Interesting. I've never heard of this before, but according to

wikipedia, simple harmonic motion is given by:

$\displaystyle x(t)=A\cos{(2\pi ft+\phi)}$

where $\displaystyle x(t)$ is displacement, $\displaystyle t$ is time, $\displaystyle A$ is amplitude, $\displaystyle f$ is frequency, and $\displaystyle \phi$ is phase.

We also note that period $\displaystyle T$ is given by:

$\displaystyle T=\frac{1}{f}$

This means that $\displaystyle f=\frac{3}{\pi}$. So:

$\displaystyle x(t)=A\cos{(2\pi \frac{3}{\pi}t+\phi)}=A\cos{(6t+\phi)}$

To find velocity, we take the derivative:

$\displaystyle v(t)=x'(t)=-6A\sin{(6t+\phi)}$

To find velocity extrema, we take the derivative of $\displaystyle v(t)$ and set it equal to zero:

$\displaystyle v'(t)=a(t)=-36A\cos{(6t+\phi)}$

$\displaystyle -36A\cos{(6t+\phi)}=0$

$\displaystyle 6t+\phi=\frac{(2n-1)\pi}{2}$

$\displaystyle t=\frac{(2n-1)\pi-2\phi}{12}$

Let's plug that into our velocity function:

$\displaystyle -6A\sin{[6\frac{(2n-1)\pi-2\phi}{12}+\phi]}=5$

$\displaystyle A=-\frac{5}{6\sin{[\frac{(2n-1)\pi}{2}]}}$

$\displaystyle A=\{-\frac{5}{6\sin{[\frac{\pi}{2}]}},-\frac{5}{6\sin{[\frac{3\pi}{2}]}}\}$

$\displaystyle A=\{-\frac{5}{6},\frac{5}{6}\}$

Since the amplitude must be a positive value, we can just say:

$\displaystyle A=\frac{5}{6}$

Now, the center of oscillation is another way of saying $\displaystyle x(t)=0$. So:

$\displaystyle x(t)=\frac{5}{6}\cos{(6t+\phi)}=0$

$\displaystyle \cos{(6t+\phi)}=0$

$\displaystyle 6t+\phi=\frac{(2n-1)\pi}{2}$

$\displaystyle t=\frac{(2n-1)\pi-2\phi}{12}$

Let's let $\displaystyle t=0$ and $\displaystyle n=1$:

$\displaystyle 0=\frac{([2(1)-1]\pi-2\phi}{12}$

$\displaystyle \phi=\frac{\pi}{2}$

And we plug that into our velocity function:

$\displaystyle v(t)=-5\sin{(6t+\frac{\pi}{2})}$

Then plug in $\displaystyle t$:

$\displaystyle v(0.2)=-5\sin{(6[0.2]+\frac{\pi}{2})}\approx-1.81$

But of course we know that speed is relative, and so the value is actually:

$\displaystyle v(t_0+0.2)\approx\pm1.81\;|\;x(t_0)=0$