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Damped Harmonic Oscillator/Resonance

  1. Mar 5, 2013 #1

    CAF123

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    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. jcsd
  3. Mar 5, 2013 #2

    ehild

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    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
     
  4. Mar 5, 2013 #3

    CAF123

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    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
  5. Mar 5, 2013 #4

    ehild

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    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
     
  6. Mar 5, 2013 #5

    CAF123

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    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?
     
  7. Mar 5, 2013 #6

    ehild

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    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
     
  8. Mar 5, 2013 #7

    CAF123

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    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}}?##
     
  9. Mar 5, 2013 #8

    ehild

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    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
     
  10. Mar 6, 2013 #9

    CAF123

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    ##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}##.
     
  11. Mar 6, 2013 #10

    ehild

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

    ehild
     
  12. Mar 6, 2013 #11

    CAF123

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    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?
     
  13. Mar 6, 2013 #12

    ehild

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    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
     
  14. Mar 6, 2013 #13

    CAF123

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    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
  15. Mar 6, 2013 #14

    ehild

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    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
     
  16. Mar 6, 2013 #15

    CAF123

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    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
  17. Mar 6, 2013 #16

    ehild

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

    ehild
     
  18. Mar 6, 2013 #17

    CAF123

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    I have scanned my full workings.
    Thanks.
     

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  19. Mar 6, 2013 #18

    CAF123

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    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
  20. Mar 6, 2013 #19

    ehild

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

    ehild
     
  21. Mar 6, 2013 #20

    CAF123

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    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.
     
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