Damped Harmonic Oscillator/Resonance

[CAF123]

Homework Statement

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.

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.

 

[ehild]

Homework Statement

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.

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 $x_p = C \cos (nt) + D \sin(nt)$.

Yes, it is correct.

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.

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

 

[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 $\omega = w$:$$ \sqrt{\frac{g^2[4k^2n^2 + (w^2 - n^2)^2]}{(w^2(w^2 - n^2) - 4k^2n^2 - n^2(w^2 - n^2))^2}}$$

 

[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

 

[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?

 

[ehild]

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?

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

 

[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 $k \neq \frac{w}{\sqrt{2}}?$

 

[ehild]

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?

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).

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 $k \neq \frac{w}{\sqrt{2}}?$

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

ehild

 

[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 $n_{1,2}$.

 

[ehild]

$n^2$ 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 $n_{1,2}$.

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

ehild

 

[CAF123]

Yes, n is positive, neglect the negative radical.

And the conditions for the existence of a maximum is that the term under the radical does not vanish?

n1,2 are the angular frequencies where the squared amplitude is half the maximum amplitude.

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?

 

[ehild]

And the conditions for the existence of a maximum is that the term under the radical does not vanish?

It can be neither zero nor negative.

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 $A^2$.

Divide the square of the maximum amplitude by 2 and equate it with the expression for A².

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?

You have to solve for n. it can not stay on the right hand side.
The width of the resonance is |n1-n2|

ehild

 

[CAF123]

Divide the square of the maximum amplitude by 2 and equate it with the expression for A².

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

The width of the resonance is |n1-n2|

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.

 

[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

 

[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?

 

[ehild]

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?

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

ehild

 

[CAF123]

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

ehild

I have scanned my full workings.
Thanks.

 

[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 $\frac{w(\sqrt{2} - 4k)}{\sqrt{2}}$.

 

[ehild]

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

ehild

 

[CAF123]

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

ehild

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.

 

[ehild]

The parameters of the oscillator are given, you need to find the frequency of the driving force where the amplitude is maximum. And you did it already, and found the resonant frequency n: n²=ω²-2k². Substitute back for n² into the expression of A².

ehild

 

[CAF123]

The parameters of the oscillator are given, you need to find the frequency of the driving force where the amplitude is maximum. And you did it already, and found the resonant frequency n: n²=ω²-2k². Substitute back for n² into the expression of A².

ehild

Thats what I did and I get a quadratic in $n^2$: $$n^4 + n^2(4k^2 - 2w^2 - 8k^2n^2) + w^4 - 8k^4 = 0$$ which gives $$n_{1,2} = \sqrt{\frac{-(-4k^2 - 2w^2) \pm \sqrt{(-4k^2 - 2w^2)^2 - (4(w^4-8k^4))}}{2}}$$

I then took one away from the other and got an expression of the form $$\frac{\sqrt{-a + \sqrt{b}} - \sqrt{-a - \sqrt{b}}}{\sqrt{2}},\,\,\,a = (-4k^2 - 2w^2),\,\,b=8k^2(2w^2 - 4k^2)$$

 

[CAF123]

And if $k^2 <<w^2$, my expression reduces to $\frac{w(\sqrt{2}-4k}{\sqrt{2}}$

 

[ehild]

Thats what I did

You did it wrong, and got an impossible result. The expression of the maximum amplitude is valid for the resonant frequency n_r of the driving force. A_{max}^2=\frac{g^2}{4k^2(k^2+n_r^2)}.
At n=n_r the amplitude is maximum. At n=n₁ or n=n₂, the amplitude is A_max/√2
n_r can not be equal to n_1,2.
Express the maximum amplitude with the parameters of the oscillator, x and k.

ehild

 

[CAF123]

So I equate $1/2(A^2_{max})$ with my expression for $A^2$:
$$\frac{g^2}{8k^2(k^2+n_r^2)} = \frac{g^2}{(w^2 -n^2)^2 + 4k^2n^2)}$$, subbing in $n_r^2 = w^2 - 2k^2$ and then solving for n?

 

[ehild]

Yes, it would be better.

ehild