Damped Oscillator equation - Energy

[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

 

[Dick]

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.

 

[rude man]

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]

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]

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]

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.

 

[rude man]

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?

 

[Dick]

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.

 

[rude man]

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.