Science:Math Exam Resources/Courses/MATH110/December 2017/Question 06 (c)

MATH110 December 2017

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Question 06 (c)

A small ball is attached to a vertical spring hanging from the ceiling. Suppose you pull the ball (vertically) down towards the ground. At ${\textstyle t=0}$ you let go and the ball begins to oscillate following the stretching and compression of the spring. Let $$h(t)=0.5\sin(2\pi (t-1))+2$$ be the position of the ball (in metres) at time ${\textstyle t}$ (in seconds) since it was released. Let the ground level be 0 m. (c) What is the acceleration of the ball at ${\textstyle t=1}$?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The acceleration is the rate of change of the velocity, hence the second derivative of the displacement (position).

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. By the Hint, we need to find ${\textstyle h''(1)}$. We first recall from part (b) that $$h'(t)=\pi \cos(2\pi (t-1)).$$ Using the chain rule with ${\textstyle F(t)=\cos(t)}$, ${\textstyle G(t)=2\pi (t-1)}$, ${\textstyle F'(t)=-\sin(t)}$, ${\textstyle G'(t)=2\pi }$, we have $${\begin{aligned}h''(t)&=\pi (F(G(t)))'\\&=\pi F'(G(t))G'(t)\\&=\pi (-\sin(2\pi (t-1)))\cdot 2\pi \\&=-2\pi ^{2}\sin(2\pi (t-1)).\end{aligned}}$$ At time ${\textstyle t=1}$, we have $${\begin{aligned}h''(1)&=-2\pi ^{2}\sin(2\pi (1-1))\\&=-2\pi ^{2}\sin(0)\\&=0.\end{aligned}}$$ Therefore, the acceleration at time ${\textstyle t=1}$ is ${\textstyle h''(1)=0}$. Answer: the acceleration at ${\textstyle t=1}$ is ${\textstyle {\color {blue}0}}$ (metre per second squared).