spring mass undamped motion

A mass weighing 10 pounds stretches a spring 1/4 foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point 1/3 foot above the equilibrium position with a downward velocity of 5/4 ft/s.

At what time does the mass attain a displacement below the equilibrium position numerically equal to 1/2 the amplitude?

my position function is $\displaystyle x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)

$

I have tried to put it into the form $\displaystyle x(t)=Asin(\omega t+\phi)

$

and I am getting $\displaystyle x(t)=\frac{1}{3}sin(160t+\frac{\pi}{2})

$

The time I am getting is $\displaystyle \frac{\pi}{480}s$ initially, but I don't know if this is right. I'm also having trouble writing the answer as a series. Can anyone help me? Thanks