# Work done by damping, harmonic oscillator, help?

1. Feb 26, 2012

### Lengalicious

Ok here's the question:
A body m is attached to a spring with spring constant k. While the body executes oscillations it also experiences a damping force F = -βv where 'v' is time derivative of displacement of the body from its equilibrium position.

I believe equation of motion is F = m*d^2x/dt^2 + kx + β*dx/dt.

(i) Let us completely ignore the damping term F = -βv. Assuming that the initial displacement is x0 and the initial velocity is 0, write down the expressions for x(t) and velocity as function of time.

I'm confused here because it says to completely ignore the damping term? In which case wouldn't the mass just function as an undamped oscillator where x(t) = Acos(ωt) and v(t) = -ωAsin(ωt)?

(ii) Now let us assume that the damping coefficient 'β' is non zero but very small, so that over one period of oscillation we may take x(t) and v(t) to be of the same form found in part (i) above. With this assumption, find the work done by the damping force during one time period of oscillation. Express your answer in terms of k, m, x0, and β.

I think the integral of the force with respect to the x variable would give the work done for that force? So i'de get -βvx0 = W? And now i'm lost, I don't think the equations I wrote in (ii) are correct since the variables have nothing to do with this part.

Any help appreciated thanks.

Last edited: Feb 26, 2012
2. Feb 26, 2012

### fluidistic

That looks correct to me.
Looks correct to me.

Please provide the expression of x(t) and v(t) of "part (ii)" above.

Edit: Oops, vela is right.

Last edited: Feb 26, 2012
3. Feb 26, 2012

### vela

Staff Emeritus
Since F is defined to be -βv, you're saying
$$-\beta \frac{dx}{dt} = m\frac{d^2x}{dt^2}+kx+\beta\frac{dx}{dt}$$ I don't think that's what you meant.

4. Feb 26, 2012

### Lengalicious

My bad, the -βdx/dt = Fdamped, not just F.

EDIT: Fluidistic, I'm confused with what you mean? I already did provide those expressions no? 'x(t) = Acos(ωt) and v(t) = -ωAsin(ωt)?'

Last edited: Feb 26, 2012
5. Feb 26, 2012

### vela

Staff Emeritus
So what's F supposed to stand for in your equation then?

For (ii), you indeed want to calculate the integral you mentioned, but how did you end up with that answer?

6. Feb 26, 2012

### Lengalicious

INT} Fdamped dx = INT} -βv dx = -βvx but i'm guessing the limits would be between x0 and 0 or something in order to get x0 into the equation, but then how would I relate this to k and m variables, and also i don't see where the equations from (i) come into play here?

EDIT: I think the F in equation of motion stands for the resultant force no?

7. Feb 26, 2012

### vela

Staff Emeritus
According to Newton's second law, the term $m\frac{d^2x}{dt^2}$ is the net force, right?

The work is indeed given by
$$W = \int_{x_0}^x (-\beta v)\,dx,$$ but you can't treat v as a constant because it varies as x changes. Try using the fact that $v = dx/dt$ to change the variable of integration from x to t.

8. Feb 27, 2012

### Lengalicious

Ok so I would get -βdx^2/dt, but then how do I go about integrating this since there is now both a time element and displacement element? Confused

9. Feb 27, 2012

### vela

Staff Emeritus
You might find it easier to do this way: calculate the power dissipated by the force and integrate that over one cycle.

10. Feb 27, 2012

### Lengalicious

That sounds like a possibility but we haven't been given any power relationships in lectures up till now so I asssume the lecturer wants me to work it out by integrating the force. Also I have to use those relationships in (i) for V(t) and x(t).

11. Feb 27, 2012

### vela

Staff Emeritus
I'm sure you've covered power already. It's usually done when you're introduced to work and energy concepts.

12. Feb 27, 2012

### Lengalicious

Well I know power = workdone / time, but i need the work done to find the power out =/

13. Feb 27, 2012

### Lengalicious

Ok so Fd = -βv

v = -ωAsin(ωt)

so Fd = βωAsin(ωt)

ω = √k/m, and the period is T = 2√m/k

so Fd = β(√k/m)Asin(2)

W = {/INT x^x_0} β(√k/m)Asin(2) dx

am I on the right lines here?

Now i'm stuck though because the amplitude is changing right? So I cant integrate by taking out as constant? Is there a way to get rid of the amplitude because then I would have an equation for work including all variables mentioned.

EDIT: actually the A is A_0 initial amplitude, which due to negligible damping of the system would remain constant no? Therefore I can just take it out? And I imagine you assume the amplitude at its maximum is 1? Which would get rid of it or am I making stuff up?

Last edited: Feb 27, 2012
14. Feb 28, 2012

### vela

Staff Emeritus
Fine up to here.

Why are you setting t=T?

For the undamped harmonic oscillator, you have
\begin{align*}
x(t) &= A\cos \omega t \\
v(t) &= -A\omega \sin \omega t
\end{align*} The amplitude A and the frequency ω are constants. In part (ii), you're assuming these two expressions still hold to a good approximation.

15. Feb 28, 2012

### Lengalicious

Ok sorry, now i'm really just lost, any big hints you could give me?

16. Feb 28, 2012

### vela

Staff Emeritus
v = dx/dt. Solve for dx.