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Ratio of the periods of a damped and undamped oscillator

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Given: The amplitude of a damped harmonic oscillator drops to 1/e of its initial value after n complete cycles. Show that the ratio of period of the oscillation to the period of the same oscillator with no damping is given by

    [tex]\frac {T_d} {T_0} = \sqrt {1+ \frac {1} {4\pi^2n^2}} = 1+\frac {1} {8\pi^2n^2}[/tex]

    where the approximation in the last expression is valid if n is large.

    2. Relevant equations
    I think

    [tex]\frac {\Delta E} {E} = \frac {T_d} {\tau}[/tex]

    3. The attempt at a solution

    So the energy of the damped oscillator is equal to

    [tex]E=\frac {kA^2} {2}[/tex]

    And the change in energy over n cycles is equal to

    [tex]\Delta E= \frac {kA^2} {2} - \frac {k(\frac {A} {e})^2} {2} = \frac {kA^2} {2} (1-\frac {1} {e^2})[/tex]

    So

    [tex]\frac {\frac {kA^2} {2} (1-\frac {1} {e^2})} {\frac {kA^2} {2}} = \frac {nT_d} {\tau}[/tex]

    This is as far as I get, meaningfully. I'm struggling to find a way to relate the period of the damped and undamped oscillator that makes e drop out. Any thoughts?
     
  2. jcsd
  3. Sep 19, 2016 #2

    Orodruin

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    Why would this hold?

    I suggest that you instead look at the actual solution to the damped oscillator.
     
  4. Sep 19, 2016 #3

    Sorry, I guess I did give that route a shot.

    [tex]x = Ae^{-\gamma t}[/tex]

    [tex]\frac {A} {e} = Ae^{-\gamma t}[/tex] so the exponent must equal 1 after n cycles. From there on I'm stuck. I'm not sure how to put n into the exponent to make it give the desired behavior. I must be missing an equation or two..
     
  5. Sep 19, 2016 #4
    I guess I'm just confused on how to pull the period of oscillation from that equation.
     
  6. Sep 19, 2016 #5

    Orodruin

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    This is not the solution, it is just the amplitude. You need the full solution to the damped oscillator differential equation.

    It is correct that ##\gamma t## must be equal to one after ##n## periods, but to know the period you must know the oscillatory part of the solution. You cannot do this looking at the amplitude only.
     
  7. Sep 19, 2016 #6

    ehild

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  8. Sep 19, 2016 #7
    I appreciate the help. I've got it now I think. Here's what I found.

    The equation is

    [tex]\frac {A} {e}=Ae^{-\gamma T_d}cos(\omega T_d + \phi)[/tex]

    In order for this equation to equal A/e,

    [tex]
    \gamma T_d=1[/tex]

    [tex]\omega_d T_d=2\pi n[/tex]

    [tex]
    T_d= \frac {2 \pi n} {\omega_d}= \frac {1} {\gamma}[/tex]

    Which makes the damped frequency equal to

    [tex]\omega_d = 2\pi n \gamma[/tex]

    We can then relate the frequency of the damped oscillator to the frequency of the undamped through the equation

    [tex]
    \omega_d = \sqrt{\omega_0^2 - \gamma^2}[/tex]

    [tex]2\pi n \gamma = \sqrt{\omega_0^2 - \gamma^2}[/tex]

    [tex]\omega_0 = \gamma \sqrt{1+4pi^2n^2}[/tex]

    Then the ratio is:

    [tex]\frac {T_d} {T_0} = \frac {\frac {2\pi} {\omega_d}} {\frac {2\pi} {\omega_0}}=\frac {w_0} {w_d}= \frac {\sqrt{1+4\pi^2n^2}} {2\pi n}=\sqrt{1+ \frac {1} {4\pi^2n^2}}[/tex]
     
    Last edited: Sep 19, 2016
  9. Sep 19, 2016 #8

    Orodruin

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    Right. I assume you also understand the working behind the approximation
    $$
    \sqrt{1+\frac{1}{4\pi^2 n^2}} \simeq 1 + \frac{1}{8\pi^2 n^2},
    $$
    which was part of the question (and is an approximation, not an equality, valid for large ##n##).
     
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