electrostatic field.
Those who have taken second-term first year calculus may recognize this process as integration. Those who
have taken second year calculus will recognize this more specifically as a volume integral.

As you may expect this process is in general very difficult, and Physics 102 you will only need to worry
about point charges like in the previous examples. But there exists a special technique for calculating
the electric field of more complicated charge distributions more easily in situations with lots of
symmetry. This technique uses Gauss's Law.

Gauss's Law is a very general law of electrostatics and is in fact one of the four famous Maxwell Equations,
which when taken together desribe the entirely of classical electromagnetism. It is possible to derive
Guass's Law from Coulomb's Law, but this requires a complicated calculation which requires integral
calculus. It is however very easy to derive Coulomb's Law from Guass's Law, and as a result Guass's Law is
usually considered to be more fundamental, even though Coulombs's Law was discovered through experiment.

Guass's Law states that the electric flux through any closed surface is proportional to the net charge
enclosed by the surface. More specifically:

<math>\oint \mathbf{E} \cdot d\mathbf{a} = 4\pi k Q_{enc}</math> [[File:eflux2.gif|right]]

Where this denotes a surface integral. A surface integral consists of splitting a surface
into many small, approximately flat and square pieces. For each piece, we define a vector
perpendicular to the surface and pointing outwards with magnitude equal to the area of the
square. Then taking the dot product with the electric field effectively multiples the
component of the electric field perpendicular to the surface by the area of the small piece.
Taking the sum of all these products and taking the limit as the squares get very small
gives us a surface integral, which is in some sense a measure of how much the electric field
points out of the surface, as it ignores the component that is points along the surface.

One useful way to think about a surface integral is to imagine that the electric field is actually
representing the flow of water, which the magnitude and direction of the electric field represent the
speed and direction of water flowing through that point. In this case the electric flux through a
surface is how much water flows out of the surface per unit time. A positive electric flux means
water is flowing out, a negative electric flux means water is flowing in, and a electric flux of zero
means that the total flow in equals the total flow out.

Continuing with the water flow analogy, we can consider the implication of Gauss's Law, which states
that the electric flux through a closed surface is proportional to the enclosed charge. This is equivalent
to saying (in out water analogy) that if there is a net flow of water out of a closed surface, then
there must be some source (such as a faucet constantly pouring out water), and that if there is more
outward flow then there must either be more faucets or they must be pouring out water faster. Similarly,
if there is a net flow of water into a surface, then there must be a sink constantly sucking up water,
and that if there is more inward flow then there must be more sinks or they must be sucking up water
faster. In terms of the electric field positive charges play the role of faucets and negative charges
play the role of sinks. If there is a positive electric flux through a closed surface, then there
must be a net positive charge inside, and similarly a negative flux through a closed surface means
there must be a net negative charge inside.

The fascinating property of Gauss's Law is that it does not depend on where inside the surface the
charges are locate, or even how the surface is shaped. If two completely different closed surfaces
contain the same net charge, the electric flux through them will be the same. If the charges are
distributed in completely different configurations, so long as the total net charge is the same,they
will have the same electric flux. We can use this property to easily calculate the electric field
due to charge distributions in which is it easy to calculate the electric flux. We will see some
examples.

====Examples====

'''Q''': Derive Coulombs Law using Gauss's Law. [[File:spherefluxpic3.jpg|right]]

'''A''': Suppose there is a point charge <math>q</math>. Imagine a surface surrounding the point with
radius <math>r</math>. We must calculate the electric flux through this surface from from the point
charge. First we note that the electric field on the surface of the sphere must always be perpendicular
to the sphere. We can say this not because we secretly already know Coulomb's Law and thus know the
electric field must point radially out from the point charge, but by appealing to symmetry. It is worth
labouring a bit on the concept of symmetry because it is an important tool in physics: Suppose that
the electric field at the surface were not perpendicular to the surface. Then it must be pointing in
some particular direction, say slightly to the left. But then we could simply walk around to the other
side of the sphere and it would look like the field was pointing slightly to the right. In both views
however, the physical situation was identical since the point charge still is at the center
of the sphere. If the charge were not in the center we could not make this argument, since when you moved
to the other side, the electric field would change direction, but the charge would have flipped sides as
well. It is a general principle of physics that the Laws of Physics should be true no matter
what angle you look at the situation. Thus we can conclude that the above situation cannot be true if the
charge is at the center and that the electric field must be perpendicular to the surface everywhere.

Once we've done that we can also conclude by symmetry that the electric field must have the same
magnitude at every point. (Suppose this were not the case, and that the electric field was stronger on
one side. Then if you looked at it from the other side, we would have the same geometric situation but
the electric field would now be stronger on the right). With these assumptions, we can simplify Gauss's Law:

<math>\oint\mathbf{E}\cdot d\mathbf{a} = E_r\oint d\mathbf{a} = E_r A</math>

We can take the electric field out of the integral because we know that it only has a radial component
and is the same magnitude everywhere on the surface. By using the formula for the surface area of a
sphere and applying Gauss's Law, we can conclude:

<math>E_r 4 \pi r^2 = 4 \pi k q~</math>

<math>E_r = \frac{kq}{r^2}</math>

Which is Coulomb's Law!

Electrostatics

2010-03-15T01:36:09Z

RaymondGoerke: /* Gauss's Law */

The electric field is a fundamental property of the universe. It is a vector field defined over all of space, which means that at every
point in space it has a value that is a vector with a direction and magnitude. Roughly speaking, this value represents the force per
unit charge that a point charge would experience if placed at that point.

The study of electrostatics is the study of the electric field in situations where charges are no longer permitted to move, or are so
small that their motions do not effect the electric field. Eventually you will learn about magnetostatics, which is the study of
constant, steady flows of charge in closed loops or long curves, and then if you study physics at a higher level you will learn about
electrodynamics, which deals with the generalized interactions between moving charges. For now though, it is useful to learn about the
simpler field of electrostatics.

The foundational principle of electrostatics which was derived for experiment from which we can derive all of electrostatics is the
definition of the electric field of a point particle of charge <math>q</math> at a point a distance <math>r</math> from the particle
in the direction <math>\hat{r}</math>

:<math>E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}</math>

We will see how this equation can be used to derive many more interesting properties of electrostatics.

==The Principle of Superposition==

The most important idea in electrostatics is the principle of superposition. This principle states that if you know the
electric field due to a specific configuration of charge and the electric field of some other configuration of charge, then the electric
field due to putting both of those configurations on top of each other is simply the sum of the electric fields of each individual
configuration.

The simplest posible case where we can use the principle of superposition is to calculate the electrostatic
field from a collection of point charges.

====Example====
[[File:efield2.jpg|right]]
'''Q''':A point charge <math>+q</math> is located a distance 1 from the origin along the positive x-axis.
Another point charge <math>-q</math> is located a distance 1 from the origin along the negative x-axis.
What is the electric field at a distance 1 from the origin along the y-axis (point A in the diagram)?

'''A''':We can calculate the electric field from each individual charge using Coulomb's Law. Then using
the principle of superposition, we know that the total electric field is the sum of the field from each
charge. [[File:efield1.jpg|right]]

Since point A is a total distance <math>\sqrt{2}</math> from each charge we use Coulomb's Law:

<math>E_+ = \frac{kq}{(\sqrt{2})^2} = \frac{kq}{2}</math>

<math>E_- = \frac{k(-q)}{(\sqrt{2})^2} = \frac{-kq}{2}</math>

Where <math>E_+</math> and <math>E_-</math> are the contributions from the positive and negative point
charges respectively.

When we add the contributions together we must remember that they are vectors. The value for the electric field
at contribution at A from the positive charge is a vector pointing along the direction between the positive charge
and A. Since the value is positive, the vector points towards A. The same is true for the contribution from the
negative charge, except that it is negative, and thus points away from A. As depicted in the diagram, the vertical
components of the vectors cancel each other and the horizontal components add. The result is an electric field
vector pointing to the left with magnitude (with <math>\sin(\pi /4) = \cos(\pi /4) = 1/\sqrt{2}~</math>):

<math> E = \frac{kq}{2}\frac{1}{\sqrt{2}} + \frac{kq}{2}\frac{1}{\sqrt{2}} = \frac{kq}{\sqrt{2}}</math>

==Gauss's Law==

In principle, we can use the principle of superposition to calculate the electrostatic field of any charge
distribution, even solid objects with complicated shapes and some (not necessarily constant) charge density
(amount of charge per unit volume). The way this is acheived is by splitting up the object into very small
pieces which act alomst like a point charge and then adding their contributions. If we take the limit as the
pieces get smaller and smaller, we will get closer and closer to the real value of the electrostatic field.
Those who have taken second-term first year calculus may recognize this process as integration. Those who
have taken second year calculus will recognize this more specifically as a volume integral.

As you may expect this process is in general very difficult, and Physics 102 you will only need to worry
about point charges like in the previous examples. But there exists a special technique for calculating
the electric field of more complicated charge distributions more easily in situations with lots of
symmetry. This technique uses Gauss's Law.

Gauss's Law is a very general law of electrostatics and is in fact one of the four famous Maxwell Equations,
which when taken together desribe the entirely of classical electromagnetism. It is possible to derive
Guass's Law from Coulomb's Law, but this requires a complicated calculation which requires integral
calculus. It is however very easy to derive Coulomb's Law from Guass's Law, and as a result Guass's Law is
usually considered to be more fundamental, even though Coulombs's Law was discovered through experiment.

Guass's Law states that the electric flux through any closed surface is proportional to the net charge
enclosed by the surface. More specifically:

<math>\oint \mathbf{E} \cdot d\mathbf{a} = 4\pi k Q_{enc}</math> [[File:eflux2.gif|right]]

Where this denotes a surface integral. A surface integral consists of splitting a surface
into many small, approximately flat and square pieces. For each piece, we define a vector
perpendicular to the surface and pointing outwards with magnitude equal to the area of the
square. Then taking the dot product with the electric field effectively multiples the
component of the electric field perpendicular to the surface by the area of the small piece.
Taking the sum of all these products and taking the limit as the squares get very small
gives us a surface integral, which is in some sense a measure of how much the electric field
points out of the surface, as it ignores the component that is points along the surface.

One useful way to think about a surface integral is to imagine that the electric field is actually
representing the flow of water, which the magnitude and direction of the electric field represent the
speed and direction of water flowing through that point. In this case the electric flux through a
surface is how much water flows out of the surface per unit time. A positive electric flux means
water is flowing out, a negative electric flux means water is flowing in, and a electric flux of zero
means that the total flow in equals the total flow out.

Continuing with the water flow analogy, we can consider the implication of Gauss's Law, which states
that the electric flux through a closed surface is proportional to the enclosed charge. This is equivalent
to saying (in out water analogy) that if there is a net flow of water out of a closed surface, then
there must be some source (such as a faucet constantly pouring out water), and that if there is more
outward flow then there must either be more faucets or they must be pouring out water faster. Similarly,
if there is a net flow of water into a surface, then there must be a sink constantly sucking up water,
and that if there is more inward flow then there must be more sinks or they must be sucking up water
faster. In terms of the electric field positive charges play the role of faucets and negative charges
play the role of sinks. If there is a positive electric flux through a closed surface, then there
must be a net positive charge inside, and similarly a negative flux through a closed surface means
there must be a net negative charge inside.

The fascinating property of Gauss's Law is that it does not depend on where inside the surface the
charges are locate, or even how the surface is shaped. If two completely different closed surfaces
contain the same net charge, the electric flux through them will be the same. If the charges are
distributed in completely different configurations, so long as the total net charge is the same,they
will have the same electric flux. We can use this property to easily calculate the electric field
due to charge distributions in which is it easy to calculate the electric flux. We will see some
examples.

====Examples====

'''Q''': Derive Coulombs Law using Gauss's Law. [[File:spherefluxpic3.jpg|right]]

'''A''': Suppose there is a point charge <math>q</math>. Imagine a surface surrounding the point with
radius <math>r</math>. We must calculate the electric flux through this surface from from the point
charge. First we note that the electric field on the surface of the sphere must always be perpendicular
to the sphere. We can say this not because we secretly already know Coulomb's Law and thus know the
electric field must point radially out from the point charge, but by appealing to symmetry. It is worth
labouring a bit on the concept of symmetry because it is an important tool in physics: Suppose that
the electric field at the surface were not perpendicular to the surface. Then it must be pointing in
some particular direction, say slightly to the left. But then we could simply walk around to the other
side of the sphere and it would look like the field was pointing slightly to the right. In both views
however, the physical situation was identical since the point charge still looks like it is in the center
of the sphere. It is a general principle of physics that the Laws of Physics should be true no matter
what angle you look at the situation. Thus we can conclude that the above situation cannot be true and
that the electric field must be perpendicular to the surface everywhere.

Once we've done that we can also conclude by symmetry that the electric field must have the same
magnitude at every point. With these assumptions, we can simplify Gauss's Law:

in progress[[User:RaymondGoerke|RaymondGoerke]] 01:36, 15 March 2010 (UTC)

File:Spherefluxpic3.jpg

2010-03-15T00:41:55Z

RaymondGoerke:

File:Eflux2.gif

2010-03-15T00:15:33Z

RaymondGoerke:

Electrostatics

2010-03-14T23:59:51Z

RaymondGoerke: /* Example */

File:Efield2.jpg

2010-03-14T23:59:17Z

RaymondGoerke:

File:Efield1.jpg

2010-03-14T23:59:07Z

RaymondGoerke:

File:Efiled2.jpg

2010-03-14T23:14:51Z

RaymondGoerke:

File:Eflield1.jpg

2010-03-14T23:14:35Z

RaymondGoerke:

Electrostatics

2010-03-09T20:28:23Z

RaymondGoerke: /* The Principle of Superposition */

File:Diskcharge.jpg

2010-02-08T03:47:17Z

RaymondGoerke:

File:Ringcharge.jpg

2010-02-08T03:35:59Z

RaymondGoerke: uploaded a new version of "File:Ringcharge.jpg"

File:Ringcharge.jpg

2010-02-08T03:34:09Z

RaymondGoerke:

File:Linecharge.jpg

2010-02-08T03:15:08Z

RaymondGoerke: uploaded a new version of "File:Linecharge.jpg"

File:Linecharge.jpg

2010-02-08T03:13:17Z

RaymondGoerke:

PhysicsHelp

2010-02-01T03:59:20Z

RaymondGoerke:

[[Image:Hand_on_Water.jpg|right|frame|A cool picture vaguely related to physics]]

== Plan to add to this wiki? ==
You'll need to [http://wiki.ubc.ca/Main_Page login with your CWL] first .

Also please read this: [[Editing Math Equations using TeX]]

TeX equations look very nice and are very easy to use. If you plan on editing the wiki please make use of them.

== Physics Tutoring Resources ==
[[http://www.physics.ubc.ca/index.phtml UBC Dept of Physics and Astronomy]]: seminars, links to resources

====UBC Physics Courses:====

[[How to write a good exam ]]

=====Physics 101: Energy and Waves=====

[[Buoyancy, Pressure, Bernoulli's Equation]]

[[Thermodynamics and Heat Transfer]]

[[Simple and Damped Harmonic Motion]]

[[Waves and the Doppler Effect]]

=====Physics 102: Electricity, Light and Radiation=====

[[Nuclear Physics]]

[[Electrostatics]]

[[Capacitors]]

[[Resistors]]

[[Electric Circuits]]

====Quicklinks to Challenging topics in First Year Physics:====

[[Motions and Mechanics]]

[[Acceleration Velocity Position]]

[[Uncertainty and Error]]

[[Wedges]]

[[Centripedal Acceleration]]

[[Temperature]]

[[Pressure]]

[[Bernoullis Equation]]

[[Waves]]

[[Light Waves]]

[[Bouyancy]]

[[Light Bulbs]]

====Mathematics for Physics====

[[Differential Equations]]

[[Isotopes Half Life]]

====Student Societies:====

[[http://www.physics.ubc.ca/~physsoc/ UBC Physics Society]]: private tutors, exam packs, links to resources

[[http://www.phas.ubc.ca/~fizz/ FIZZ]]: Engineering Physics Student Society Homepage: exam database, forums.

[[http://www.physics.ubc.ca/~biophys/ Biophysics Student Society]]: Society Homepage

[[http://www.tutoring.ams.ubc.ca AMS Tutoring Services]]: Information about Drop-In, Online, and Appointment tutoring with AMS Tutors.

====Resources====

[[http://www.quantum-physics.polytechnique.fr/en/index.html Ecole Polytechnic's]] Quantum Physics online animations/ java applets

[[http://jersey.uoregon.edu/vlab/ University of Oregon Department of Physics]] vitual lab resources (animations, etc) for learning astrophysics, energy and environment, thermodynamics, mechanics

====People to talk to====

[[Category:Tutoring Wikis]]

PhysicsHelp

2010-02-01T03:58:40Z

RaymondGoerke:

[[Image:Hand_on_Water.jpg|right|frame]]]

== Plan to add to this wiki? ==
You'll need to [http://wiki.ubc.ca/Main_Page login with your CWL] first .

Also please read this: [[Editing Math Equations using TeX]]

TeX equations look very nice and are very easy to use. If you plan on editing the wiki please make use of them.

== Physics Tutoring Resources ==
[[http://www.physics.ubc.ca/index.phtml UBC Dept of Physics and Astronomy]]: seminars, links to resources

====UBC Physics Courses:====

[[How to write a good exam ]]

=====Physics 101: Energy and Waves=====

[[Buoyancy, Pressure, Bernoulli's Equation]]

[[Thermodynamics and Heat Transfer]]

[[Simple and Damped Harmonic Motion]]

[[Waves and the Doppler Effect]]

=====Physics 102: Electricity, Light and Radiation=====

[[Nuclear Physics]]

[[Electrostatics]]

[[Capacitors]]

[[Resistors]]

[[Electric Circuits]]

====Quicklinks to Challenging topics in First Year Physics:====

[[Motions and Mechanics]]

[[Acceleration Velocity Position]]

[[Uncertainty and Error]]

[[Wedges]]

[[Centripedal Acceleration]]

[[Temperature]]

[[Pressure]]

[[Bernoullis Equation]]

[[Waves]]

[[Light Waves]]

[[Bouyancy]]

[[Light Bulbs]]

====Mathematics for Physics====

[[Differential Equations]]

[[Isotopes Half Life]]

====Student Societies:====

[[http://www.physics.ubc.ca/~physsoc/ UBC Physics Society]]: private tutors, exam packs, links to resources

[[http://www.phas.ubc.ca/~fizz/ FIZZ]]: Engineering Physics Student Society Homepage: exam database, forums.

[[http://www.physics.ubc.ca/~biophys/ Biophysics Student Society]]: Society Homepage

[[http://www.tutoring.ams.ubc.ca AMS Tutoring Services]]: Information about Drop-In, Online, and Appointment tutoring with AMS Tutors.

====Resources====

[[http://www.quantum-physics.polytechnique.fr/en/index.html Ecole Polytechnic's]] Quantum Physics online animations/ java applets

[[http://jersey.uoregon.edu/vlab/ University of Oregon Department of Physics]] vitual lab resources (animations, etc) for learning astrophysics, energy and environment, thermodynamics, mechanics

====People to talk to====

[[Category:Tutoring Wikis]]

PhysicsHelp

2010-02-01T03:58:00Z

RaymondGoerke: /* Physics Tutoring Resources */

[[Image:Hand_on_Water.jpg|right|frame|uploaded to Flickr by [http://www.flickr.com/photos/cdm/53197139/sizes/m/]]]

== Plan to add to this wiki? ==
You'll need to [http://wiki.ubc.ca/Main_Page login with your CWL] first .

Also please read this: [[Editing Math Equations using TeX]]

TeX equations look very nice and are very easy to use. If you plan on editing the wiki please make use of them.

== Physics Tutoring Resources ==
[[http://www.physics.ubc.ca/index.phtml UBC Dept of Physics and Astronomy]]: seminars, links to resources

====UBC Physics Courses:====

[[How to write a good exam ]]

=====Physics 101: Energy and Waves=====

[[Buoyancy, Pressure, Bernoulli's Equation]]

[[Thermodynamics and Heat Transfer]]

[[Simple and Damped Harmonic Motion]]

[[Waves and the Doppler Effect]]

=====Physics 102: Electricity, Light and Radiation=====

[[Nuclear Physics]]

[[Electrostatics]]

[[Capacitors]]

[[Resistors]]

[[Electric Circuits]]

====Quicklinks to Challenging topics in First Year Physics:====

[[Motions and Mechanics]]

[[Acceleration Velocity Position]]

[[Uncertainty and Error]]

[[Wedges]]

[[Centripedal Acceleration]]

[[Temperature]]

[[Pressure]]

[[Bernoullis Equation]]

[[Waves]]

[[Light Waves]]

[[Bouyancy]]

[[Light Bulbs]]

====Mathematics for Physics====

[[Differential Equations]]

[[Isotopes Half Life]]

====Student Societies:====

[[http://www.physics.ubc.ca/~physsoc/ UBC Physics Society]]: private tutors, exam packs, links to resources

[[http://www.phas.ubc.ca/~fizz/ FIZZ]]: Engineering Physics Student Society Homepage: exam database, forums.

[[http://www.physics.ubc.ca/~biophys/ Biophysics Student Society]]: Society Homepage

[[http://www.tutoring.ams.ubc.ca AMS Tutoring Services]]: Information about Drop-In, Online, and Appointment tutoring with AMS Tutors.

====Resources====

[[http://www.quantum-physics.polytechnique.fr/en/index.html Ecole Polytechnic's]] Quantum Physics online animations/ java applets

[[http://jersey.uoregon.edu/vlab/ University of Oregon Department of Physics]] vitual lab resources (animations, etc) for learning astrophysics, energy and environment, thermodynamics, mechanics

====People to talk to====

[[Category:Tutoring Wikis]]

Electricity

2010-02-01T03:57:36Z

RaymondGoerke: moved Electricity to Electrostatics: More accurate title

#REDIRECT [[Electrostatics]]

Electrostatics

2010-02-01T03:57:36Z

RaymondGoerke: moved Electricity to Electrostatics: More accurate title

Electrostatics

2010-02-01T03:57:16Z

RaymondGoerke:

Nuclear Physics

2010-02-01T01:39:00Z

RaymondGoerke: Created page with 'Nuclear physics is the study of the nucleus. The nucleon model is a successful model wherein the smallest fundamental elements of nucleus are the proton and neutron. Although it…'

Nuclear physics is the study of the nucleus. The nucleon model is
a successful model wherein the smallest fundamental elements of
nucleus are the proton and neutron. Although it is now known that
protons and neutrons can be built up from smaller elements called
quarks, the nucleon model works very well to describe nuclear
reactions and interactions.

The force that binds nucleons together into the nucleus is called
the strong nuclear force, or simply the strong force. It is the
most powerful known force, although it has a limited range.
The balance between the strong force and the electromagnetic force(and also the weak nuclear force, not discussed in physics 102)
that determine the stability and types of decays in nuclei.

==Binding Energy==

The famous equation describing mass-energy equivalence is
<math>E = mc^2</math>. On macroscopic scales, masses of systems
are mostly constant and mechanical energies of systems are
typically much smaller than the energy-equivalent of the masses
involved. For this reason mass-energy equivalence is not usually
an important factor to consider when describing systems. As an
example, consider that the energy equivalence of a 1 kilogram
object (the mass of an average textbook) is
<math>9\times 10^{16} J</math>. In order for the same object to
have that much kinetic energy, it would need to be traveling
faster than the speed of light.

However, on nuclear scales, mass-energy equivalence is an
important factor since the typical energies of interactions are
comparable to the mass-energies of the objects involved.

Consider the deuterium nucleus, which is composed of one proton
and one neutron. The measured masses of a deuterium nucleus
(<math>m_d</math>), a proton (<math>m_p</math>) and a neutron
(<math>m_n</math>), are:

:<math>m_d = 2.0141 u</math>

:<math>m_p = 1.0073 u</math>

:<math>m_n = 1.0087 u</math>

where <math>u</math> is the atomic mass unit, defined to be one
twelfth the mass of carbon 12, or <math>1.6605 \times 10^{-24} kg</math>.

We see that the total mass of a proton and a neutron is
<math> 2.0160 u</math>, slightly less than the mass of a deuterium
nucleus. We can calculate the energy equivalent of this difference
in mass:

:<math>E = \Delta m c^2 = (2.0160u - 2.0141u)(931.5 \frac{MeV/c^2}{u}) = 42.7559 MeV/c^2</math>

This energy is often called the ''mass defect'' or binding energy
of the nucleus. Like compressing a string, squishing neutrons and
protons together requires energy because of the repulsive forces
they apply on each other. This energy is stored as binding energy
and is subtracted from the mass energy of each part. In the
process of nuclear fission, nuclei are split apart, which releases
this stored energy.

==Radioactive Decay==

When the strong force is overcome by the repulsive forces between
nucleons, nuclei undergo various types of decay. The three major
types are ''alpha'', ''beta'', and ''gamma'' decay.

PhysicsHelp

2010-01-17T23:24:50Z

RaymondGoerke: /* Physics 102: Electricity, Light and Radiation */

[[Image:Hand_on_Water.jpg|right|frame|uploaded to Flickr by [http://www.flickr.com/photos/cdm/53197139/sizes/m/]]]

== Plan to add to this wiki? ==
You'll need to [http://wiki.ubc.ca/Main_Page login with your CWL] first .

Also please read this: [[Editing Math Equations using TeX]]

TeX equations look very nice and are very easy to use. If you plan on editing the wiki please make use of them.

== Physics Tutoring Resources ==
[[http://www.physics.ubc.ca/index.phtml UBC Dept of Physics and Astronomy]]: seminars, links to resources

====UBC Physics Courses:====

[[How to write a good exam ]]

=====Physics 101: Energy and Waves=====

[[Buoyancy, Pressure, Bernoulli's Equation]]

[[Thermodynamics and Heat Transfer]]

[[Simple and Damped Harmonic Motion]]

[[Waves and the Doppler Effect]]

=====Physics 102: Electricity, Light and Radiation=====

[[Nuclear Physics]]

[[Electricity]]

[[Capacitors]]

[[Resistors]]

[[Electric Circuits]]

====Quicklinks to Challenging topics in First Year Physics:====

[[Motions and Mechanics]]

[[Acceleration Velocity Position]]

[[Uncertainty and Error]]

[[Wedges]]

[[Centripedal Acceleration]]

[[Temperature]]

[[Pressure]]

[[Bernoullis Equation]]

[[Waves]]

[[Light Waves]]

[[Bouyancy]]

[[Light Bulbs]]

====Mathematics for Physics====

[[Differential Equations]]

[[Isotopes Half Life]]

====Student Societies:====

[[http://www.physics.ubc.ca/~physsoc/ UBC Physics Society]]: private tutors, exam packs, links to resources

[[http://www.phas.ubc.ca/~fizz/ FIZZ]]: Engineering Physics Student Society Homepage: exam database, forums.

[[http://www.physics.ubc.ca/~biophys/ Biophysics Student Society]]: Society Homepage

[[http://www.tutoring.ams.ubc.ca AMS Tutoring Services]]: Information about Drop-In, Online, and Appointment tutoring with AMS Tutors.

====Resources====

[[http://www.quantum-physics.polytechnique.fr/en/index.html Ecole Polytechnic's]] Quantum Physics online animations/ java applets

[[http://jersey.uoregon.edu/vlab/ University of Oregon Department of Physics]] vitual lab resources (animations, etc) for learning astrophysics, energy and environment, thermodynamics, mechanics

====People to talk to====

[[Category:Tutoring Wikis]]

Simple and Damped Harmonic Motion

2009-11-23T00:59:11Z

RaymondGoerke: /* Doppler Effect */

Return to: [[PhysicsHelp]]
==Simple Harmonic Motion==

Many simple systems can be approximated or even accurately described by Simple Harmonic Motion.
The motion of a pendulum or spring, of waves on the ocean or waves of sound all have similar traits.

Simple harmonic motion refers to motion that can be modeled by the following equation:

<math>x = A \cos(\omega t + \phi)~</math>

which solves the differential equation:

<math>F = m\frac{d^2 x}{dt^2} = -kx</math>

<math>x</math> is the position in metres (<math>m</math>).

<math>k</math> is a constant of proportionality, normally called the spring constant, in kilograms
per second squared (<math>kg/s^2</math>).

<math>A</math> is the amplitude in metres (<math>m</math>).

<math>\omega</math> is the angular frequency in radians per second (<math>s^{-1}</math>).

<math>t</math> is the time in seconds (<math>s</math>).

<math>\phi~</math> is a constant phase shift in radians.

One can show, by differentiating the first equation twice, that <math>\omega = \sqrt{\frac{k}{m}}</math>.

Note that although <math>\omega</math> is written in radians per second, radians have no physical
units, and so <math>\omega</math> just has units of <math>s^{-1}</math>. The if one wants to talk
about the revolutions or rotations per second of an oscillation, we usually write <math>f</math> and
note that since there are <math>2\pi</math> radians in one full rotation,

<math> f = \frac{\omega}{2\pi}</math>

This frequence <math>f</math> also has units of <math>s^{-1}</math>, but in this case we call them
Hertz, and denote <math>Hz</math>.

Physically, the simple harmonic oscillator represents an object, such as a mass on a spring, moving
back and forth from a minimum to a maximum position with neglegible resistance. The minimum and
maximum positions correspond to <math>-A</math> and <math>A</math> in the first equation. We can
talk about the period of the system, meaning the time it takes an object to start from <math>A</math>,
move to <math>-A</math> and then get back to <math>A</math> again. We usually denote the period
<math>T</math> and it is clear from the definitions that:

<math>T = \frac{1}{f} = \frac{2\pi}{\omega}</math>

The constant <math>\phi~</math> in our first equation represents the initial state of the system.
Given any particular initial position and velocity, we can determine a <math>\phi</math> that will
match the objects motion for all <math>t</math>. We can see from elementary triginometry that the
first equation can be re-written:

<math>x = A \sin(\omega t + \phi_2)~</math>

Where we have simply <math> \phi_2 = \phi + \frac{\pi}{2}</math>.

We can take the derivative of the first expression to find the velocity of the object as a function
of time:

<math>v = \frac{dx}{dt} = -\omega A \sin(\omega t + \phi)</math>

One thing to note about these expressions for position and velocity is that when the position is
equal to zero, which occurs whenever <math>\omega t + \phi = \frac{n \pi}{2}</math>, where
<math>n</math> is any integer, the velocity is at a maximum magnitude, either <math>A\omega</math>
or <math>-A\omega</math>. This means that the object is moving faster when it is close to the middle
of its motion. Correspondingly, when the position is at a maximum or minimum (<math>\pm A</math>), which occurs
whenever <math> \omega t + \phi = n\pi~</math>, where <math>n</math> is any integer, the velocity is
at a minimum magnitude, <math>0</math>. Thus the object stops instantaneously as it reaches its
maximum and minimum positions.

===Calculating phi===

Many people often have trouble calculating <math>\phi</math>. Let's assume we have a problem where we
have already calculated or were given <math>A</math> and <math>\omega</math>. Most problems will tell
you, in some way or another, the amplitude at a time <math>t</math>. For instance, one might say that
the object is released at some maximum amplitude <math>A</math> at time <math>t = 0</math>. This is a
very simple case. We can see immediately from the first equation that if <math> t = 0</math> then
<math> A \cos(\phi) = A~</math> and so <math>\phi = 0~</math> is the solution. It turns out that
<math>\phi = 2\pi~</math> will also work, and in general if you add any multiple of <math>2\pi</math>
to <math>\phi~</math> you will still have a valid answer. By convention, we usually like to talk
about the solutions that is somewhere between <math>0</math> and <math>2\pi</math>.

In the more general case, a problem will tell you that at some time <math>t = t_1</math>, the object
is at position of <math>x_1</math> or has a velocity <math>v_1</math>. Here we have to be more
careful to find the correct answer. The best approach is to just do lots of examples to get used to
the ideas.

'''Q''': Consider a mass on a spring with amplitude <math>3cm</math> and frequency <math>2\pi</math>
radians per second. If you know that at time <math>t = 0.5s</math> the position is
<math>-3/\sqrt{2}cm</math>, and that the object is moving downwards, then what is <math>\phi~</math>?

'''A''': Look at a graph of the objects position and over time.
[[File:shm.png|left|thumb|A graph of two possible equations with amplitude -3/sqrt(2) at t=0.5; note that the
derivatives of each curve differ in sign.]]
The dashed line is at <math>x = -3/\sqrt{2}</math>. Note that given any solution <math>\phi_0~</math>
(corresponding to the solid curve), we can find another solution <math>\phi_1~</math> (corresponding
to the dotted curve), which passes through the same point at the same time. Since both of these
curves pass through the correct point at the correct time, we have to be able to tell them apart.
Luckiliy, by some handy properties of
triginomotry, the velocity in each of these solutions will always differ in sign, so it is enough to
know which direction the object is travelling in to find the correct answer. Thus we can solve our
first equation for <math>\phi~</math>:

<math>\phi = \cos^{-1}(\frac{x}{A}) - \omega t = \cos^{-1}(-\frac{1}{\sqrt{2}}) - \pi = -\frac{\pi}{4}</math>

Plug this into the equation for the velocity, and we get:

<math>v = -\omega A \sin(\omega t + \phi) = -2\pi (3) \sin(\pi -\frac{\pi}{4}) = \frac{1}{\sqrt{2}}cm/s</math>

But this velocity is positive, and we were told that the object is travelling downwards. We need to
find the other solution of <math>cos^{-1}</math> (remember that it is a multi-valued function) that is
within the same period. You may remember from high school trigonometry that we can get another
solution to cosine by subtracting the first solution from <math>2\pi</math> and solving for
<math>\phi_2~</math>.

<math>\omega t + \phi_2 = 2\pi - (\omega t + \phi)~</math>

<math>\phi_2 = 2\pi - 2\omega t - \phi = \frac{\pi}{4}</math>

Plugging this into our equation for velocity as above gives us a result of
<math>-\frac{1}{\sqrt{2}}cm/s</math>, which works.
Thus our answer is <math>\phi = \frac{\pi}{4}</math>

A very similar type of problem is one where the velocity is specified instead of the position. In
this case the problem is solved in the exact same way: We solve the velocity equation to find an
initial answer for <math>\phi~</math> and check if that corresponds with what we know about the
position. If not, we must find another solution to sine, which, again from high school trigonometry,
is <math>\pi</math> minus the first solutions. Hence:

<math>\omega t + \phi_2 = \pi - (\omega t + \phi)~</math>

<math>\phi_2 = \pi - 2\omega - \phi~</math>

==Damped Harmonic Motion==

In real systems, masses on springs don't continue to oscillate forever at the same amplitude;
eventually the oscillations die away and the object stops. In order to better model these kinds of
systems we can talk about the damped harmonic oscillator, which is the soltion to the differential
equation:

<math>F = m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt}</math>

Where we have taken the diffential equation for the simple harmonic oscillator and added a damping
term, <math>-b \frac{dx}{dt}</math>, where <math>b</math> is called the damping constant or drag
coefficient. Since this damping term acts in the opposite direction of motion and is proportional to
velocity, it causes objects with high vlocity to slow down quickly. We can solve the damped harmonic
oscillator equation by using techniques that you will learn if you take a differential equaitons
course. The solutions are of the form:

<math>x = Ae^{-\frac{b}{2m}t}cos(\omega t + \phi)</math>

where <math> \omega = \sqrt{\omega_0^2 - \frac{b^2}{4m^2}}</math>

and <math>\omega_0 = \sqrt{k/m}</math>

Observe that <math>\omega_0</math> is just the frequency of oscillation of a simple harmonic
oscillator. Thus we can see that if we add damping to a simple harmonic oscillator, the frequency
will change and the amplitude of the oscillations will exponentially decay with time.

===Forced / Driven Oscillations===

For forced oscillations (also known as driven oscillations) you cannot usually
solve the position of the oscillator as a function of time except in steady
state without using more advanced techniques with differential equations. What
this means is that usually the solution for a given set of initial conditions of
a damped oscillator has a complicated intial behaviour, called transients, which
are quickly damped away as the system approaches what is called the steady
state solution, which is the long-term periodic behaviour of the system.

For example, if we were to take a normal damped mass-on-a-spring system and we
were to "drive" it by pushing it back and forth continuously at some particular
frequency, i.e.

<math>F = A \cos(\omega t)~</math>

the solution (the steady state behaviour of the system) would take the form:

<math>x = \frac{A}{Z_m} \cos(\omega t + \phi)</math>

<math>Z_m = \sqrt{b^2 - {\left(\omega m - \frac{k}{\omega}\right)}^2}</math>

<math>x</math> is the displacement in meters (<math>m</math>).

<math>A</math> is the amplitude of the driving force, in Newtons (<math>N</math>).

<math>\omega</math> is the frequency of the driving oscillations in radians per
seconds (<math>s^{-1}</math>).

<math>\phi~</math> is a phase constant in radians.

The most important thing to note from this is that the amplitude is maximum when
<math>Z_m</math> is a minimum. This occurs when <math>wm = \frac{k}{\omega}</math> or
<math> \omega = \sqrt{\frac{k}{m}}</math>, that is to say the driving frequency is the
same as the natural frequency for a spring system. This idea that maximum
amplitude occurs when the system is driven at its natural frequency occurs for
all damped driven systems.

====Problem Solving====

'''Q''' On your first trip to Planet X you happen to take along a
<math>200 g</math> mass, a <math>40 cm</math> long spring, a meter stick, and
a stopwatch. You're curious about the free-fall acceleration on Planet X, where
ordinary tasks seem easier than on earth, but you can't find this information in
your Visitor's Guide. One night you suspend the spring from the ceiling in your
room and hang the mass from it. You find that the mass stretches the spring by
<math>27.4 cm</math>. You then pull the mass down <math>10.1 cm</math> and
release it. With the stopwatch you find that 10 oscillations take
<math>17.8 s</math>. Use this information to solve for the acceleration due to
gravity, math>g</math> on Planet X.

'''A''' First we use the identity <math>f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}</math>
to find the spring constant <math>k</math>. <math>f</math> is the oscillations
per second, and we know that ten oscillations take <math>17.8s</math>, so we can
solve

<math>f = \frac{10}{17.8s} \approx 0.5617 Hz</math>

and then

<math>k = {(2\pi f)}^2 m = {(2\pi 0.5617)}^2 (0.200kg) \approx 2.492 kg/s^2</math>

We know that when the mass is at rest at <math>27.4 cm</math> (or <math>0.274m</math>)
from its initial equilibrium position, the gravitational force balances out the force of the spring, i.e.

<math>F = ky = mg</math>

With <math>m = 200g = 0.200 kg</math>, <math>k = 2.492 kg/s^2</math>, <math>y = 0.274m</math>
we can find the gravity g:

<math>g = \frac{ky}{m} = \frac{(2.492 kg/s^2)(0.274 m)}{(0.200kg)} \approx 3.414 m/s^2</math>

Return to [[PhysicsHelp]]

Simple and Damped Harmonic Motion

2009-11-23T00:58:56Z

RaymondGoerke: /* Travelling Waves */

Buoyancy, Pressure, Bernoulli's Equation

2009-11-23T00:54:56Z

RaymondGoerke: /* Problem Solving */

== Buoyancy ==

Back to: [[PhysicsHelp]]

The buoyant force tends is a force that acts in the upward direction when an object is partially or fully submerged in water.

The equation for the force is:

: <math> F_b = V_s~\rho _L~g </math>

<math>F_b</math> is the buoyant force, pointing upwards, in Newtons (<math>N</math>).

<math>V_s</math> is the volume submerged in metres cubed (<math>m^3</math>).

<math>\rho _L</math> is the density of the liquid (or gas) that the object is being submerged in kilograms per metre cubed (<math>kg/m^3</math>).

<math>g</math> is the force of gravity in Newtons per kilogram (<math>N/kg</math>).

'''Note:''' if an object is only partially submerged, then <math>V_s</math> is only the volume of the submerged section.

Consider a floating object. The floating object has two forces acting on it, that of gravity and that of buoyancy. If the object is not sinking or rising, then these two forces will be equal. The force of gravity generally the form:

: <math>F_g = m g = \rho V g~</math>

<math>F_g</math> is the force of gravity in Newtons (<math>N</math>).

<math>m</math> is the mass in (<math>kg</math>).

<math>g</math> is the acceleration due to gravity in Newtons per kilogram (<math>N/kg</math>).

<math>\rho</math> is the density of the object in kilograms per metre cubed (<math>kg/m^3</math>).

<math>V</math> is the volume of the object in metres cubed (<math>m^3</math>).

This equation will be equal to the above one in static equilibrium, and depending what variables are given, you can use these to solve for the remaining ones.

===Problem Solving===

The key to solving these problems is to equate the mass of the object and the water being displaced. For example:

Q: A <math>5.20 cm</math>-tall cylinder floats in water with its axis perpendicular to the surface. The length of the cylinder above water is 2.00cm . What is the cylinder's mass density?

A: The volume of water displaced is the volume of cylinder in the water:
: <math> V = h \pi r^2 = (0.032m) \pi r^2~</math>
The density of water is <math>1000 kg/m^3</math>, so we have mass of water displaced:
: <math> m = \rho _L V = (1000kg/m^3)(0.032m) \pi r^2 = 32 \pi r^2 kg~</math>
We know that the mass of the object equals the mass of the water displaced, so we solve for the density:
: <math> \rho = \frac{m}{V} = \frac{32 \pi r^2}{0.052 \pi r^2} \approx 615.38 kg/m^3</math>

== Pressure ==

Pressure is a force per unit area. Thus, the force on an object is the pressure applied times the area over which it is applied:

: <math> F = PA~ </math>

<math>F</math> is the total force in Newtons (<math>N</math>).
<math>P</math> is the pressure in Pascals (<math>Pa</math>).
<math>A</math> is the area in metre squared (<math>m^2</math>).

So if we imagine the window of a submarine, the pressure inside is <math>1 atm</math> (or <math>101.3 kPa</math>) and the pressure outside could be <math>500 Pa</math>. The pressure difference times the area of the window (<math>0.05m</math>) would give the total force from the water on the submarine window.

:<math>F = (P_{outside} - P_{inside}) A = (101300Pa - 500Pa) (0.05m) = 5040 N~</math>

We can also calculate the pressure at some depth. For this we use the equation:

: <math>P_2 = P_1 + \rho _L g (h_2 - h_1)~</math>

<math>P_2</math> is the pressure at height <math>h_2</math> in Pascals (<math>Pa</math>).
<math>P_1</math> is the pressure at height <math>h_1</math> in Pascals (<math>Pa</math>).
<math>Rho_L</math> is the density of the liquid, in kilograms per metre cubed (<math>kg/m^3</math>).
<math>g</math> is the force of gravity, in Newtons per kilogram (<math>N/kg</math>).
<math>h_2</math> is the height for the pressure <math>P_2</math> in meters (<math>m</math>).
<math>h_1</math> is the height for the pressure <math>P_1</math> in meters (<math>m</math>).

=== Problem Solving ===

The key to solving hydraulic lift problems is to remember that the pressure at equal heights of an incompressible fluid is equal.

Q: Consider a <math>70.0kg</math> physics student standing on an oil-filled hydraulic lift to hold up four <math>100 kg</math> football players at the same height. If the physics student's piston is <math>17.0m</math> in diameter, what is the diameter of the football players' piston?

A: We know the pressure at equal heights in the fluid must be equal. We can calculate the pressure on the football side:

: <math> P_1 = \frac{F}{A} = \frac{(100kg)*4}{\pi r^2}</math>

On the physics student's side, we have:

: <math> P_2 = \frac{F}{A} = \frac{70kg}{\pi (8.5m)^2}</math>

By equating these two pressures we can solve for the radius of the piston on the football players' side:

: <math>P_1 = P_2~</math>

: <math>\rightarrow \frac{(100kg)*4}{\pi r^2} = \frac{70kg}{\pi (8.5m)^2}</math>

: <math>\rightarrow r^2 = \frac{(100kg)*4*\pi (8.5m)^2}{\pi(70kg)}</math>

: <math>\rightarrow r = \sqrt{\frac{(100kg)*4*\pi (8.5m)^2}{\pi(70kg)}} \approx 20.319m </math>

Thus the diameter of the football players' piston is:

: <math> d = 2r \approx 40.638m </math>

[[File:hydro1.png|right|thumb]]

Q: A <math>70kg</math> physics student balances a <math>1200kg</math> elephant using a hydraulic lift filled with oil (<math>\rho = 900 kg/m^3</math>). (a): The elephant's piston is <math>1m</math> in radius, what is the radius of the students piston? (b): If a second <math>70kg</math> physics student joins the first, how high will they lift the elephant?

A(a): Again we can equate the pressure on both sides to find the radius of the student's piston:

: <math> P_1 = P_2~ </math>

: <math> \frac{(70kg)g}{\pi r^2} = \frac{(1200kg)g}{\pi (1m)^2} </math>

: <math>\rightarrow r^2 = \frac{(70kg)(1m)^2}{(1200kg)}</math>

: <math>\rightarrow r = \sqrt{\frac{(70kg)(1m)^2}{(1200kg)}} \approx 0.2415m </math>

A(b): When the second student gets on the lift, the system is no longer in equilibrium and will move. The students' piston will move down and the elephant's piston will move up. When the lift has finished moving, the elephant will be some height <math>h</math> above the students. The important concept to realize here is that there are three different distanced to consider in this problem: The total height difference between the students and the elephant, the distance moved by the students, and the distance moved by the elephant. The two sides won't move the
same distance because their pistons are different sizes.

[[File:hydro2.png|right|thumb]]

In this new configuration, we still know that the pressure of the fluid at the same height is equal, so we can consider the pressure at the height of the students and compare it to the pressure at the same height on the elephant side. But the pressure on the elephant's side is not just the pressure due to the mass of the elephant, but also the pressure due to the mass of the oil above that height. We know that the mass of the oil is <math>\rho V</math> and the volume of oil can be written <math>V = hA_2</math>, where h is the total height difference and A_2 is the area of the elephant's piston, so we can write the total force of gravity from the oil above the piston as:

:<math> m = \rho h A_2 g~ </math>

Then the total force at that height on the elephant's piston is the sum of the force due to gravity of the elephant and the oil:

:<math> F = (1200kg)g + \rho h A_2 g~ </math>

The total pressure at that point is simply the force divided by the area of the piston:

:<math> P = \frac{(1200kg)g + \rho h A_2 g}{A_2} </math>

Clearly the <math>A_2</math>'s in the second term cancel and we're left with:

:<math> P = P_2 + \rho g h~</math>

Where <math>P_2</math> is the pressure just due to the mass of the elephant. We can now set this equal to <math>P_1</math>, the pressure due to the students, and solve for <math>h</math>. We have:

:<math> P_1 = P_2 + \rho g h~ </math>

:<math> \frac{(70kg + 70kg)g}{\pi (0.2415m)^2} = \frac{(1200kg)g}{\pi (1m)^2} + (900kg/m^3)g h </math>

:<math> \rightarrow h = \frac{\frac{(70kg + 70kg)}{\pi (0.2415m)^2} - \frac{(1200kg)}{\pi (1m)^2}}{(900kg/m^3)} \approx 0.4246m</math>

But remember that we're not quite done: we know the elephant's total height above the students, but because the students side dropped a bit we still need to work out how high the elephant was lifted about its initial height. To do this we must make two crucial observations. First, we must realize that clearly the sum of the distance moved by the students' side and the elephants side must equal the total difference in height <math>h</math>, since they started off at the same height. Thus:

:<math> d_1 + d_2 = h </math>

Where <math>d_1</math> is the distance moved by the students' side and <math>d_2</math> is the distance moved by the elephant's side. Secondly we know that the total volume of oil displaced on each side must be equal, otherwise the system would have gained or lost oil. But we know that the volume displaced is just the distance moved times the area, so we have:

:<math> d_1A_1 = d_2A_2 </math>

If we solve one of these equations for <math>d_1</math> and substitute into the other, we can solve for <math>d_2</math> to get:

:<math> d_2 = \frac{h}{1+\frac{A_2}{A_1}}</math>

Then we can plug in our numbers to get a value for <math>d_2</math>:

:<math>d_2 = \frac{0.4246m}{1+\frac{\pi (1m)^2}{\pi (0.2415m)^2}} \approx 0.0234m</math>

Thus the elephant was lifted <math>2.34cm</math>.

One strategy for these types of problems is to simple memorize the most relevant formulas, <math>P_1 = P_2 + \rho g h</math> and <math>d_2 = \frac{h}{1+\frac{A_2}{A_1}}</math>, but notice how the whole solution of the problem was based solely on the idea that pressure at equal heights of an incompressible fluid is equal. If you remember this fact, you should be able to use simple reasoning and algebra to adapt to many kinds of problems without memorizing which formulas to use.

==Bernoulli's Equation==

Bernoulli's Equation relates the pressure of a fluid with its height and velocity, allowing you to predict the properties as a fluid at a general point in a complicated system pipes based on your knowledge of it a some particular point.
The full equation takes the following form:

: <math>v^2 \frac{\rho}{2} + \rho g h + P = \mathsf{const}</math>

<math>v~</math> is the velocity of the fluid in metres per second (<math>m/s</math>)

<math>g~</math> is the gravitational constant in metres per second squared (<math>m/s^2</math>)

<math>h~</math> is the height of the fluid in meteres (<math>m</math>)

<math>p~</math> is the pressure of the fluid in Pascals (<math>Pa</math>)

<math>\rho~</math> is the density of the liquid in kilograms per metre cubed (<math>kg/m^3</math>)

<math>\mathsf{const}</math> is a constant.

Another useful fact to remember related to bernoulli's equation is the fact that
in a steady flow of an incompressible liquid, the total rate of volume flow per
unit time at any point is constant. This is easy to understand because it is
basically a statement that since the liquid is incompressible, it can't "bunch
up" or "spread out" at any point (the density of the liquid is constant). This
means that if you have liquid flowing through a pipe with a shrinking diameter,
the velocity of the liquid much increase: Consider that after a small time
<math>dt</math> a small bunch of liquid travels a distance <math>dx = v dt</math>,
where <math>v</math> is the velocity. Then the total volume that has been
displaced is <math>dV = Adx</math>, where A is the area of the pipe at this
point. In other words: <math> dV = A (v dt)</math>, and thus we say:

:<math>\frac{dV}{dt} = Av = \mathsf{const}</math>

===Problem Solving===

'''Q''': A horizontal circular pipe 10.0 cm in diameter has a smooth reduction in diameter
to a pipe 5.00 cm in diameter but remains at a constant height. If the pressure
of the water in the larger pipe is 8.00 x 10^4 Pa and the pressure in the
smaller pipe is 6.00 x 10^4 Pa, at what rate does water flow through the pipe?

'''A''': Using Bernoulli's equation, we can say:

:<math>\frac{{v_1}^2\rho}{2} + P_1 = \frac{{v_2}^2\rho}{2} + P_2</math>

Where we have cancelled the <math>h</math> terms since the height is constant.

Next we can say that since the volume flow rate is constant:

:<math>~A_1v_1 = A_2v_2</math>

:<math>v_1 = \frac{A_2}{A_1} v_2</math>

Plugging this into the previous line give us:

:<math>\frac{{\left(\frac{A_2}{A_1}\right)}^2 {v_2}^2 \rho}{2} + P_1 = \frac{{v_2}^2\rho}{2} + P_2</math>

Which we can solve for <math>v_2</math>

:<math>v_2 = \sqrt{\frac{2\left(P_1 - P_2\right)}{\rho \left(1 - {\left(\frac{A_2}{A_1}\right)}^2\right)}}</math>

Plugging in the numbers from the question gives us:

<math>~v_2 = 6.532 m/s</math>

Now, we want to find the flow rate of the water, so which is just the area times
the velocity at some point in the flow. So we've solve for the velocity in the
smaller pipe, so we can calculate the flow rate using the area of the smaller
pipe:

<math>\frac{dV}{dt} = A_2v_2 = 0.0128 m^3/s</math>

*Back to [[PhysicsHelp]]

PhysicsHelp

2009-11-23T00:45:58Z

RaymondGoerke: /* Plan to add to this wiki? */

[[Image:Hand_on_Water.jpg|right|frame|uploaded to Flickr by [http://www.flickr.com/photos/cdm/53197139/sizes/m/]]]

== Plan to add to this wiki? ==
You'll need to [http://wiki.ubc.ca/Main_Page login with your CWL] first .

Also please read this: [[Editing Math Equations using TeX]]

TeX equations look very nice and are very easy to use. If you plan on editing the wiki please make use of them.

== Physics Tutoring Resources ==
[[http://www.physics.ubc.ca/index.phtml UBC Dept of Physics and Astronomy]]: seminars, links to resources

====UBC Physics Courses:====

[[How to write a good exam ]]

=====Physics 101: Energy and Waves=====

[[Buoyancy, Pressure, Bernoulli's Equation]]

[[Thermodynamics and Heat Transfer]]

[[Simple and Damped Harmonic Motion]]

[[Waves and the Doppler Effect]]

=====Physics 102: Electricity, Light and Radiation=====

[[Electricity]]

[[Capacitors]]

[[Resistors]]

[[Electric Circuits]]

====Quicklinks to Challenging topics in First Year Physics:====

[[Motions and Mechanics]]

[[Acceleration Velocity Position]]

[[Uncertainty and Error]]

[[Wedges]]

[[Centripedal Acceleration]]

[[Temperature]]

[[Pressure]]

[[Bernoullis Equation]]

[[Waves]]

[[Light Waves]]

[[Bouyancy]]

[[Light Bulbs]]

====Mathematics for Physics====

[[Differential Equations]]

[[Isotopes Half Life]]

====Student Societies:====

[[http://www.physics.ubc.ca/~physsoc/ UBC Physics Society]]: private tutors, exam packs, links to resources

[[http://www.phas.ubc.ca/~fizz/ FIZZ]]: Engineering Physics Student Society Homepage: exam database, forums.

[[http://www.physics.ubc.ca/~biophys/ Biophysics Student Society]]: Society Homepage

[[http://www.tutoring.ams.ubc.ca AMS Tutoring Services]]: Information about Drop-In, Online, and Appointment tutoring with AMS Tutors.

====Resources====

[[http://www.quantum-physics.polytechnique.fr/en/index.html Ecole Polytechnic's]] Quantum Physics online animations/ java applets

[[http://jersey.uoregon.edu/vlab/ University of Oregon Department of Physics]] vitual lab resources (animations, etc) for learning astrophysics, energy and environment, thermodynamics, mechanics

====People to talk to====

[[Category:Tutoring Wikis]]

Help:Editing Math Equations using TeX

2009-11-23T00:45:36Z

RaymondGoerke:

For future editors of the PhysicsHelp page: This is how you edit math equations
using the TeX syntax to make nice looking equations. Please use TeX when writing
math. Trying to put equations directly into the text doesn't look very nice and TeX is very easy to learn.

If you already use TeX, then all you need to know is that your normal syntax
must be surrounded by tags:

<nowiki><math> syntax </math></nowiki>

If you've never used TeX before, here's a crash course:

===Fractions===

To make a fraction use:

\frac{foo}{bar}

These render as: <math>\frac{foo}{bar}</math>

===Superscript and Subscript===

To make superscripts and subscripts, use:

x^2
y_0

These render as: <math>x^2~y_0</math>

===Greek Letters===

To make greek letters, you just need to know their names. Use a capital for the
capital letter, a lower-case for the lower-case letter:

\pi
\theta
\omega
\Omega
\gamma
\Gamma
\alpha
\beta

These render as like: <math>\pi~\theta~\omega~\Omega~\gamma~\Gamma~\alpha~\beta</math>

===Trig Stuff===

You can use the following to render trig functions without italics so it looks
nicer:

\cos(\theta)
\sin(\theta)
\tan(\theta)

These render as: <math>\cos(\theta)~\sin(\theta)~\tan(\theta)</math>, as opposed
to:

cos(\theta)
sin(\theta)
tan(\theta)

Which look like: <math>cos(\theta)~sin(\theta)~tan(\theta)</math>

In general, if you want to make words not in italics in math, use

\text{foo bar}

Which looks like <math>\text{foo bar}</math> instead of <math>foo bar</math>

===Big Brackets===

Usually, a normal parenthesis or bracket will do fine:

x^2 (2x + y)

Renders as: <math>x^2 (2x + y)</math>

But if you have big stuff like fractions, it doesn't always look as nice:

(\frac{\pi}{2})

renders as: <math>(\frac{\pi}{2})</math>

Instead, use:

x^2 \left(2x + y\right)
\left( \frac{\pi}{2} \right)

Which looks like: <math> x^2 \left(2x + y\right)~\left( \frac{\pi}{2} \right)</math>

Notice that these kinds of brackets are always the right size. They also work
with square brackets:

\left[ \frac{\pi}{2} \right]
\left[ x^2 (2x + y) \right]

Renders as: <math> \left[ \frac{\pi}{2} \right]~\left[ x^2 (2x + y) \right]</math>

===Other===

These are some other things that may be useful. In general, if you want to know
how to make a type of symbol, you can find many usefuly lists by searching for TeX or LaTeX math
symbols.

\sqrt{foo}

<math>\sqrt{foo}</math>

\int_a^b f(x)dx

<math>\int_a^b f(x)dx</math>

\pm

<math>\pm</math>

\mp

<math>\mp</math>

\approx

<math>\approx</math>

One thing you may notice is that adding spaces between symbols will not add more space in the rendered
output.

a_0 a_1
a_0 a_1

These will both render as:
<math>a_0 a_1</math>
<math>a_0 a_1</math>

If you want to force spacing between symbols, you have to use the "~" symbol:

<nowiki>a_0~~~~~~~~~a_1</nowiki>

renders as: <math>a_0~~~~~~~~~a_1</math>

Also, it can be useful to note that there as two kinds of math font that the wiki will try to use.
One is a smaller font which fits better into a line of normal text, and the other is a larger font
which looks nicer when you have an equation on its own line. Whenever you make somthing "big" like
a fraction or square root symbol, the wiki will automatically use bigger font. Sometimes you may
want to force a line to be bigger because it looks nicer. I have found no nice way to do this other
than it just so happens that if you put a "~" at the end of a line that line will be rendered in the
bigger font; since the "~" is at the end you won't notice that there is technically an extra blank
space there.

A\cos(\omega t)</math>
A\cos(\omega t)~</math>

These will render as:

<math>A\cos(\omega t)</math>
<math>A\cos(\omega t)~</math>

Lastly, don't be afraid to nest things together to make really complicated looking expressions:

v_2 = \sqrt{\frac{2\left(P_1 - P_2\right)}{\rho \left(1 - {\left(\frac{A_2}{A_1}\right)}^2\right)}}

will render as:
<math>v_2 = \sqrt{\frac{2\left(P_1 - P_2\right)}{\rho \left(1 - {\left(\frac{A_2}{A_1}\right)}^2\right)}}</math>

Editing Math Equations usign TeX

2009-11-23T00:38:34Z

RaymondGoerke: moved Editing Math Equations usign TeX to Editing Math Equations using TeX

#REDIRECT [[Editing Math Equations using TeX]]

Help:Editing Math Equations using TeX

2009-11-23T00:38:34Z

RaymondGoerke: moved Editing Math Equations usign TeX to Editing Math Equations using TeX

For future editors of the PhysicsHelp page: This is how you edit math equations
using the TeX syntax to make nice looking equations. Please '''ALWAYS''' use TeX when writing
math. Trying to put equations directly into the text looks very ugly and TeX is not very hard
to use.

If you already use TeX, then all you need to know is that your normal syntax
must be surrounded by tags:

<nowiki><math> syntax </math></nowiki>

If you've never used TeX before, here's a crash course:

===Fractions===

To make a fraction use:

\frac{foo}{bar}

These render as: <math>\frac{foo}{bar}</math>

===Superscript and Subscript===

To make superscripts and subscripts, use:

x^2
y_0

These render as: <math>x^2~y_0</math>

===Greek Letters===

To make greek letters, you just need to know their names. Use a capital for the
capital letter, a lower-case for the lower-case letter:

\pi
\theta
\omega
\Omega
\gamma
\Gamma
\alpha
\beta

These render as like: <math>\pi~\theta~\omega~\Omega~\gamma~\Gamma~\alpha~\beta</math>

===Trig Stuff===

You can use the following to render trig functions without italics so it looks
nicer:

\cos(\theta)
\sin(\theta)
\tan(\theta)

These render as: <math>\cos(\theta)~\sin(\theta)~\tan(\theta)</math>, as opposed
to:

cos(\theta)
sin(\theta)
tan(\theta)

Which look like: <math>cos(\theta)~sin(\theta)~tan(\theta)</math>

In general, if you want to make words not in italics in math, use

\text{foo bar}

Which looks like <math>\text{foo bar}</math> instead of <math>foo bar</math>

===Big Brackets===

Usually, a normal parenthesis or bracket will do fine:

x^2 (2x + y)

Renders as: <math>x^2 (2x + y)</math>

But if you have big stuff like fractions, it doesn't always look as nice:

(\frac{\pi}{2})

renders as: <math>(\frac{\pi}{2})</math>

Instead, use:

x^2 \left(2x + y\right)
\left( \frac{\pi}{2} \right)

Which looks like: <math> x^2 \left(2x + y\right)~\left( \frac{\pi}{2} \right)</math>

Notice that these kinds of brackets are always the right size. They also work
with square brackets:

\left[ \frac{\pi}{2} \right]
\left[ x^2 (2x + y) \right]

Renders as: <math> \left[ \frac{\pi}{2} \right]~\left[ x^2 (2x + y) \right]</math>

===Other===

These are some other things that may be useful. In general, if you want to know
how to make a type of symbol, you can find many usefuly lists by searching for TeX or LaTeX math
symbols.

\sqrt{foo}

<math>\sqrt{foo}</math>

\int_a^b f(x)dx

<math>\int_a^b f(x)dx</math>

\pm

<math>\pm</math>

\mp

<math>\mp</math>

\approx

<math>\approx</math>

One thing you may notice is that adding spaces between symbols will not add more space in the rendered
output.

a_0 a_1
a_0 a_1

These will both render as:
<math>a_0 a_1</math>
<math>a_0 a_1</math>

If you want to force spacing between symbols, you have to use the "~" symbol:

<nowiki>a_0~~~~~~~~~a_1</nowiki>

renders as: <math>a_0~~~~~~~~~a_1</math>

Also, it can be useful to note that there as two kinds of math font that the wiki will try to use.
One is a smaller font which fits better into a line of normal text, and the other is a larger font
which looks nicer when you have an equation on its own line. Whenever you make somthing "big" like
a fraction or square root symbol, the wiki will automatically use bigger font. Sometimes you may
want to force a line to be bigger because it looks nicer. I have found no nice way to do this other
than it just so happens that if you put a "~" at the end of a line that line will be rendered in the
bigger font; since the "~" is at the end you won't notice that there is technically an extra blank
space there.

A\cos(\omega t)</math>
A\cos(\omega t)~</math>

These will render as:

<math>A\cos(\omega t)</math>
<math>A\cos(\omega t)~</math>

Lastly, don't be afraid to nest things together to make really complicated looking expressions:

v_2 = \sqrt{\frac{2\left(P_1 - P_2\right)}{\rho \left(1 - {\left(\frac{A_2}{A_1}\right)}^2\right)}}

will render as:
<math>v_2 = \sqrt{\frac{2\left(P_1 - P_2\right)}{\rho \left(1 - {\left(\frac{A_2}{A_1}\right)}^2\right)}}</math>

Help:Editing Math Equations using TeX

2009-11-23T00:38:24Z

RaymondGoerke:

Help:Editing Math Equations using TeX

2009-11-22T23:08:28Z

RaymondGoerke:

For future editors of the PhysicsHelp page: This is how you edit math equations
using the TeX syntax to make nice looking equations.

If you already use TeX, then all you need to know is that your normal syntax
must be surrounded by tags:

<nowiki><math> syntax </math></nowiki>

If you've never used TeX before, here's a crash course:

===Fractions===

To make a fraction use:

\frac{foo}{bar}

It looks like this: <math>\frac{foo}{bar}</math>

===Superscript and Subscript===

To make superscripts and subscripts, use:

x^2
y_0

It looks like: <math>x^2 y_0</math>

===Greek Letters===

To make greek letters, you just need to know their names. Use a capital for the
capital letter, a lower-case for the lower-case letter:

\pi
\theta
\omega
\Omega
\gamma
\Gamma
\alpha
\beta

It looks like: <math>\pi \theta \omega \Omega \gamma \Gamma \alpha \beta</math>

===Trig Stuff===

You can use the following to render trig functions without italics so it looks
nicer:

\cos(\theta)
\sin(\theta)
\tan(\theta)

It looks like: <math>\cos(\theta) \sin(\theta) \tan(\theta)</math>, as opposed
to:

cos(\theta)
sin(\theta)
tan(\theta)

Which looks like: <math>cos(\theta) sin(\theta) tan(\theta)</math>

In general, if you want to make words not in italics in math, use

\text{foo bar}

Which looks like <math>\text{foo bar}</math> instead of <math>foo bar</math>

===Big Brackets===

Usually, a normal parenthesis or bracket will do fine:

x^2 (2x + y)

Renders as: <math>x^2 (2x + y)</math>

But if you have big stuff like fractions, it doesn't always look as nice:

(\frac{\pi}{2})

Renders as: <math>(\frac{\pi}{2})</math>

Instead, use:

x^2 \left(2x + y\right)
\left( \frac{\pi}{2} \right)

Which looks like: <math> x^2 \left(2x + y\right) \left( \frac{\pi}{2} \right)</math>

Notice that these kinds of brackets are always the right size. They also work
with square brackets:

\left[ \frac{\pi}{2} \right]
\left[ x^2 (2x + y) \right]

Renders as: <math> \left[ \frac{\pi}{2} \right] \left[ x^2 (2x + y) \right]</math>

===Other===

The above will cover the majority of the things you will use most often.
What follows is a more miscellaeous list of useful things.

\sqrt{foo}
\int_a^b f(x)
\sin
\cos

Editing Math Equations with LaTeX

2009-11-22T23:08:01Z

RaymondGoerke: moved Editing Math Equations with LaTeX to Editing Math Equations usign TeX

#REDIRECT [[Editing Math Equations usign TeX]]

Help:Editing Math Equations using TeX

2009-11-22T23:08:01Z

RaymondGoerke: moved Editing Math Equations with LaTeX to Editing Math Equations usign TeX

For future editors of the PhysicsHelp page: This is how you edit math equations
using the LaTeX syntax to make nice looking equations.

If you already use LaTeX, then all you need to know is that your normal syntax
must be surrounded by tags:

<nowiki><math> syntax </math></nowiki>

If you've never used LaTeX before, here's a crash course:

===Fractions===

To make a fraction use:

\frac{foo}{bar}

It looks like this: <math>\frac{foo}{bar}</math>

===Superscript and Subscript===

To make superscripts and subscripts, use:

x^2
y_0

It looks like: <math>x^2 y_0</math>

===Greek Letters===

To make greek letters, you just need to know their names. Use a capital for the
capital letter, a lower-case for the lower-case letter:

\pi
\theta
\omega
\Omega
\gamma
\Gamma
\alpha
\beta

It looks like: <math>\pi \theta \omega \Omega \gamma \Gamma \alpha \beta</math>

===Trig Stuff===

You can use the following to render trig functions without italics so it looks
nicer:

\cos(\theta)
\sin(\theta)
\tan(\theta)

It looks like: <math>\cos(\theta) \sin(\theta) \tan(\theta)</math>, as opposed
to:

cos(\theta)
sin(\theta)
tan(\theta)

Which looks like: <math>cos(\theta) sin(\theta) tan(\theta)</math>

In general, if you want to make words not in italics in math, use

\text{foo bar}

Which looks like <math>\text{foo bar}</math> instead of <math>foo bar</math>

===Big Brackets===

Usually, a normal parenthesis or bracket will do fine:

x^2 (2x + y)

Renders as: <math>x^2 (2x + y)</math>

But if you have big stuff like fractions, it doesn't always look as nice:

(\frac{\pi}{2})

Renders as: <math>(\frac{\pi}{2})</math>

Instead, use:

x^2 \left(2x + y\right)
\left( \frac{\pi}{2} \right)

Which looks like: <math> x^2 \left(2x + y\right) \left( \frac{\pi}{2} \right)</math>

Notice that these kinds of brackets are always the right size. They also work
with square brackets:

\left[ \frac{\pi}{2} \right]
\left[ x^2 (2x + y) \right]

Renders as: <math> \left[ \frac{\pi}{2} \right] \left[ x^2 (2x + y) \right]</math>

===Other===

The above will cover the majority of the things you will use most often.
What follows is a more miscellaeous list of useful things.

\sqrt{foo}
\int_a^b f(x)
\sin
\cos

Buoyancy, Pressure, Bernoulli's Equation

2009-11-16T02:39:38Z

RaymondGoerke: /* Bernoulli's Equation */

== Buoyancy ==

Back to: [[PhysicsHelp]]

The buoyant force tends is a force that acts in the upward direction when an object is partially or fully submerged in water.

The equation for the force is:

: <math> F_b = V_s~\rho _L~g </math>

<math>F_b</math> is the buoyant force, pointing upwards, in Newtons (<math>N</math>).

<math>V_s</math> is the volume submerged in metres cubed (<math>m^3</math>).

<math>\rho _L</math> is the density of the liquid (or gas) that the object is being submerged in kilograms per metre cubed (<math>kg/m^3</math>).

<math>g</math> is the force of gravity in Newtons per kilogram (<math>N/kg</math>).

'''Note:''' if an object is only partially submerged, then <math>V_s</math> is only the volume of the submerged section.

Consider a floating object. The floating object has two forces acting on it, that of gravity and that of buoyancy. If the object is not sinking or rising, then these two forces will be equal. The force of gravity generally the form:

: <math>F_g = m g = \rho V g~</math>

<math>F_g</math> is the force of gravity in Newtons (<math>N</math>).

<math>m</math> is the mass in (<math>kg</math>).

<math>g</math> is the acceleration due to gravity in Newtons per kilogram (<math>N/kg</math>).

<math>\rho</math> is the density of the object in kilograms per metre cubed (<math>kg/m^3</math>).

<math>V</math> is the volume of the object in metres cubed (<math>m^3</math>).

This equation will be equal to the above one in static equilibrium, and depending what variables are given, you can use these to solve for the remaining ones.

===Problem Solving===

The key to solving these problems is to equate the mass of the object and the water being displaced. For example:

Q: A <math>5.20 cm</math>-tall cylinder floats in water with its axis perpendicular to the surface. The length of the cylinder above water is 2.00cm . What is the cylinder's mass density?

A: The volume of water displaced is the volume of cylinder in the water:
: <math> V = h \pi r^2 = (0.032m) \pi r^2~</math>
The density of water is <math>1000 kg/m^3</math>, so we have mass of water displaced:
: <math> m = \rho _L V = (1000kg/m^3)(0.032m) \pi r^2 = 32 \pi r^2 kg~</math>
We know that the mass of the object equals the mass of the water displaced, so we solve for the density:
: <math> \rho = \frac{m}{V} = \frac{32 \pi r^2}{0.052 \pi r^2} \approx 615.38 kg/m^3</math>

== Pressure ==

Pressure is a force per unit area. Thus, the force on an object is the pressure applied times the area over which it is applied:

: <math> F = PA~ </math>

<math>F</math> is the total force in Newtons (<math>N</math>).
<math>P</math> is the pressure in Pascals (<math>Pa</math>).
<math>A</math> is the area in metre squared (<math>m^2</math>).

So if we imagine the window of a submarine, the pressure inside is <math>1 atm</math> (or <math>101.3 kPa</math>) and the pressure outside could be <math>500 Pa</math>. The pressure difference times the area of the window (<math>0.05m</math>) would give the total force from the water on the submarine window.

:<math>F = (P_{outside} - P_{inside}) A = (101300Pa - 500Pa) (0.05m) = 5040 N~</math>

We can also calculate the pressure at some depth. For this we use the equation:

: <math>P_2 = P_1 + \rho _L g (h_2 - h_1)~</math>

<math>P_2</math> is the pressure at height <math>h_2</math> in Pascals (<math>Pa</math>).
<math>P_1</math> is the pressure at height <math>h_1</math> in Pascals (<math>Pa</math>).
<math>Rho_L</math> is the density of the liquid, in kilograms per metre cubed (<math>kg/m^3</math>).
<math>g</math> is the force of gravity, in Newtons per kilogram (<math>N/kg</math>).
<math>h_2</math> is the height for the pressure <math>P_2</math> in meters (<math>m</math>).
<math>h_1</math> is the height for the pressure <math>P_1</math> in meters (<math>m</math>).

=== Problem Solving ===

The key to solving hydraulic lift problems is to remember that the pressure at equal heights of an incompressible fluid is equal.

Q: Consider a <math>70.0kg</math> physics student standing on an oil-filled hydraulic lift to hold up four <math>100 kg</math> football players at the same height. If the physics student's piston is <math>17.0m</math> in diameter, what is the diameter of the football players' piston?

A: We know the pressure at equal heights in the fluid must be equal. We can calculate the pressure on the football side:

: <math> P_1 = \frac{F}{A} = \frac{(100kg)*4}{\pi r^2}</math>

On the physics student's side, we have:

: <math> P_2 = \frac{F}{A} = \frac{70kg}{\pi (8.5m)^2}</math>

By equating these two pressures we can solve for the radius of the piston on the football players' side:

: <math>P_1 = P_2~</math>

: <math>\rightarrow \frac{(100kg)*4}{\pi r^2} = \frac{70kg}{\pi (8.5m)^2}</math>

: <math>\rightarrow r^2 = \frac{(100kg)*4*\pi (8.5m)^2}{\pi(70kg)}</math>

: <math>\rightarrow r = \sqrt{\frac{(100kg)*4*\pi (8.5m)^2}{\pi(70kg)}} \approx 20.319m </math>

Thus the diameter of the football players' piston is:

: <math> d = 2r \approx 40.638m </math>

[[File:hydro1.png|right]]

Q: A <math>70kg</math> physics student balances a <math>1200kg</math> elephant using a hydraulic lift filled with oil (<math>\rho = 900 kg/m^3</math>). (a): The elephant's piston is <math>1m</math> in radius, what is the radius of the students piston? (b): If a second <math>70kg</math> physics student joins the first, how high will they lift the elephant?

[[File:hydro2.png|right]]

A(a): Again we can equate the pressure on both sides to find the radius of the student's piston:

: <math> P_1 = P_2~ </math>

: <math> \frac{(70kg)g}{\pi r^2} = \frac{(1200kg)g}{\pi (1m)^2} </math>

: <math>\rightarrow r^2 = \frac{(70kg)(1m)^2}{(1200kg)}</math>

: <math>\rightarrow r = \sqrt{\frac{(70kg)(1m)^2}{(1200kg)}} \approx 0.2415m </math>

A(b): When the second student gets on the lift, the system is no longer in equilibrium and will move. The students' piston will move down and the elephant's piston will move up. When the lift has finished moving, the elephant will be some height <math>h</math> above the students. The important concept to realize here is that there are three different distanced to consider in this problem: The total height difference between the students and the elephant, the distance moved by the students, and the distance moved by the elephant. The two sides won't move the
same distance because their pistons are different sizes.

In this new configuration, we still know that the pressure of the fluid at the same height is equal, so we can consider the pressure at the height of the students and compare it to the pressure at the same height on the elephant side. But the pressure on the elephant's side is not just the pressure due to the mass of the elephant, but also the pressure due to the mass of the oil above that height. We know that the mass of the oil is <math>\rho V</math> and the volume of oil can be written <math>V = hA_2</math>, where h is the total height difference and A_2 is the area of the elephant's piston, so we can write the total force of gravity from the oil above the piston as:

:<math> m = \rho h A_2 g~ </math>

Then the total force at that height on the elephant's piston is the sum of the force due to gravity of the elephant and the oil:

:<math> F = (1200kg)g + \rho h A_2 g~ </math>

The total pressure at that point is simply the force divided by the area of the piston:

:<math> P = \frac{(1200kg)g + \rho h A_2 g}{A_2} </math>

Clearly the <math>A_2</math>'s in the second term cancel and we're left with:

:<math> P = P_2 + \rho g h~</math>

Where <math>P_2</math> is the pressure just due to the mass of the elephant. We can now set this equal to <math>P_1</math>, the pressure due to the students, and solve for <math>h</math>. We have:

:<math> P_1 = P_2 + \rho g h~ </math>

:<math> \frac{(70kg + 70kg)g}{\pi (0.2415m)^2} = \frac{(1200kg)g}{\pi (1m)^2} + (900kg/m^3)g h </math>

:<math> \rightarrow h = \frac{\frac{(70kg + 70kg)}{\pi (0.2415m)^2} - \frac{(1200kg)}{\pi (1m)^2}}{(900kg/m^3)} \approx 0.4246m</math>

But remember that we're not quite done: we know the elephant's total height above the students, but because the students side dropped a bit we still need to work out how high the elephant was lifted about its initial height. To do this we must make two crucial observations. First, we must realize that clearly the sum of the distance moved by the students' side and the elephants side must equal the total difference in height <math>h</math>, since they started off at the same height. Thus:

:<math> d_1 + d_2 = h </math>

Where <math>d_1</math> is the distance moved by the students' side and <math>d_2</math> is the distance moved by the elephant's side. Secondly we know that the total volume of oil displaced on each side must be equal, otherwise the system would have gained or lost oil. But we know that the volume displaced is just the distance moved times the area, so we have:

:<math> d_1A_1 = d_2A_2 </math>

If we solve one of these equations for <math>d_1</math> and substitute into the other, we can solve for <math>d_2</math> to get:

:<math> d_2 = \frac{h}{1+\frac{A_2}{A_1}}</math>

Then we can plug in our numbers to get a value for <math>d_2</math>:

:<math>h = \frac{0.4246m}{1+\frac{\pi (1m)^2}{\pi (0.2415m)^2}} \approx 0.0234m</math>

Thus the elephant was lifted <math>2.34cm</math>.

One strategy for these types of problems is to simple memorize the most relevant formulas, <math>P_1 = P_2 + \rho g h</math> and <math>d_2 = \frac{h}{1+\frac{A_2}{A_1}}</math>, but notice how the whole solution of the problem was based solely on the idea that pressure at equal heights of an incompressible fluid is equal. If you remember this fact, you should be able to use simple reasoning and algebra to adapt to many kinds of problems without memorizing which formulas to use.

==Bernoulli's Equation==

Bernoulli's Equation relates the pressure of a fluid with its height and velocity, allowing you to predict the properties as a fluid at a general point in a complicated system pipes based on your knowledge of it a some particular point.
The full equation takes the following form:

: <math>v^2 \frac{\rho}{2} + \rho g h + P = \mathsf{const}</math>

<math>v~</math> is the velocity of the fluid in metres per second (<math>m/s</math>)

<math>g~</math> is the gravitational constant in metres per second squared (<math>m/s^2</math>)

<math>h~</math> is the height of the fluid in meteres (<math>m</math>)

<math>p~</math> is the pressure of the fluid in Pascals (<math>Pa</math>)

<math>\rho~</math> is the density of the liquid in kilograms per metre cubed (<math>kg/m^3</math>)

<math>\mathsf{const}</math> is a constant.

Another useful fact to remember related to bernoulli's equation is the fact that
in a steady flow of an incompressible liquid, the total rate of volume flow per
unit time at any point is constant. This is easy to understand because it is
basically a statement that since the liquid is incompressible, it can't "bunch
up" or "spread out" at any point (the density of the liquid is constant). This
means that if you have liquid flowing through a pipe with a shrinking diameter,
the velocity of the liquid much increase: Consider that after a small time
<math>dt</math> a small bunch of liquid travels a distance <math>dx = v dt</math>,
where <math>v</math> is the velocity. Then the total volume that has been
displaced is <math>dV = Adx</math>, where A is the area of the pipe at this
point. In other words: <math> dV = A (v dt)</math>, and thus we say:

:<math>\frac{dV}{dt} = Av = \mathsf{const}</math>

===Problem Solving===

'''Q''': A horizontal circular pipe 10.0 cm in diameter has a smooth reduction in diameter
to a pipe 5.00 cm in diameter but remains at a constant height. If the pressure
of the water in the larger pipe is 8.00 x 10^4 Pa and the pressure in the
smaller pipe is 6.00 x 10^4 Pa, at what rate does water flow through the pipe?

'''A''': Using Bernoulli's equation, we can say:

:<math>\frac{{v_1}^2\rho}{2} + P_1 = \frac{{v_2}^2\rho}{2} + P_2</math>

Where we have cancelled the <math>h</math> terms since the height is constant.

Next we can say that since the volume flow rate is constant:

:<math>~A_1v_1 = A_2v_2</math>

:<math>v_1 = \frac{A_2}{A_1} v_2</math>

Plugging this into the previous line give us:

:<math>\frac{{\left(\frac{A_2}{A_1}\right)}^2 {v_2}^2 \rho}{2} + P_1 = \frac{{v_2}^2\rho}{2} + P_2</math>

Which we can solve for <math>v_2</math>

:<math>v_2 = \sqrt{\frac{2\left(P_1 - P_2\right)}{\rho \left(1 - {\left(\frac{A_2}{A_1}\right)}^2\right)}}</math>

Plugging in the numbers from the question gives us:

<math>~v_2 = 6.532 m/s</math>

Now, we want to find the flow rate of the water, so which is just the area times
the velocity at some point in the flow. So we've solve for the velocity in the
smaller pipe, so we can calculate the flow rate using the area of the smaller
pipe:

<math>\frac{dV}{dt} = A_2v_2 = 0.0128 m^3/s</math>

*Back to [[PhysicsHelp]]

Buoyancy, Pressure, Bernoulli's Equation

2009-11-16T02:32:35Z

RaymondGoerke: /* Bernoulli's Equation */

== Buoyancy ==

Back to: [[PhysicsHelp]]

The buoyant force tends is a force that acts in the upward direction when an object is partially or fully submerged in water.

The equation for the force is:

: <math> F_b = V_s~\rho _L~g </math>

<math>F_b</math> is the buoyant force, pointing upwards, in Newtons (<math>N</math>).

<math>V_s</math> is the volume submerged in metres cubed (<math>m^3</math>).

<math>\rho _L</math> is the density of the liquid (or gas) that the object is being submerged in kilograms per metre cubed (<math>kg/m^3</math>).

<math>g</math> is the force of gravity in Newtons per kilogram (<math>N/kg</math>).

'''Note:''' if an object is only partially submerged, then <math>V_s</math> is only the volume of the submerged section.

Consider a floating object. The floating object has two forces acting on it, that of gravity and that of buoyancy. If the object is not sinking or rising, then these two forces will be equal. The force of gravity generally the form:

: <math>F_g = m g = \rho V g~</math>

<math>F_g</math> is the force of gravity in Newtons (<math>N</math>).

<math>m</math> is the mass in (<math>kg</math>).

<math>g</math> is the acceleration due to gravity in Newtons per kilogram (<math>N/kg</math>).

<math>\rho</math> is the density of the object in kilograms per metre cubed (<math>kg/m^3</math>).

<math>V</math> is the volume of the object in metres cubed (<math>m^3</math>).

This equation will be equal to the above one in static equilibrium, and depending what variables are given, you can use these to solve for the remaining ones.

===Problem Solving===

The key to solving these problems is to equate the mass of the object and the water being displaced. For example:

Q: A <math>5.20 cm</math>-tall cylinder floats in water with its axis perpendicular to the surface. The length of the cylinder above water is 2.00cm . What is the cylinder's mass density?

A: The volume of water displaced is the volume of cylinder in the water:
: <math> V = h \pi r^2 = (0.032m) \pi r^2~</math>
The density of water is <math>1000 kg/m^3</math>, so we have mass of water displaced:
: <math> m = \rho _L V = (1000kg/m^3)(0.032m) \pi r^2 = 32 \pi r^2 kg~</math>
We know that the mass of the object equals the mass of the water displaced, so we solve for the density:
: <math> \rho = \frac{m}{V} = \frac{32 \pi r^2}{0.052 \pi r^2} \approx 615.38 kg/m^3</math>

== Pressure ==

Pressure is a force per unit area. Thus, the force on an object is the pressure applied times the area over which it is applied:

: <math> F = PA~ </math>

<math>F</math> is the total force in Newtons (<math>N</math>).
<math>P</math> is the pressure in Pascals (<math>Pa</math>).
<math>A</math> is the area in metre squared (<math>m^2</math>).

So if we imagine the window of a submarine, the pressure inside is <math>1 atm</math> (or <math>101.3 kPa</math>) and the pressure outside could be <math>500 Pa</math>. The pressure difference times the area of the window (<math>0.05m</math>) would give the total force from the water on the submarine window.

:<math>F = (P_{outside} - P_{inside}) A = (101300Pa - 500Pa) (0.05m) = 5040 N~</math>

We can also calculate the pressure at some depth. For this we use the equation:

: <math>P_2 = P_1 + \rho _L g (h_2 - h_1)~</math>

<math>P_2</math> is the pressure at height <math>h_2</math> in Pascals (<math>Pa</math>).
<math>P_1</math> is the pressure at height <math>h_1</math> in Pascals (<math>Pa</math>).
<math>Rho_L</math> is the density of the liquid, in kilograms per metre cubed (<math>kg/m^3</math>).
<math>g</math> is the force of gravity, in Newtons per kilogram (<math>N/kg</math>).
<math>h_2</math> is the height for the pressure <math>P_2</math> in meters (<math>m</math>).
<math>h_1</math> is the height for the pressure <math>P_1</math> in meters (<math>m</math>).

=== Problem Solving ===

The key to solving hydraulic lift problems is to remember that the pressure at equal heights of an incompressible fluid is equal.

Q: Consider a <math>70.0kg</math> physics student standing on an oil-filled hydraulic lift to hold up four <math>100 kg</math> football players at the same height. If the physics student's piston is <math>17.0m</math> in diameter, what is the diameter of the football players' piston?

A: We know the pressure at equal heights in the fluid must be equal. We can calculate the pressure on the football side:

: <math> P_1 = \frac{F}{A} = \frac{(100kg)*4}{\pi r^2}</math>

On the physics student's side, we have:

: <math> P_2 = \frac{F}{A} = \frac{70kg}{\pi (8.5m)^2}</math>

By equating these two pressures we can solve for the radius of the piston on the football players' side:

: <math>P_1 = P_2~</math>

: <math>\rightarrow \frac{(100kg)*4}{\pi r^2} = \frac{70kg}{\pi (8.5m)^2}</math>

: <math>\rightarrow r^2 = \frac{(100kg)*4*\pi (8.5m)^2}{\pi(70kg)}</math>

: <math>\rightarrow r = \sqrt{\frac{(100kg)*4*\pi (8.5m)^2}{\pi(70kg)}} \approx 20.319m </math>

Thus the diameter of the football players' piston is:

: <math> d = 2r \approx 40.638m </math>

[[File:hydro1.png|right]]

Q: A <math>70kg</math> physics student balances a <math>1200kg</math> elephant using a hydraulic lift filled with oil (<math>\rho = 900 kg/m^3</math>). (a): The elephant's piston is <math>1m</math> in radius, what is the radius of the students piston? (b): If a second <math>70kg</math> physics student joins the first, how high will they lift the elephant?

[[File:hydro2.png|right]]

A(a): Again we can equate the pressure on both sides to find the radius of the student's piston:

: <math> P_1 = P_2~ </math>

: <math> \frac{(70kg)g}{\pi r^2} = \frac{(1200kg)g}{\pi (1m)^2} </math>

: <math>\rightarrow r^2 = \frac{(70kg)(1m)^2}{(1200kg)}</math>

: <math>\rightarrow r = \sqrt{\frac{(70kg)(1m)^2}{(1200kg)}} \approx 0.2415m </math>

A(b): When the second student gets on the lift, the system is no longer in equilibrium and will move. The students' piston will move down and the elephant's piston will move up. When the lift has finished moving, the elephant will be some height <math>h</math> above the students. The important concept to realize here is that there are three different distanced to consider in this problem: The total height difference between the students and the elephant, the distance moved by the students, and the distance moved by the elephant. The two sides won't move the
same distance because their pistons are different sizes.

In this new configuration, we still know that the pressure of the fluid at the same height is equal, so we can consider the pressure at the height of the students and compare it to the pressure at the same height on the elephant side. But the pressure on the elephant's side is not just the pressure due to the mass of the elephant, but also the pressure due to the mass of the oil above that height. We know that the mass of the oil is <math>\rho V</math> and the volume of oil can be written <math>V = hA_2</math>, where h is the total height difference and A_2 is the area of the elephant's piston, so we can write the total force of gravity from the oil above the piston as:

:<math> m = \rho h A_2 g~ </math>

Then the total force at that height on the elephant's piston is the sum of the force due to gravity of the elephant and the oil:

:<math> F = (1200kg)g + \rho h A_2 g~ </math>

The total pressure at that point is simply the force divided by the area of the piston:

:<math> P = \frac{(1200kg)g + \rho h A_2 g}{A_2} </math>

Clearly the <math>A_2</math>'s in the second term cancel and we're left with:

:<math> P = P_2 + \rho g h~</math>

Where <math>P_2</math> is the pressure just due to the mass of the elephant. We can now set this equal to <math>P_1</math>, the pressure due to the students, and solve for <math>h</math>. We have:

:<math> P_1 = P_2 + \rho g h~ </math>

:<math> \frac{(70kg + 70kg)g}{\pi (0.2415m)^2} = \frac{(1200kg)g}{\pi (1m)^2} + (900kg/m^3)g h </math>

:<math> \rightarrow h = \frac{\frac{(70kg + 70kg)}{\pi (0.2415m)^2} - \frac{(1200kg)}{\pi (1m)^2}}{(900kg/m^3)} \approx 0.4246m</math>

But remember that we're not quite done: we know the elephant's total height above the students, but because the students side dropped a bit we still need to work out how high the elephant was lifted about its initial height. To do this we must make two crucial observations. First, we must realize that clearly the sum of the distance moved by the students' side and the elephants side must equal the total difference in height <math>h</math>, since they started off at the same height. Thus:

:<math> d_1 + d_2 = h </math>

Where <math>d_1</math> is the distance moved by the students' side and <math>d_2</math> is the distance moved by the elephant's side. Secondly we know that the total volume of oil displaced on each side must be equal, otherwise the system would have gained or lost oil. But we know that the volume displaced is just the distance moved times the area, so we have:

:<math> d_1A_1 = d_2A_2 </math>

If we solve one of these equations for <math>d_1</math> and substitute into the other, we can solve for <math>d_2</math> to get:

:<math> d_2 = \frac{h}{1+\frac{A_2}{A_1}}</math>

Then we can plug in our numbers to get a value for <math>d_2</math>:

:<math>h = \frac{0.4246m}{1+\frac{\pi (1m)^2}{\pi (0.2415m)^2}} \approx 0.0234m</math>

Thus the elephant was lifted <math>2.34cm</math>.

One strategy for these types of problems is to simple memorize the most relevant formulas, <math>P_1 = P_2 + \rho g h</math> and <math>d_2 = \frac{h}{1+\frac{A_2}{A_1}}</math>, but notice how the whole solution of the problem was based solely on the idea that pressure at equal heights of an incompressible fluid is equal. If you remember this fact, you should be able to use simple reasoning and algebra to adapt to many kinds of problems without memorizing which formulas to use.

==Bernoulli's Equation==

Bernoulli's Equation relates the pressure of a fluid with its height and velocity, allowing you to predict the properties as a fluid at a general point in a complicated system pipes based on your knowledge of it a some particular point.
The full equation takes the following form:

: <math>v^2 \frac{\rho}{2} + \rho g h + P = \mathsf{const}</math>

<math>v~</math> is the velocity of the fluid in metres per second (<math>m/s</math>)

<math>g~</math> is the gravitational constant in metres per second squared (<math>m/s^2</math>)

<math>h~</math> is the height of the fluid in meteres (<math>m</math>)

<math>p~</math> is the pressure of the fluid in Pascals (<math>Pa</math>)

<math>\rho~</math> is the density of the liquid in kilograms per metre cubed (<math>kg/m^3</math>)

<math>\mathsf{const}</math> is a constant.

Another useful fact to remember related to bernoulli's equation is the fact that
in a steady flow of an incompressible liquid, the total rate of volume flow per
unit time at any point is constant. This is easy to understand because it is
basically a statement that since the liquid is incompressible, it can't "bunch
up" or "spread out" at any point (the density of the liquid is constant). This
means that if you have liquid flowing through a pipe with a shrinking diameter,
the velocity of the liquid much increase: Consider that after a small time
<math>dt</math> a small bunch of liquid travels a distance <math>dx = v dt</math>,
where <math>v</math> is the velocity. Then the total volume that has been
displaced is <math>dV = Adx</math>, where A is the area of the pipe at this
point. In other words: <math> dV = A (v dt)</math>, and thus we say:

:<math>\frac{dV}{dt} = Av = \mathsf{const}</math>

===Problem Solving===

'''Q''': A horizontal circular pipe 10.0 cm in diameter has a smooth reduction in diameter
to a pipe 5.00 cm in diameter but remains at a constant height. If the pressure
of the water in the larger pipe is 8.00 x 10^4 Pa and the pressure in the
smaller pipe is 6.00 x 10^4 Pa, at what rate does water flow through the pipe?

'''A''': Using Bernoulli's equation, we can say:

:<math>\frac{{v_1}^2\rho}{2} + P_1 = \frac{{v_2}^2\rho}{2} + P_2</math>

Where we have cancelled the <math>h</math> terms since the height is constant.

Next we can say that since the volume flow rate is constant:

:<math>A_1v_1 = A_2v_2</math>

:<math>v_1 = \frac{A_2}{A_1} v_2</math>

Plugging this into the previous line give us:

:<math>\frac{{\left(\frac{A_2}{A_1}\right)}^2 {v_2}^2 \rho}{2} + P_1 = \frac{{v_2}^2\rho}{2} + P_2</math>

Which we can solve for <math>v_2</math>

:<math>v_2 = \sqrt{\frac{2\left(P_1 - P_2\right)}{\rho \left(1 - {\left(\frac{A_2}{A_1}\right)}^2\right)}}</math>

Plugging in the numbers from the question gives us:

<math>~v_2 = 6.532 m/s</math>

Now, we want to find the flow rate of the water, so which is just the area times
the velocity at some point in the flow. So we've solve for the velocity in the
smaller pipe, so we can calculate the flow rate using the area of the smaller
pipe:

<math>\frac{dV}{dt} = A_2v_2 = 0.0128 m^3/s</math>

*Back to [[PhysicsHelp]]

PhysicsHelp

2009-11-15T23:34:13Z

RaymondGoerke: /* Resources */

[[Image:Hand_on_Water.jpg|right|frame|uploaded to Flickr by [http://www.flickr.com/photos/cdm/53197139/sizes/m/]]]

== Plan to add to this wiki? ==
You'll need to [http://wiki.ubc.ca/Main_Page login with your CWL] first .

== Physics Tutoring Resources ==
[[http://www.physics.ubc.ca/index.phtml UBC Dept of Physics and Astronomy]]: seminars, links to resources

====UBC Physics Courses:====

[[How to write a good exam ]]

=====Physics 101: Energy and Waves=====

[[Buoyancy, Pressure, Bernoulli's Equation]]

[[Thermodynamics and Heat Transfer]]

[[Simple and Damped Harmonic Motion]]

[[Waves and the Doppler Effect]]

=====Physics 102: Electricity, Light and Radiation=====

[[Electricity]]

[[Capacitors]]

[[Resistors]]

[[Electric Circuits]]

====Quicklinks to Challenging topics in First Year Physics:====

[[Motions and Mechanics]]

[[Acceleration Velocity Position]]

[[Uncertainty and Error]]

[[Wedges]]

[[Centripedal Acceleration]]

[[Temperature]]

[[Pressure]]

[[Bernoullis Equation]]

[[Waves]]

[[Light Waves]]

[[Bouyancy]]

[[Light Bulbs]]

====Mathematics for Physics====

[[Differential Equations]]

[[Isotopes Half Life]]

====Student Societies:====

[[http://www.physics.ubc.ca/~physsoc/ UBC Physics Society]]: private tutors, exam packs, links to resources

[[http://www.phas.ubc.ca/~fizz/ FIZZ]]: Engineering Physics Student Society Homepage: exam database, forums.

[[http://www.physics.ubc.ca/~biophys/ Biophysics Student Society]]: Society Homepage

[[http://www.tutoring.ams.ubc.ca AMS Tutoring Services]]: Information about Drop-In, Online, and Appointment tutoring with AMS Tutors.

====Resources====

[[http://www.quantum-physics.polytechnique.fr/en/index.html Ecole Polytechnic's]] Quantum Physics online animations/ java applets

[[http://jersey.uoregon.edu/vlab/ University of Oregon Department of Physics]] vitual lab resources (animations, etc) for learning astrophysics, energy and environment, thermodynamics, mechanics

====People to talk to====

[[Category:Tutoring Wikis]]

PhysicsHelp

2009-11-15T23:33:40Z

RaymondGoerke: /* Physics 101: Energy and Waves */

[[Image:Hand_on_Water.jpg|right|frame|uploaded to Flickr by [http://www.flickr.com/photos/cdm/53197139/sizes/m/]]]

== Plan to add to this wiki? ==
You'll need to [http://wiki.ubc.ca/Main_Page login with your CWL] first .

== Physics Tutoring Resources ==
[[http://www.physics.ubc.ca/index.phtml UBC Dept of Physics and Astronomy]]: seminars, links to resources

====UBC Physics Courses:====

[[How to write a good exam ]]

=====Physics 101: Energy and Waves=====

[[Buoyancy, Pressure, Bernoulli's Equation]]

[[Thermodynamics and Heat Transfer]]

[[Simple and Damped Harmonic Motion]]

[[Waves and the Doppler Effect]]

=====Physics 102: Electricity, Light and Radiation=====

[[Electricity]]

[[Capacitors]]

[[Resistors]]

[[Electric Circuits]]

====Quicklinks to Challenging topics in First Year Physics:====

[[Motions and Mechanics]]

[[Acceleration Velocity Position]]

[[Uncertainty and Error]]

[[Wedges]]

[[Centripedal Acceleration]]

[[Temperature]]

[[Pressure]]

[[Bernoullis Equation]]

[[Waves]]

[[Light Waves]]

[[Bouyancy]]

[[Light Bulbs]]

====Mathematics for Physics====

[[Differential Equations]]

[[Isotopes Half Life]]

====Student Societies:====

[[http://www.physics.ubc.ca/~physsoc/ UBC Physics Society]]: private tutors, exam packs, links to resources

[[http://www.phas.ubc.ca/~fizz/ FIZZ]]: Engineering Physics Student Society Homepage: exam database, forums.

[[http://www.physics.ubc.ca/~biophys/ Biophysics Student Society]]: Society Homepage

[[http://www.tutoring.ams.ubc.ca AMS Tutoring Services]]: Information about Drop-In, Online, and Appointment tutoring with AMS Tutors.

====Resources====

[[http://www.quantum-physics.polytechnique.fr/en/index.html Ecole Polytechnic's]] Quantum Physics online animations/ java applets

[[http://jersey.uoregon.edu/vlab/ University of Oregon Department of Physics]] vitual lab resources (animations, etc) for learning astrophysics, energy and environment, thermodynamics, mechanics

====People to talk to====

[[Category:Tutoring Wikis]]

PhysicsHelp

2009-11-15T23:33:11Z

RaymondGoerke: /* Physics Tutoring Resources */

[[Image:Hand_on_Water.jpg|right|frame|uploaded to Flickr by [http://www.flickr.com/photos/cdm/53197139/sizes/m/]]]

== Plan to add to this wiki? ==
You'll need to [http://wiki.ubc.ca/Main_Page login with your CWL] first .

== Physics Tutoring Resources ==
[[http://www.physics.ubc.ca/index.phtml UBC Dept of Physics and Astronomy]]: seminars, links to resources

====UBC Physics Courses:====

[[How to write a good exam ]]

=====Physics 101: Energy and Waves=====

[[Buoyancy, Pressure, Bernoulli's Equation]]

[[Simple and Damped Harmonic Motion]]

[[Waves and the Doppler Effect]]

[[Thermodynamics and Heat Transfer]]

=====Physics 102: Electricity, Light and Radiation=====

[[Electricity]]

[[Capacitors]]

[[Resistors]]

[[Electric Circuits]]

====Quicklinks to Challenging topics in First Year Physics:====

[[Motions and Mechanics]]

[[Acceleration Velocity Position]]

[[Uncertainty and Error]]

[[Wedges]]

[[Centripedal Acceleration]]

[[Temperature]]

[[Pressure]]

[[Bernoullis Equation]]

[[Waves]]

[[Light Waves]]

[[Bouyancy]]

[[Light Bulbs]]

====Mathematics for Physics====

[[Differential Equations]]

[[Isotopes Half Life]]

====Student Societies:====

[[http://www.physics.ubc.ca/~physsoc/ UBC Physics Society]]: private tutors, exam packs, links to resources

[[http://www.phas.ubc.ca/~fizz/ FIZZ]]: Engineering Physics Student Society Homepage: exam database, forums.

[[http://www.physics.ubc.ca/~biophys/ Biophysics Student Society]]: Society Homepage

[[http://www.tutoring.ams.ubc.ca AMS Tutoring Services]]: Information about Drop-In, Online, and Appointment tutoring with AMS Tutors.

====Resources====

[[http://www.quantum-physics.polytechnique.fr/en/index.html Ecole Polytechnic's]] Quantum Physics online animations/ java applets

[[http://jersey.uoregon.edu/vlab/ University of Oregon Department of Physics]] vitual lab resources (animations, etc) for learning astrophysics, energy and environment, thermodynamics, mechanics

====People to talk to====

[[Category:Tutoring Wikis]]

Waves and the Doppler Effect

2009-11-15T23:13:19Z

RaymondGoerke: /* Doppler Effect */

Return to [[PhysicsHelp]]

==Travelling Waves==

In phyics, we often encouter waves which, in addition to having an amplitude
at each point in space, also appear to move in some direction over time.
Such objects are called travelling waves. A simple sinusoidal travelling wave
can be represented by the equation:

<math>y = cos(kx - \omega t)~</math>

Where we call <math>k</math> the wavenumber, which has units of radians per
metre (<math>m^{-1}</math>, and <math>\omega</math> is the angular
frequency with units of radians per second <math>s^{-1}</math>. At any
particular time, we can think of <math>-\omega t</math> as a phase shift
<math>\phi~</math>, and then this is the eqation of a normal wave defined at
every point along the x-axis:

<math>y = cos(kx + \phi)~</math>

If we remember from our previous analysis of waves that <math>\phi~</math>
represents a shift of the waveform to the left or the right, it makes sense that
when <math>\phi~</math> depends on time, the wave with be continuously shifting
to the left or the right.

Recall also that a positive <math>\phi~</math> corresponds to the wave shifting
to the left, and a negative <math>\phi~</math> corresponds to the wave shifting
to the right. Thinking about that, we can see why we use <math>cos(kx - \omega t)</math>
and not <math>cos(kx + \omega t)</math>, because in this case whenever <math>
k</math> and <math>\omega</math> are both positive, the wave moves to the right,
which is the direction we usually associate with positive numbers.

===Direction of Travelling Waves===

[[File:travelingWave2.png|left|thumb|Two snapshots of a travelling wave at t=0
and t=1, with k=1 and w=1. Note that when the time changes by dt, the wave
shifts to the right by the quantity w/k*dt.]]

It is pretty easy to tell by glancing at the eqution of a travelling wave which
direction it is moving. To do this we need to recall the relationships between
various numerical parameters of a wave. Recall that:

<math>k = \frac{2\pi}{\lambda}</math>

and

<math>\omega = 2\pi f~</math>

Where <math>\lambda</math> is the wavelength and <math>f</math> is the
frequency in Hertz, or number of complete wavelengths per second. Since the
wavelength is defined to be the length in metres of one wavelength, and the
frequency is the number of complete wavelengths per second, it follows that
the quantity <math>f\lambda</math> has units of metres per second and is the
velocity of the wave. We can re-express <math>f</math> and <math>\lambda</math>
in terms of the parameters in the equation by a travelling wave by using the
eqations above:

<math>v = f\lambda = \frac{\omega}{2\pi} \frac{2\pi}{k} = \frac{\omega}{k}</math>

All this really means is that when you see the equation of a wave, say:

<math>cos(2x + 3t)~</math>

in order to determine which direction is is moving, it is sufficient to read off
the values for <math>k</math> and <math>\omega</math>:

<math>k = 2</math>

<math>\omega = -3</math>

thus:

<math>v = \frac{\omega}{k} = \frac{-3}{2} = -1.5 m/s</math>

The only possibly confusing part is to remember that there is a negative sign
in front of the <math>\omega</math> in the general equation for the travelling
wave, so you have to flip the sign of the number in front of <math>t</math>.

==Doppler Effect==

[[image:Doppler.jpg|left|thumb|The Doppler Effect]]

The Doppler Effect is observed when you have sources of light, sound, or other
kinds of waves being emitted from a source moving with respect to the observer.
If you stand still and a source is emitting waves at a frequency <math>f</math>,
you will observe the frequency <math>f</math>. If you or the source moves, then
the frequency will shift based on your relative motion. The most common example
of the doppler effect is a car with a siren speeding down the road as you stand
on the sidewalk. The siren changes tone as it passed you from high to low
(corresponding to the change in frequency of the sound waves from hight to low).
The general equation for the observed frequency when a source and observer are
moving is:

<math>f_o = f_s \left(\frac{v \pm v_o}{v \mp v_s}\right)</math>

<math>f_o~</math> is the frequency measured by the observer in Hertz (<math>Hz</math>).

<math>f_s~</math> is the emitted frequency (from the source) in Hertz (<math>Hz</math>).

<math>v~</math> is the speed of the wave (for sound, the speed of sound) in
metres per second (<math>m/s</math>).

<math>v_o~</math> is the speed of the observer in meters per second (<math>m/s</math>).

<math>v_s~</math> is the speed of the source in meters per second (<math>m/s</math>).

The only confusing part of the doppler effect is choosing the right sign for
<math>v_o</math> and <math>v_s</math>. There are is a simple convention which is
described by two statements:

If the object and the observer are moving towards each other, <math>v_o</math> must be positive.

The sign for <math>v_s</math> is chosen to be consistent with the direction for <math>v_o</math>
(i.e. if the observer is moving in the same direction as the source,
then they will have the same sign. If they're travelling in opposite directions,
they will have opposite sign). This sign convention is chosen because when the
source and observer are moving towards each other, the observed frequency is higher than
the emitted frequency. Similarly:

If the source and the observer are moving away from each other, <math>v_o</math> must be negative.

Again, the sign of <math>v_s</math> is chosen to be consistent with the sign of
<math>v_o</math>. This convention ensures that when the observer and the source
are moving away from each other the observed frequency is less than the emitted
frequency.

Note, as alluded to above, that the source and the observer might be moving in the
same direction, in which case you must determined their relative velocity to
decide whether they will eventually meet or if they will drift apart. (In these
cases <math>v_o</math> and <math>v_s</math> will have the same sign.

Return to [[PhysicsHelp]]

Help:Editing Math Equations using TeX

2009-11-09T03:55:51Z

RaymondGoerke: Created page with 'For future editors of the PhysicsHelp page: This is how you edit math equations using the LaTeX syntax to make nice looking equations. If you already use LaTeX, then all you nee…'

Waves and the Doppler Effect

2009-11-09T03:14:13Z

RaymondGoerke: /* Doppler Effect */

Return to [[PhysicsHelp]]

==Travelling Waves==

In phyics, we often encouter waves which, in addition to having an amplitude
at each point in space, also appear to move in some direction over time.
Such objects are called travelling waves. A simple sinusoidal travelling wave
can be represented by the equation:

<math>y = cos(kx - \omega t)~</math>

Where we call <math>k</math> the wavenumber, which has units of radians per
metre (<math>m^{-1}</math>, and <math>\omega</math> is the angular
frequency with units of radians per second <math>s^{-1}</math>. At any
particular time, we can think of <math>-\omega t</math> as a phase shift
<math>\phi~</math>, and then this is the eqation of a normal wave defined at
every point along the x-axis:

<math>y = cos(kx + \phi)~</math>

If we remember from our previous analysis of waves that <math>\phi~</math>
represents a shift of the waveform to the left or the right, it makes sense that
when <math>\phi~</math> depends on time, the wave with be continuously shifting
to the left or the right.

Recall also that a positive <math>\phi~</math> corresponds to the wave shifting
to the left, and a negative <math>\phi~</math> corresponds to the wave shifting
to the right. Thinking about that, we can see why we use <math>cos(kx - \omega t)</math>
and not <math>cos(kx + \omega t)</math>, because in this case whenever <math>
k</math> and <math>\omega</math> are both positive, the wave moves to the right,
which is the direction we usually associate with positive numbers.

===Direction of Travelling Waves===

[[File:travelingWave2.png|left|thumb|Two snapshots of a travelling wave at t=0
and t=1, with k=1 and w=1. Note that when the time changes by dt, the wave
shifts to the right by the quantity w/k*dt.]]

It is pretty easy to tell by glancing at the eqution of a travelling wave which
direction it is moving. To do this we need to recall the relationships between
various numerical parameters of a wave. Recall that:

<math>k = \frac{2\pi}{\lambda}</math>

and

<math>\omega = 2\pi f~</math>

Where <math>\lambda</math> is the wavelength and <math>f</math> is the
frequency in Hertz, or number of complete wavelengths per second. Since the
wavelength is defined to be the length in metres of one wavelength, and the
frequency is the number of complete wavelengths per second, it follows that
the quantity <math>f\lambda</math> has units of metres per second and is the
velocity of the wave. We can re-express <math>f</math> and <math>\lambda</math>
in terms of the parameters in the equation by a travelling wave by using the
eqations above:

<math>v = f\lambda = \frac{\omega}{2\pi} \frac{2\pi}{k} = \frac{\omega}{k}</math>

All this really means is that when you see the equation of a wave, say:

<math>cos(2x + 3t)~</math>

in order to determine which direction is is moving, it is sufficient to read off
the values for <math>k</math> and <math>\omega</math>:

<math>k = 2</math>

<math>\omega = -3</math>

thus:

<math>v = \frac{\omega}{k} = \frac{-3}{2} = -1.5 m/s</math>

The only possibly confusing part is to remember that there is a negative sign
in front of the <math>\omega</math> in the general equation for the travelling
wave, so you have to flip the sign of the number in front of <math>t</math>.

==Doppler Effect==

[[image:Doppler.jpg|left|thumb|The Doppler Effect]]

The Doppler Effect is observed when you have sources of light, sound, or other
kinds of waves being emitted from a source moving with respect to the observer.
If you stand still and a source is emitting waves at a frequency <math>f</math>,
you will observe the frequency <math>f</math>. If you or the source moves, then
the frequency will shift based on your relative motion. The most common example
of the doppler effect is a car with a siren speeding down the road as you stand
on the sidewalk. The siren changes tone as it passed you from high to low
(corresponding to the change in frequency of the sound waves from hight to low).
The general equation for the observed frequency when a source and observer are
moving is:

<math>f_o = f_s \left(\frac{v + v_o}{v + v_s}\right)</math>

<math>f_o~</math> is the frequency measured by the observer in Hertz (<math>Hz</math>).

<math>f_s~</math> is the emitted frequency (from the source) in Hertz (<math>Hz</math>).

<math>v~</math> is the speed of the wave (for sound, the speed of sound) in
metres per second (<math>m/s</math>).

<math>v_o~</math> is the speed of the observer in meters per second (<math>m/s</math>).

<math>v_s~</math> is the speed of the source in meters per second (<math>m/s</math>).

The only confusing part of the doppler effect is choosing the right sign for
<math>v_o</math> and <math>v_s</math>. There are is a simple convention which is
described by two statements:

If the object and the observer are moving towards each other, <math>v_o</math> must be positive.

The sign for <math>v_s</math> is chosen to be consistent with the direction for <math>v_o</math>
(i.e. if the observer is moving in the same direction as the source,
then they will have the same sign. If they're travelling in opposite directions,
they will have opposite sign). This sign convention is chosen because when the
source and observer are moving towards each other, the observed frequency is higher than
the emitted frequency. Similarly:

If the source and the observer are moving away from each other, <math>v_o</math> must be negative.

Again, the sign of <math>v_s</math> is chosen to be consistent with the sign of
<math>v_o</math>. This convention ensures that when the observer and the source
are moving away from each other the observed frequency is less than the emitted
frequency.

Note, as alluded to above, that the source and the observer might be moving in the
same direction, in which case you must determined their relative velocity to
decide whether they will eventually meet or if they will drift apart. (In these
cases <math>v_o</math> and <math>v_s</math> will have the same sign.

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Waves and the Doppler Effect

2009-11-09T01:27:52Z

RaymondGoerke: /* Doppler Effect */

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==Travelling Waves==

In phyics, we often encouter waves which, in addition to having an amplitude
at each point in space, also appear to move in some direction over time.
Such objects are called travelling waves. A simple sinusoidal travelling wave
can be represented by the equation:

<math>y = cos(kx - \omega t)~</math>

Where we call <math>k</math> the wavenumber, which has units of radians per
metre (<math>m^{-1}</math>, and <math>\omega</math> is the angular
frequency with units of radians per second <math>s^{-1}</math>. At any
particular time, we can think of <math>-\omega t</math> as a phase shift
<math>\phi~</math>, and then this is the eqation of a normal wave defined at
every point along the x-axis:

<math>y = cos(kx + \phi)~</math>

If we remember from our previous analysis of waves that <math>\phi~</math>
represents a shift of the waveform to the left or the right, it makes sense that
when <math>\phi~</math> depends on time, the wave with be continuously shifting
to the left or the right.

Recall also that a positive <math>\phi~</math> corresponds to the wave shifting
to the left, and a negative <math>\phi~</math> corresponds to the wave shifting
to the right. Thinking about that, we can see why we use <math>cos(kx - \omega t)</math>
and not <math>cos(kx + \omega t)</math>, because in this case whenever <math>
k</math> and <math>\omega</math> are both positive, the wave moves to the right,
which is the direction we usually associate with positive numbers.

===Direction of Travelling Waves===

[[File:travelingWave2.png|left|thumb|Two snapshots of a travelling wave at t=0
and t=1, with k=1 and w=1. Note that when the time changes by dt, the wave
shifts to the right by the quantity w/k*dt.]]

It is pretty easy to tell by glancing at the eqution of a travelling wave which
direction it is moving. To do this we need to recall the relationships between
various numerical parameters of a wave. Recall that:

<math>k = \frac{2\pi}{\lambda}</math>

and

<math>\omega = 2\pi f~</math>

Where <math>\lambda</math> is the wavelength and <math>f</math> is the
frequency in Hertz, or number of complete wavelengths per second. Since the
wavelength is defined to be the length in metres of one wavelength, and the
frequency is the number of complete wavelengths per second, it follows that
the quantity <math>f\lambda</math> has units of metres per second and is the
velocity of the wave. We can re-express <math>f</math> and <math>\lambda</math>
in terms of the parameters in the equation by a travelling wave by using the
eqations above:

<math>v = f\lambda = \frac{\omega}{2\pi} \frac{2\pi}{k} = \frac{\omega}{k}</math>

All this really means is that when you see the equation of a wave, say:

<math>cos(2x + 3t)~</math>

in order to determine which direction is is moving, it is sufficient to read off
the values for <math>k</math> and <math>\omega</math>:

<math>k = 2</math>

<math>\omega = -3</math>

thus:

<math>v = \frac{\omega}{k} = \frac{-3}{2} = -1.5 m/s</math>

The only possibly confusing part is to remember that there is a negative sign
in front of the <math>\omega</math> in the general equation for the travelling
wave, so you have to flip the sign of the number in front of <math>t</math>.

==Doppler Effect==

[[image:Doppler.jpg|left|thumb|The Doppler Effect]]
If you stand still and a source is emitting waves at a frequency '''F''', then you will hear them at '''F'''. If you or the source moves then the frequency you hear the waves can be written as:'''F' = F*(V +- V_o)/(V -+ V_s)'''

'''F'''' is the new frequency you see, in __Hz__
'''F''' is the emitted frequency, in __Hz__
'''V''' is the speed of the wave (for sound, the speed of sound), in __meters / second__
'''V_o''' is the speed of the object, in __meters / second__
'''V_s''' is the speed of the source (emitter of the waves), in __meters / second__

The most complicated part about this is the sign of '''V_o''' and '''V_s''', and they obey the following rules:
If the object is moving towards the source, then '''V_o''' is positive.
If the object is moving away from the source, then '''V_o''' is negative.
If the source is moving towards the object, then '''V_s''' is negative.
If the source is moving away from the object, then '''V_s''' is positive.

The best way to remember this is that, in general, if the object or the source are moving towards each other (either one moving) then '''f' > f''', and if the objects are moving away from each other then '''f'< f'''. This is an easy way to remember the correct sign.

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