A Damped Oscillator and Negative Damping Force

[Nanofan01]

A damped oscillator is described by the equation m(x'') + b(x') + kx = 0, where the damping force is given by F = -b(x'). Show that the rate of change of the total energy of the oscillator is equal to the (negative) rate at which the damping force dissipates energy.

 

[dacruick]

Hi there,

to me, damping intrinsically means that the force opposes motion. So if you have a negative damping force, it's either confusing, or redundant.

This also seems like a homework question, in which case it's not posted in the right forum.

It seems to me like you could use common sense for this question then match the math up to it. If there is no damping, that means that the oscillation will be periodically identical (like a sine wave). No energy will be lost. Once you add that damping, that is the only cause of dissipating energy.

If you calculate the total change in energy of that equation of motion, then do the same without the damping force, you should find that potential + kinetic will be constant. Therefore you can conclude that the damping term is solely responsible for the loss of energy.

 

[AlephZero]

to me, damping intrinsically means that the force opposes motion. So if you have a negative damping force, it's either confusing, or redundant.

The force does oppose the motion (assuming b > 0).

m(x'') + b(x') + kx = 0
m(x'') + kx = - b(x') = F

 

[Nanofan01]

It makes sense that the rate of change of the oscillator is equal to the negative damping force but how do you mathematically derive [-b(x')] from [m(x'') + b(x') + kx]?

 

[Nanofan01]

Also, which would be the right forum?

 

[dacruick]

It makes sense that the rate of change of the oscillator is equal to the negative damping force but how do you mathematically derive [-b(x')] from [m(x'') + b(x') + kx]?

the kinetic energy is described by the m(x'') term, and the potential energy is described by the kx term. That leaves one term out right? Which just so happens to be your answer.

And there is a specific forum for homework. I don't think you'll have too much trouble finding it if you look.