# Homework Help: Mean Input Power & Q value , Damped Harmonic Motion

1. Feb 16, 2016

### RJLiberator

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I'm working on part a.

The numerical value of Q.

I have an equation stating that Q = ω_0/ϒ.

I dont really know what ϒ is, in other places (http://farside.ph.utexas.edu/teaching/315/Waves/node13.html) it seems like the frequency.

But, I also do not have w_0. So perhaps this is the wrong equation or perhaps I am looking at it wrong.

I also have: P_max = (1/2) * (Q*F_0^2)/(m*w_0)
but I do not think this helps directly.

2. Feb 16, 2016

### BvU

Correct. What would help is an expression for P as a function of $\omega$. Because from the picture you can see that P halves when you are $\omega/50$ away from $\omega_0$ ....

Time to fill in the blank under "2. Relevant equations"

3. Feb 16, 2016

### RJLiberator

Sometimes with foreign topics it is hard to know what is a relevant equation and what is irrelevant :p.

P(w) = F_0^2*w_0/(2*K*Q) * [1/((w_0/w-w/w_0)^2 + 1/Q^2)]

At w = w_0, we get it to be maximized.

P(w) = F_0^2*w_0*Q/2K

i'm not entirely sure how this helps, tho.
As we do not know F_0, other then that it is held constant.
w_0 is unknown, but you are saying that P halves when we are w/50 away from w_0. Why 50?

If we take (1/2)*P(w) = F_0^2*w_0*Q/4K

That doesn't seem to help.

I am guessing that I am not making a connecting between F_0, K, and w_0?

4. Feb 16, 2016

### BvU

From the picture we see $P(\omega_0 + \omega_0/50) = {1\over 2} P(\omega_0)$. A lot of the factors that bother you divide out when you work out this equation....and who knows, you end up with something containing $Q$ ...

5. Feb 16, 2016

### Staff: Mentor

It would be instructive to look up the definition of the Q factor. There is a definition that involves bandwidth (-3 dB, or half-power points) that will be of particular interest.

6. Feb 16, 2016

### RJLiberator

Definition of Q factor : https://en.wikipedia.org/wiki/Q_factor

So, reformatting my previous equation we see:

2*K*P(w_0)/w_0*F_0^2 = Q

How would the definition of Q help me here? I would think that we are looking for Q so we do not want to replace it with anything.

Maybe this is it: P(max) = Q*F_0^2/(2m*w_0)
If I input this in for P(w_0) then we see

k*Q/(m*w_0^2) = Q
But then I cancel out Q and i'm stuck.

k/(m*w_0^2) = 1

7. Feb 16, 2016

### Staff: Mentor

Sure you would! Q has a definition in terms of the bandwidth. The bandwidth is defined in terms of the half-power points. You are given the half power points on the diagram in terms of the resonant frequency $\omega_o$. So, what's the bandwidth on your diagram?

8. Feb 16, 2016

### RJLiberator

0.04w_0 = bandwidth = Q ?

9. Feb 16, 2016

### Staff: Mentor

Go back to the Wikipedia page and look at the definition of Q in terms of the bandwidth!

10. Feb 16, 2016

### RJLiberator

Aha.,

f_0/The change in frequency = Q

So, w_0/0.04*w_0 = 1/0.04 = 25 is the value for Q!

11. Feb 16, 2016

### Staff: Mentor

Huzzah! Yes!

12. Feb 16, 2016

### RJLiberator

Solved ! Thank you kindly.