What is the rate of energy for a damped oscillator?

AI Thread Summary
The discussion focuses on deriving the rate of energy loss for a damped oscillator, expressed as dE/dt = -bv^2, where b is the damping coefficient. The user initially struggles with differentiating the energy equation E = 1/2 mv^2 + 1/2 kx^2 and incorrectly calculates dE/dt. A respondent clarifies the correct differentiation process, emphasizing the need to use mv*dv/dt for the kinetic energy term and kx*dx/dt for the potential energy term. The user acknowledges the correction and expresses gratitude for the guidance received. This exchange highlights the importance of accurate differentiation in solving physics problems related to damped oscillators.
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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 alot
 
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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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