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?
The velocity is given by $v=12\cos\left(2t+\pi/2\right)$, which is maximised when the cosine of the part in brackets is 1. This happens when $2t+\pi/2$ is a whole multiple of $2\pi$. The first two such values of $t$ are $3\pi/4$ and $7\pi/4$.