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Average potential and kinetic energy of SHO

by mewmew
Tags: average, energy, kinetic, potential
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Oct17-05, 11:21 PM
P: 115
I am using Frenches book on waves and have a question. You have a standard driven damped oscillator and I am suppose to find the average potential and kinetic energy of it. They are both similar so I will just use the potential for example. I took [tex] \frac{1}{T}\int^T_0 \frac{1}{2}KX^2 dt[/tex] Where [tex]X=Cos(\omega t-\rho)[/tex] I solved the resulting integral with half angle formulas but found my answer to be a bit messy, is that the correct way of getting the average kinetic energy or am I missing something here? Thanks
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Oct18-05, 03:47 AM
P: 130
There's a trick to doing integrals of cos²(t). First, you should know that:

sin²(t)+cos²(t) = 1 for all t.


[tex]\int _a ^b (\cos ^2 (t) + \sin ^2 (t))dt = b-a = \int _a ^b \cos ^2 (t)dt + \int _a ^b \sin ^2 (t)dt[/tex]

Also, you should know that

[tex]\int \limits_T \cos ^2 (t)dt = \int \limits_T \sin ^2 (t)dt[/tex]

where T is one period. Therefore, since your average potential energy will be taken over one period (T = b-a), you can conclude that:

[tex]\int _a ^b (\cos ^2 (t) + \sin ^2 (t))dt = \int \limits_T (\cos ^2 (t) + \sin ^2 (t))dt = \int \limits_T (\cos ^2 (t) + \cos ^2 (t))dt = 2\int \limits_T \cos ^2 (t)dt=T[/tex]
[tex]\int _a ^b \cos ^2 (t)dt = (1/2)T[/tex]

Now about your question concerning average kinetic energy. Your system is driven, so you have to account for the added energy from the driver and also the loss of energy from the damping. What I think you should do is take your differential equation of motion and find the function of velocity wrt time and find the average kinetic energy using a similar method.

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