Question

e. at what times and positions is the velocity a maximum?
a. t = 0, 2π, 4π,...; the position is (0, - 36), (2π, - 36), (4π, - 36),...
b. t = 3π/2, 5π/2, 7π/2,...; the position is (3π/2, - 72), (5π/2, 0), (7π/2, - 72)
c. t = π, 3π, 5π,...; the position is (π,0), (3π,0), (5π,0),...
d. t = π/2, 3π/2, 5π/2,...; the position is (π/2,0), (3π/2, - 72), (5π/2,0)
f. the acceleration of the oscillator is a(t)=v(t). find and graph the acceleration function
a(t) =

Explanation:

Step1: Recall the relationship between position, velocity and acceleration in simple - harmonic motion

In simple - harmonic motion, the velocity $v(t)$ is the derivative of the position function $x(t)$ and the acceleration $a(t)$ is the derivative of the velocity function $v(t)$ (i.e., $a(t)=v^{\prime}(t)$). The velocity of a simple - harmonic oscillator is maximum when the acceleration is zero.

Step2: Analyze the given acceleration function and find when it is zero

For a simple - harmonic oscillator, the acceleration function is often of the form $a(t)= - A\omega^{2}\cos(\omega t+\varphi)$. The acceleration $a(t) = 0$ when $\cos(\omega t+\varphi)=0$. This occurs when $\omega t+\varphi=(2n + 1)\frac{\pi}{2}$, where $n = 0,1,2,\cdots$. In the case of a standard simple - harmonic motion starting from the equilibrium position, when $a(t)=0$, the velocity is maximum.
For a simple - harmonic oscillator, the general form of the motion has acceleration $a(t)$ being zero at $t=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\cdots$.

Answer:

D. $t=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\cdots$; the position is $(\frac{\pi}{2},0),(\frac{3\pi}{2}, - 72),(\frac{5\pi}{2},0)\cdots$