# Damped Harmonic Oscillator/Resonance

1. Mar 5, 2013

### CAF123

1. The problem statement, all variables and given/known data
A damped oscillator is subjected to a simple harmonic force, satisfying $$\ddot{x}(t) + 2k\dot{x}(t) + \omega^2x(t) = g \cos (nt),$$where $g, k, \omega, n +ve.$

1) Show that for $t >>1/k$ the position x(t) has the form $A \cos (nt - \phi)$, and find A and $\phi$ in terms of k,n,\omega, g.

2) Find the resonance frequency. Under what conditions is there a maximum?

3) For $k^2 << \omega^2$ find the width of the resonance, defined by the distance between the two frequencies $n_{1,2}$ where the squared amplitude goes down to half its orginal value.

3. The attempt at a solution

1)No condition on k and ω is given so I think I have three separate homogeneous solns, each for the conditions $k^2 = \omega^2, k^2 < \omega^2$ and $k^2 > \omega^2$. In order to find the particular solution, I made the trial function $x_p = C \cos (nt) + D \sin (nt)$, differentaited twice and subbed into the eqn given. This give me very ugly and messy expressions for C and D. Am I correct in saying that in the case t >>1/k, then the expressions for the homogeneous solutions (in all three cases outlined above) simply vanish? If so, this would just leave the particular soln $x_p = C \cos (nt) + D \sin(nt)$. The expressions I got for C and D are too messy to put up here , so I might scan my work instead.

In order to find A and phi in terms of those quantities, should I just sub $A \cos (nt - \phi)$ into the eqn and get A and $\phi$ that way? This would make 2) easier and I am yet to attempt 3).
Many thanks.

2. Mar 5, 2013

### ehild

Yes, it is correct.

The sum of a sine and a cosine of the same frequency is equivalent with a single cosine or sine function, that includes a phase constant. The problem tells you to search the particular solution in the form Acos(nt-Φ)

ehild

3. Mar 5, 2013

### CAF123

I managed to obtain $\phi$ quite easily, however my expression for A just looks really complicated:$$C cos(nt) + D sin(nt) = A cos(nt-\phi) \Rightarrow A cos \phi = C, A \sin \phi = D$$. This gives $$A^2 = D^2 + C^2 = \frac{C^2[4k^2n^2 + (\omega^2 - n^2)^2]}{(\omega^2 - n^2)^2},$$where $$C = \frac{g(\omega^2 - n^2}{\omega^2(\omega^2 - n^2) - 4k^2 n^2 - n^2(\omega^2 - n^2)}.$$

Sqauaring that and substituting into the eqn for $A^2$ is going to give a rather messy expression. Is it the way to proceed?

(I was able to obtain $\phi$ easy because I noticed that there was a single C in my expression for D and given that $tan \phi = D/C,$it just cancelled; no luck here though)

Edit: Retrospectively, A did not turn out so bad: (well, at least this is what I have): Let $\omega = w$:$$\sqrt{\frac{g^2[4k^2n^2 + (w^2 - n^2)^2]}{(w^2(w^2 - n^2) - 4k^2n^2 - n^2(w^2 - n^2))^2}}$$

Last edited: Mar 5, 2013
4. Mar 5, 2013

### ehild

It is easier if you have the particular solution in the form of Acos(nt-Φ). Substitute it and its derivatives into the differential equation.
Denote nt-Φ=θ. With that notation, the right hand side is gcos(θ+Φ). Expand and compare the cos(θ) and sin(θ) terms on both sides. You get two equations. Square and add them, so as Φ cancel. It will be easy to isolate A2.

ehild

5. Mar 5, 2013

### CAF123

Do you mean simply substitute $y = A\cos(nt-\phi)$ into the diff eqn?

Have you done this calculation? Did your A resemble that in my previous post?

6. Mar 5, 2013

### ehild

Yes, just substitute that form.
Yes, I did the calculations, and your A is a bit similar to the real one, but not quite that.

ehild

7. Mar 5, 2013

### CAF123

This gives $$A = \frac{g}{\sqrt{(w^2 - n^2)^2 + 4k^2 n^2}}$$ Incidentally, what you suggested is what I tried first, but I thought the question wanted me to actually prove such a form existed. Basically, all I need to say is that since t >>1/k, the homogeneous solns vanish and the remaining homogeneous soln can be expressed as a single trig function. Is that all I need to say?

To find the resonance frequency, I found the derivative of the above expression for A. I got $n_{1,2} = \pm \sqrt{w^2 - 2k^2}$ and I presume when they say under what conditions does this maximum amplitude occur that $w^2 \neq 2k^2$ I.e $k \neq \frac{w}{\sqrt{2}}?$

8. Mar 5, 2013

### ehild

I think what you need to show is that the homogeneous part contains an exponentially decreasing factor. That the particular solution can be a single sine or cosine function instead of the sum of a sine and a cosine is not needed to prove. First, you can try any function as a particular solution if it obeys the equation. Second: It is known from trigonometry and easy to prove that the function f(θ)=asin(θ) + bcos(θ) is equivalent to a single cosine or sine Acos(θ-Φ) if A2 =a2+b2 and tanΦ=b/a (taking the signs of a and b into account).

Not only a single frequency is excluded. Can be n2 negative?

ehild

9. Mar 6, 2013

### CAF123

$n^2$ cannot be negative. Since the frequency is a physical quantity should I neglect the - radical?

Part 3) of the question seems to imply there exists two such maximum amplitudes since they use the notation $n_{1,2}$.

10. Mar 6, 2013

### ehild

Yes, n is positive, neglect the negative radical.
n1,2 are the angular frequencies where the squared amplitude is half the maximum amplitude.

ehild

11. Mar 6, 2013

### CAF123

And the conditions for the existence of a maximum is that the term under the radical does not vanish?
For 3), the value for the resonance freq I obtained, I then put into the expression for A . This gave the maxiumum amplitude. I then divided this by 2 and equated it to the expression for $A^2$. Solving for n, I obtain a quadratic in $n^2$, which gives the following expression for n:
$$n_{1,2} = \sqrt{\frac{-(g4k^2 - g2n^2w^2) \pm \sqrt{(g4k^2 - g2n^2w^2) - (4g(gw^4-4k g^2 \sqrt{w^2 - k^2}}}{2g}}$$

Now to get the width of this resonace, presumably I put these two expressions into the expression for A (!) and subtract them?

12. Mar 6, 2013

### ehild

It can be neither zero nor negative.
Divide the square of the maximum amplitude by 2 and equate it with the expression for A2.
You have to solve for n. it can not stay on the right hand side.
The width of the resonance is |n1-n2|

ehild

13. Mar 6, 2013

### CAF123

I interpreted the words to mean equate ($A^2$) to $\frac{1}{2}A_{max}$. Is this a wrong interpretation?

Why would you not sub the values for n1 and n2 into the expression for A and then subtract those?
Edit: Is this simply because the width of the resonance is defined as the distance between the two frequencies?
Thanks.

Last edited: Mar 6, 2013
14. Mar 6, 2013

### ehild

The width of the resonance is the distance between the frequencies where the amplitude is 1/sqrt(2) of the maximal.
You got n1 and n2 from the condition that A is the same for both of them, so subtracting the A-s you get zero.

ehild

15. Mar 6, 2013

### CAF123

I redid the calculation and got the following:
Let $(4k^2 - 2w^2 - 8k^2g) = \alpha$, $(4k^2 - 2w^2 - 8k^2g)^2-(4(w^4-8k^4g))) = \beta$
So, $$n_2 - n_1 = \frac{\sqrt{-\alpha + \sqrt{\beta}} - \sqrt{-\alpha - \sqrt{\beta}}}{\sqrt{2}},$$ Is it close to what you got?

Edit: The problem I see is that the second term is imaginary.. (unless of course $\alpha$ is negative - which looks correct provided $-2w^2 - 8k^2g > 4k^2$ in addition to $-\alpha > \sqrt{\beta}$)
Edit2: I noticed that the question states $k^2 << w^2$. If I take that, then $-\alpha$ tends to $2w^2$ and $(4k^2 - 2w^2 - 8k^2g)^2 - (4(w^4 - 8k^4g))$ tends to 0 so overall I end up with 0? Is this supposed to happen?

Last edited: Mar 6, 2013
16. Mar 6, 2013

### ehild

No. g should cancel. Show your working in detail.

ehild

17. Mar 6, 2013

### CAF123

I have scanned my full workings.
Thanks.

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18. Mar 6, 2013

### CAF123

I see my error.

With the same structure of solution as above, but instead let $\alpha = -4k^2 - 2w^2$ and $\beta = (-4k^2 - 2w^2)^2 - (4(w^4 - 8k^4)$

Multiplying $\beta$ out , I think this tends to $16k^2w^2$, while $\alpha$ tends to $-2w^2$. So putting these conditons in, I get that the width tends to $\frac{w(\sqrt{2} - 4k)}{\sqrt{2}}$.

Last edited: Mar 6, 2013
19. Mar 6, 2013

### ehild

I can not read your handwriting. What did you get for the maximum A^2?

ehild

20. Mar 6, 2013

### CAF123

Sorry for the faint scan. I got $$\frac{g^2}{4k^2(k^2 + n^2)}$$ for A^2 max and so divide this by 2 gives $$\frac{g^2}{8k^2(k^2+n^2)}.$$Then equate this to A^2.