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

MATH110 December 2017

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Question 06 (d)

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. (d) Will the ball ever be at 1 metre from the ground? Provide a mathematical argument that supports your answer.

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
Recall that the sine function takes values only between ${\textstyle -1}$ and ${\textstyle 1}$.

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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 know that $$-1\leq \sin(2\pi (t-1))\leq 1.$$ Now we multiply by ${\textstyle 0.5}$ and add ${\textstyle 2}$: $$-0.5\leq 0.5\sin(2\pi (t-1))\leq 0.5,$$ $$1.5\leq 0.5\sin(2\pi (t-1))+2\leq 2.5.$$ This means that $$1.5\leq h(t)\leq 2.5.$$ Therefore, the ball will never be at ${\textstyle 1}$ metre from the ground. Answer: ${\textstyle {\color {blue}{\mbox{No}}}}$, the ball will never be at ${\textstyle 1}$ metre from the ground ${\textstyle {\color {blue}{\mbox{because }}1.5\leq h(t)\leq 2.5}.}$