What is the rate of energy for a damped oscillator?

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SUMMARY

The rate of energy loss for a damped oscillator is defined by the equation dE/dT = -bv², where b represents the damping coefficient. The energy E of the system is expressed as E = 1/2 mv² + 1/2 kx². To derive the rate of energy change, one must differentiate this energy equation and substitute the motion equation -kx - b(dx/dt) = m(d²x/dt²). The correct differentiation leads to the conclusion that dE/dt = -bv², confirming the relationship between damping and energy loss.

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[SOLVED] Energy of a damped oscillator

Homework Statement


I simply need to show that the rate of energy for a damped oscillator is given by:

dE / dT = -bv^2, where b is the dampening coefficient

Homework Equations



I am instructed to differentiate the formula:

E = 1/2 mv^{2} + 1/2 kx^{2} (1)

and use the formula: -kx - b dx/dt = m d^{2}x/dt^{2} (2)

The Attempt at a Solution



I differentitiate

E = 1/2 mv^{2} + 1/2 kx^{2}

to get dE/dT = m d^{2}x/dt^{2} + k dx/dt

the only thing I can see to do here is sub in the above formula (2), to get

dE / dt = -kx - b dx/dt + k dx/dt

or

dT / dt = -kx - bv + kvI must be missing something here, or maybe I made a mistake somewhere, but this question has been bugging me since yesterday. If anyone could steer me in the right direction I would definitely appreciate it.

Thanks a lot
 
Last edited:
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You've got dE/dt completely wrong. The derivative of (1/2)mv^2 is mv*dv/dt. Do you see why? What's the derivative of (1/2)kx^2? Also what you are trying to prove should be dE/dt=-bv^2.
 
Thanks a lot for the quick reply Dick. You're right, I corrected the typo above.

If I use mv*dv/dt as the derivative of (1/2)mv^2, and kx*dx/dt as the deriviative of (1/2)kx^2, the solution is very straight forward. I suppose I need to go back in my textbook and see how you derived that.

Thanks a lot for your help, I can already tell that this forum is going to be a huge resource for me for the next few years.

Cheers!
 

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