Damped Oscillator equation - Energy

the damped oscillator equation:

(m)y''(t) + (v)y'(t) +(k)y(t)=0

Show that the energy of the system given by

E=(1/2)mx'² + (1/2)kx²

satisfies:

dE/dt = -mvx'

i have gone through this several time simply differentiating the expression for E wrt and i end up with

dE/dt = x'(-vx')

im at a brick wall. Am i doing something wrong? Any help is much appreciated! Thanks
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 Quote by Paddyod1509 the damped oscillator equation: (m)y''(t) + (v)y'(t) +(k)y(t)=0 Show that the energy of the system given by E=(1/2)mx'² + (1/2)kx² satisfies: dE/dt = -mvx' i have gone through this several time simply differentiating the expression for E wrt and i end up with dE/dt = x'(-vx') im at a brick wall. Am i doing something wrong? Any help is much appreciated! Thanks
There is some sort of problem with the equation you have been asked to prove. A damped oscillator is always losing energy. Your solution shows that is true. The given solution would say the oscillator is sometimes gaining energy if the sign of x' is correct. I don't think that's correct.
 Recognitions: Gold Member You can't just differentiate E the way you did and prove the theorem. You need to incorporate the basic diff eq representing a damped spring-mass system. The expression for E represents ANY spring-mass system, damped or not, linear or not, etc. The only thing I can think of is to solve the diff eq (it's a simple 2nd order one with constant coeff). Apply an initial condition to x = x0. Derive the solution x(t) and then substitute in E, take dE/dt and there you are. (BTW why is y used in the diff eq and x in E?) Dick's comment is well taken! Not only his, but I noticed the dimensions don't make sense. dE/dt has dimensions of FLT-1 whereas -mvx' has dimensions of MF where M = mass F = force = MLT-2 L = length T = time. Thus -mvx' has the wrong dimensions to be dE/dt.

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Damped Oscillator equation - Energy

 Quote by rude man You can't just differentiate E the way you did and prove the theorem (BTW it's probably correct). You need to incorporate the basic diff eq representing a damped spring-mass system. The expression for E represents ANY spring-mass system, damped or not, linear or not, etc. The only thing I can think of is to solve the diff eq (it's a simple 2nd order one with constant coeff). Apply an initial condition to x = x0. Derive the solution x(t) and then substitute in E, take dE/dt and there you are. (BTW why is y used in the diff eq and x in E?)
Paddyod1509 did substitute the differential equation into dE/dt to get his answer. The given answer can't be right. -vx'^2 has the correct units of J/s. -mvx' doesn't.

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 Quote by Dick Paddyod1509 did substitute the differential equation into dE/dt to get his answer. The given answer can't be right. -vx'^2 has the correct units of J/s. -mvx' doesn't.
OK, I was misled by his wording.

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 Quote by rude man OK, I was misled by his wording.
Well, he didn't actually say that that's what he did. But once you form dE/dt it's the obvious way to get to -vx'^2.

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 Quote by Dick Well, he didn't actually say that that's what he did. But once you form dE/dt it's the obvious way to get to -vx'^2.
So did he solve the d.e. for x(t) and then substitute in E, or what?

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