Damped Oscillator equation - Energy

In summary, the damped oscillator equation is given by (m)y''(t) + (v)y'(t) +(k)y(t)=0. The energy of the system is represented by E=(1/2)mx'² + (1/2)kx² and it satisfies the equation dE/dt = -mvx'. However, there seems to be a problem with the given solution as a damped oscillator always loses energy, while the given solution suggests that it can sometimes gain energy. To prove this theorem, it is necessary to incorporate the basic differential equation for a damped spring-mass system and solve for x(t), and then substitute it into E and take dE/dt to get the
  • #1
Paddyod1509
10
0
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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  • #2
Paddyod1509 said:
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.
 
  • #3
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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  • #4
rude man said:
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.
 
  • #5
Dick said:
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.
 
  • #6
rude man said:
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.
 
  • #7
Dick said:
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?
 
  • #8
rude man said:
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.
 
  • #9
Dick said:
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.
 

1. What is a damped oscillator?

A damped oscillator is a system that experiences a decrease in amplitude over time due to the presence of external forces, such as friction or air resistance.

2. What is the equation for a damped oscillator?

The equation for a damped oscillator is given by: x = x0e-btcos(ωt + φ), where x is the position of the oscillator, x0 is the initial position, b is the damping coefficient, t is time, ω is the angular frequency, and φ is the phase angle.

3. How does energy change in a damped oscillator?

In a damped oscillator, energy is continually lost due to the presence of external forces. As the amplitude decreases, so does the kinetic and potential energy of the system.

4. What is the role of damping coefficient in a damped oscillator?

The damping coefficient, b, determines the rate at which the amplitude of the oscillator decreases. A higher damping coefficient means a faster decrease in amplitude.

5. How does the frequency of a damped oscillator change with time?

In a damped oscillator, the frequency decreases with time as the amplitude decreases. This is because the damping force affects the restoring force, causing the oscillator to slow down.

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