Damped Oscillator equation - Energy


by Paddyod1509
Tags: damped, energy, equation, oscillator
Paddyod1509
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#1
Dec8-12, 09:06 PM
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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' + (1/2)kx

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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Dick
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Dec8-12, 10:12 PM
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Quote Quote by Paddyod1509 View Post
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' + (1/2)kx

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.
rude man
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Dec9-12, 11:39 AM
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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.

Dick
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Dec9-12, 11:58 AM
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Damped Oscillator equation - Energy


Quote Quote by rude man View Post
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.
rude man
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Dec9-12, 02:09 PM
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Quote Quote by Dick View Post
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.
Dick
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Dec9-12, 02:28 PM
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Quote Quote by rude man View Post
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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Dec9-12, 03:37 PM
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Quote Quote by Dick View Post
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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Dec9-12, 03:39 PM
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Quote Quote by rude man View Post
So did he solve the d.e. for x(t) and then substitute in E, or what?
No. He found dE/dt=mx'x''+kxx'=x'(mx''+kx). The DE then tells you mx''+kx=(-vx'). You don't need to actually solve the DE.
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Dec9-12, 04:32 PM
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Quote Quote by Dick View Post
No. He found dE/dt=mx'x''+kxx'=x'(mx''+kx). The DE then tells you mx''+kx=(-vx'). You don't need to actually solve the DE.
Thanks. Good job.


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