Question

f. the acceleration of the oscillator is a(t)=v(t). find and graph the acceleration function a(t)= - 36 sin t. graph the acceleration function for 0≤t≤10.

Explanation:

Step1: Recall properties of $y = A\sin(t)$

The function $a(t)= - 36\sin(t)$ is a sine - function. The general form of a sine function is $y = A\sin(t)$, where $A$ is the amplitude. For $a(t)=-36\sin(t)$, the amplitude $|A| = 36$. The sine function $y=\sin(t)$ has a period of $2\pi\approx6.28$, and it oscillates between $- 1$ and $1$. When multiplied by $-36$, the function $a(t)=-36\sin(t)$ oscillates between $-36$ and $36$.

Step2: Analyze the graphs

When $t = 0$, $a(0)=-36\sin(0)=0$. As $t$ increases from $0$ to $\frac{\pi}{2}\approx1.57$, $\sin(t)$ increases from $0$ to $1$, so $a(t)$ decreases from $0$ to $- 36$. As $t$ increases from $\frac{\pi}{2}$ to $\pi\approx3.14$, $\sin(t)$ decreases from $1$ to $0$, so $a(t)$ increases from $-36$ to $0$. As $t$ increases from $\pi$ to $\frac{3\pi}{2}\approx4.71$, $\sin(t)$ decreases from $0$ to $-1$, so $a(t)$ increases from $0$ to $36$. As $t$ increases from $\frac{3\pi}{2}$ to $2\pi$, $\sin(t)$ increases from $-1$ to $0$, so $a(t)$ decreases from $36$ to $0$.

Answer:

We need to find the graph of $y =-36\sin(t)$ for $0\leq t\leq10$. The period of $y = \sin(t)$ is $2\pi\approx6.28$, and in the interval $[0,10]$, the function $y=-36\sin(t)$ will complete about $1.6$ periods. By analyzing the key - points ($t = 0,a(0)=0$; $t=\frac{\pi}{2},a(\frac{\pi}{2})=-36$; $t = \pi,a(\pi)=0$; $t=\frac{3\pi}{2},a(\frac{3\pi}{2}) = 36$; $t = 2\pi,a(2\pi)=0$) and the behavior of the sine function, we can determine the correct graph. Without seeing the full details of the graphs A, B, C, and D, we note that the amplitude of the function $a(t)=-36\sin(t)$ is 36, meaning the function oscillates between $-36$ and $36$. If we assume the standard orientation of the $t$ - axis (horizontal) and $a(t)$ - axis (vertical), the graph should start at the origin $(0,0)$, go down to $y=-36$ at $t=\frac{\pi}{2}\approx1.57$, come back up to $y = 0$ at $t=\pi\approx3.14$, go up to $y = 36$ at $t=\frac{3\pi}{2}\approx4.71$ and back to $y = 0$ at $t = 2\pi\approx6.28$ and repeat part of the cycle until $t = 10$. If one of the graphs matches this description, that is the correct one. If we assume the graphs are properly scaled and labeled:

1. Check the amplitude: The function $a(t)=-36\sin(t)$ has an amplitude of 36, so the maximum and minimum values of the graph should be 36 and - 36 respectively.
2. Check the starting point: When $t = 0$, $a(0)=0$.

If we assume the graphs are drawn with the $t$ - axis horizontal and $a(t)$ - axis vertical, and proper scaling, the graph that starts at the origin $(0,0)$ and has an amplitude of 36 is the correct one. Without the ability to directly select from the visual graphs A, B, C, D, we can't give a specific letter - answer, but the above is the method to determine the correct graph. If we had to make a guess based on typical graph - drawing conventions, we look for a sine - wave that starts at the origin and has peaks and troughs at $y = 36$ and $y=-36$ respectively.