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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?

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    You need the velocity to be maximum. You must have gotten a minimum.2012-06-14

1 Answers 1

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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 $7\pi/4$.

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    Cosines vary between -1 and 1. So if you want them to be maximum, that means they're 1.2012-06-14