Question

HELP

Answer 1

As it's an simple harmonic oscillator the spring will continue it motion as far down as it was compressed eg. 3", this also means that the amplitude is 3". The period is the time it takes to get back eg. 0.8s. For the frequency you just use
\[f=\frac{ 1 }{ T }=\frac{ 1 }{ 0.8 s}=1.25Hz\]
Because it is an harmonic oscillator, it can be model by the following function where A is amplitude and f is frequency
\[x(t)=A*\cos(2 \pi f t)\]
So in our case
\[x(t)=3\cos(2*\pi*1.25t)\]
And to find its position at t=3 you simply evaluate the function at that point
\[x(3)=3*\cos(2*\pi*1.25*3)=-3\]