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Damped harmonic oscillator and displacement

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    "Show that the ratio of two successive maxima in the displacement of a damped harmonic oscillator is constant."


    2. Relevant equations
    x = a e^(-[tex]\upsilon[/tex]t/2) cos ([tex]\omega[/tex]t - [tex]\vartheta[/tex])


    3. The attempt at a solution

    So I want to find when this beast has its maximum values, so I take the derivative and set it = 0

    x' = -[tex]\upsilon[/tex]*a/2 * e^(-[tex]\upsilon[/tex]t/2) * cos ([tex]\omega[/tex]t - [tex]\vartheta[/tex]) - [tex]\omega[/tex] e^(-[tex]\upsilon[/tex]t/2) * a * sin([tex]\omega[/tex]t - [tex]\vartheta[/tex])

    So I set this to 0

    and I get

    [tex]\omega[/tex] * a * sin([tex]\omega[/tex]t - [tex]\vartheta[/tex]) = [tex]\upsilon[/tex]*a/2 * cos ([tex]\omega[/tex]t - [tex]\vartheta[/tex])

    tan ([tex]\omega[/tex]t - [tex]\vartheta[/tex]) = -[tex]\upsilon[/tex]/2[tex]\omega[/tex]

    The teacher mentioned to me that from this I'm suppose to realize there are 2 solutions and from that the rest is easy..

    but the rest isn't easy!

    I don't understand, where and what are my 2 solutions, and from those 2 solutions how do I use them to plug into my original x equation to show that these successive maxima are just = to a constant? And how do I know these solutions are maximums and not minimums?
     
    Last edited: Sep 8, 2009
  2. jcsd
  3. Sep 8, 2009 #2

    kuruman

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    Just look at your expression for x(t). At what value of the cosine does it reach a maximum? What is that maximum? How long must you wait until the cosine reaches its maximum value once more? What is the value of this second maximum? This should get you started.
     
  4. Sep 8, 2009 #3
    The value of cosine reaches a maximum when cos = 1 and that occurs when t = 0 and phi = 0. One then has to wait 2(pi)/w for it to reach that maximum value again. The value of this second maximum will be 1 * the new amplitude.

    Ah.. so then

    x0 = A0 when t = 0 and phi = 0

    and

    x1 = A1 e^(2*pi * -v / w)

    and x0/x1 = A0/(A1 * e^((2*pi*-v)/w)

    Which is a constant.
     
    Last edited: Sep 8, 2009
  5. Sep 8, 2009 #4

    kuruman

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    Good job, but isn't A0 = A1 = a, the amplitude?
     
  6. Sep 8, 2009 #5
    Wait but I thought in a damped harmonic oscillator the Amplitude decreases with time?
     
  7. Sep 9, 2009 #6

    kuruman

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    It does, but a is constant. Say the phase theta is zero, to make things simple. Then at time t = 0, the amplitude is

    A(0) = a*exp(0) = a

    after one period T has elapsed, the amplitude is

    A(T) = a*exp(-v T/2)

    after two periods have elapsed, the amplitude is

    A(2T) = a*exp(-v(2T)/2)

    Do you see how it works? The problem is asking "by what number do you have to multiply the old amplitude in order to get the new amplitude?"
     
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