# Simple Harmonic Motion. Why am I wrong? Why is my equation wrong more importantly?

Problem/Solution

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I am deeply confused.

B) We know that

$x = 2\sin(3\pi t)$.

$x' = 6\pi\cos(3\pi t)$

So max speed is $6\pi$

$6\pi = 6\pi \cos(3\pi t)$

$\cos(3\pi t) = 1$

$3\pi t = 2\pi$ (reject 0 since t >0)

$t = 2/3$

But it isn't 2/3s. I understand where the solution is coming from, but I don't understand why I am wrong.

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Just because the maximum speed is $6\pi\text{ cm/s}$ does not mean that $6\pi = 6\pi \cos(3\pi t)$. Keep in mind that speed is the absolute value of velocity $x'$.

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What do you mean it doesn't mean the max speed isn't $6\pi cm/s$? My equation for velocity is x', I set it equal to $6\pi cm/s$ to find when that happens. – Hawk Jan 1 '12 at 4:06
The maximum speed is $6\pi\text{ cm/s}$. But that doesn't mean the velocity is $6\pi\text{ cm/s}$. Remember the difference between speed and velocity! – David Z Jan 1 '12 at 4:35
Dear Jak, imagine that one more keyword is "sign". This seems like a homework or exam of yours so maybe people shouldn't fully solve it for you. – Luboš Motl Jan 1 '12 at 13:05
No I understand the concept now. It's the holidays still in my book! – Hawk Jan 1 '12 at 19:08