Question

I'm stuck on this physics problem about oscillation on my homework. Any help would be great.
(The question is in the attached image below "capture.jpg")
Thank You!

Answer 1

Is the question in that image!

Answer 2

Yes

Answer 3

OK

Answer 4

The equation of your damped oscillator is:
\[mx''+\beta x'+k x=F(t)\]
where as usually, m is the mass of the oscillator, beta is the damping constant.
Now, since, if I call gamma=beta/m, we have:
\[\gamma=\frac{ \beta }{ m }=\frac{ .286 }{ 5.2 }=0.055\]
and:
\[\omega _{0}=\sqrt{\frac{ k }{ m }}=8.19 \sec ^{-1}\]
so we are in the condition:

\[\gamma << \omega _{0}\]
in other words the resonance curve is very narrow, as below:

where rho is a factor on whichthe amplitude of oscillations depends.
Namely:
\[\rho ^{2}=\frac{ 1 }{ 4m ^{2}\omega _{0}^{2}[(\omega _{0}-\omega)^{2}+\gamma ^{2.}/4 }\]
So the frequency f which maximazes the oscillator response is:
\[f=\frac{ 1 }{ 2*\pi }\sqrt{\frac{ k }{ m }}=\frac{ 8.19 }{ 6.28 }=1.3 \sec ^{-1}\]