max velocity of SHM

So I think my algebra is wrong somewhere. Lets say you have an object that goes under SHM with some initial conditions (amplitude is 5m. The piston is at 5m at t = 0 and period is 20 seconds).

Okay so $x(t) = A sin(\omega_0 + \phi)$. $\phi$ for us is 0. The frequency = $0.05 Hz$, angular frequency = $\frac{\pi}{10}$ and period of the piston = $20s$. The maximum velocity is is when the first derivative $x'(t) = \omega_0 A cos(\omega_0 t)$

The maximum velocity is when the cosine function is 1 at x = 0. So the maximum velocity is $v(t) = A\omega_0$ so that means $cos(\omega_0 t)= 1$ but $\omega_0 = \frac{\pi}{10}$ so the missing value is t. But thats trivial to find, since cosine function is 1. So $t = 20s$

So here is my issue. The maximum velocity can only be at x = 0 (when kinetic energy is max, and potential energy is zero). But the value t = 20s represents the object at x = Amplitude.

So what am I doing wrong?

edit: I just realized that if $x(t) = A sin(\omega_0 + \phi)$ then the initial conditions are not satisfied. When t = 0, sin function goes to 0 and hence everything is 0 but that dosnt satisfy initial conditions. When $t = 0$, displacement should be $5m$ However, if i replace $sin$ with $cos$ then everything works. Is that okay?

-

But the value t = 20s represents the object at x = Amplitude.

Check this calculation again. Remember your formula is $x(t) = A sin(\omega_0 t)$, and you should get full credit on your homework.

(or at least mostly full credit. There are many values of $t$ that satisfy $cos(\omega_0 t)=1$, right? you've found only one, which isn't the most general answer you could give.)

-
well $t= 2n$ for any $n \in \mathbb{Z}$ but that time still corresponds to x = amplitude. NEVERMIND: the 0.1 displacement means that we have a phase constant? edit3: Infact can i replace my formula with $x(t) = A cos(\omega_0 t)$? – masfenix Jan 30 '11 at 0:32
Three things: 1. Very close on $t$, but not quite - think of it as $cos(x)=1$ for $x=n\pi$. Plug in the value of $x$ that you see in your equations and solve for the time that satisfies $x=n\pi$ 2. Yes, in your case the phase constant $\phi$ should not be zero. This means your value for $t$ is not quite right - but only off by a small bit. 3. Why is that replacement you make justified? – spencer nelson Jan 30 '11 at 0:47
Hi, in my textbook it says that the phase velocity is zero when the object is at max displacement at t = 0 (which is the case). Secondly, the second order D.E we are trying to solve is $a + \frac{kx}{m} = 0$ and if I use $x(t) = A cos(\omega_0 t)$ then it is a solution (atleast from what I think, I dont have much practice with second order DE). I am also kind of confused by your number (1). I have also edited my original post to let you know what I mean by replace $sin$ with $cos$. – masfenix Jan 30 '11 at 0:54
@masfenix Ah! I somehow misread the initial position and thought it was 0.05m, but that's the frequency. You are completely correct that the phase constant is zero - but 1. and 3. remain. – spencer nelson Jan 30 '11 at 0:57
Okay so $t=10n$ but I am still confused. If we let $n=0$ or $n=2$ then you are still at the case when the object is at its max displacement. So how can it be at max velocity when its also at max displacement? – masfenix Jan 30 '11 at 1:05

What is the solution of $\sin(\omega t)=0$ ? It is $t=n\,\frac{\pi}{\omega}$ where $n$ is 0, 1, 2, 3, ...

For your case the answer is $t=n\,10$ or $t=0,10,20\ldots$.

-