How Does Oscillator Frequency Affect the Energy Ratio K/E?

[theuniverse]

Homework Statement

Express the ratio of the average kinetic energy K to the average total energy E of the oscillator in terms of the dimensionless quantity ωo/ω.

Homework Equations

I found that:
K = (1/2)mA^2ω^2 sin^2(ωt − δ)
E = (1/2)mA^2[ω^2 sin^2(ωt − δ) + ωo^2 cos^2(ωt − δ)]

The Attempt at a Solution

I know that my answer should be 1/[1 + (ωo/ω)^2].
I also found out that the total energy at the frequency of ω=ωo is: E=(1/2)mA^2ωo^2 (resonance) and I think it somehow relates to this problem.
I'm not quite sure how to reach the answer itself, or to make it more clear - I don't know how to come up with the terms for average K and average E.

Can anyone explain it to me?
Thank you for you time.

 

[issacnewton]

hi

basically, you can average over some time period. for example

$$K_{avg}=\frac{\int_{t_1}^{t_2}K(t)\,dt}{\int_{t_1}^{t_2}dt}$$

so how do you get t₁ and t₂ ? you can see that for

$$t=\frac{\delta}{\omega}$$

K is zero and again for

$$t=\frac{\pi+\delta}{\omega}$$

K is zero again. so we can let
$$t_1=\frac{\delta}{\omega}$$

and

$$t_2=\frac{\pi+\delta}{\omega}$$

which is the next value of t when K becomes zero. similarly you can average the total energy over the same range

$$E_{avg}=\frac{\int_{t_1}^{t_2}E(t)\,dt}{\int_{t_1}^{t_2}dt}$$

and then finally take the ratios...

 

[theuniverse]

I tried to take the integral of the energy expressions I had but the substitution of the t's became very complicated, so I was wondering, does integral [sin^2(ωt − δ)] and integral [cos^2(ωt − δ)] evaluated on the interval of t2 and t1 have a specific solution?

I know that the integral of [sin^2(ωt − δ)] over any complete period of oscillation T is equal to T/2. However I am not sure how to apply this to my problem.

I'd appreciate any further input.

 

[issacnewton]

use trigonometric identities to solve the integrals

$$\sin^2(\omega t-\delta)=\frac{1}{2}\left[1-\cos\,2(\omega t-\delta)\right]$$

and similar one for the other$$\cos^2(\omega t-\delta)=\frac{1}{2}\left[1+\cos\,2(\omega t-\delta)\right]$$