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Damped Harmonic Oscillator & Mechanical Energy

  1. Mar 27, 2005 #1
    Question: A damped harmonic oscillator loses 5.0 percent of its mechanical energy per cycle. (a) By what percentage does its frequency differ from the natural frequency [itex]\omega_0 = \sqrt{k/m}[/itex]? (b) After how may periods will the amplitude have decreased to 1/e of its original value?

    So, for (a), the answer is [itex]\omega ' / \omega_0[/itex] where

    [tex]\omega ' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}[/tex]

    But that leaves me with 3 unknowns, k, m, and b requiring three equations to solve. The only equations I can think of is E = K + U (mechanical energy) and E = 0.95TE0 where T is the number of cycles and E0 is the initial mechanical energy.

    What other equation can I use? Or is there a simpler method of finding the solution?
     
  2. jcsd
  3. Mar 27, 2005 #2

    Andrew Mason

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    The amplitude is given by:

    [tex]A = A_0e^{-\gamma t}[/tex] where [itex]\gamma = b/2m[/itex] 1

    Work out the value for [itex]\gamma[/itex] given that A^2 decreases to .95A_0^2 in the first [itex]T = 2\pi /\omega '[/itex] seconds.

    Then find [itex]\omega '[/itex] in terms of [itex]\omega_0[/itex] using:
    [itex]\omega '^2 = \omega_0^2 - \gamma^2[/itex]

    AM

    [Note: 1. The solution to the damped harmonic oscillator is:

    [tex]x = A_0e^{-\gamma t}sin(\omega 't + \phi)[/tex]

    where [itex]\omega ' = \sqrt{\omega^2 - \gamma^2}[/itex] ]
     
  4. Mar 27, 2005 #3
    How do you know the square of the amplitude decreases to .95A_0^2 in the first T seconds?
     
  5. Mar 27, 2005 #4

    Andrew Mason

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    Energy is proportional to the square of the amplitude (all energy is potential energy at maximum amplitude: [itex]E = \frac{1}{2}kx^2[/itex]). If the system loses 5% of its energy in one cycle, the square of the amplitude will decrease to 95% of the square of the original amplitude.

    AM
     
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