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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. – Raskolnikov Jun 14 '12 at 10:28
up vote 3 down vote accepted

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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Could you please explain the maximised part further? I'm not sure what that means. – Jallah Jun 14 '12 at 11:01
Cosines vary between -1 and 1. So if you want them to be maximum, that means they're 1. – user22805 Jun 14 '12 at 11:05

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