Here's the question I would like help with:
A particle is moving in simple harmonic motion according to $x=6\sin \left (2t+\frac{\pi }{2} \right )$. Find the first two times when the velocity is maximum, and the position then.
Here is my working. I then let $x=0$ and did the following:
$0=6\sin \left (2t+\frac{\pi }{2} \right )$ $\pi =2t+\frac{\pi }{2} $ $\frac{\pi}{2}=2t$ $\frac{\pi}{4}=t$
According to my textbook, the answer I got is incorrect. The provided answer is $t=\frac{3\pi}{4}, \frac{7\pi}{4}$.
Could some one please identify where I went wrong and explain the proper way of solving this question?