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Energy loss in SHM

  1. Mar 5, 2006 #1
    When middle C on a piano (frequency = 262Hz) is struck, the vibration of the piano string loses half its energy after 4s.

    (i) What is the decay time for the energy?
    (ii) What is the Q-factor for this piano wire?
    (iii) What is the fractional energy loss per cycle?

    SHM has been going great until this chapter on energy loss. I'm totally lossed. I'd really appreciate if someone could explain how to attempt these questions.

    Best Regards, Jonathan.
  2. jcsd
  3. Mar 5, 2006 #2


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    Decay time is usally defined as the time taken for a value to fall to [itex]\frac{1}{e}[/itex] times the original value.
    Last edited: Mar 5, 2006
  4. Mar 5, 2006 #3
    But how do I use the information given to calculate it?

    Best Regards, Jonathan.
  5. Mar 5, 2006 #4
    i'm not sure of this ... but
    i think we can do this problem like radioactivity problems....
    for the first part
    given that half-life = 4s. Now calculate deacy constant which is ln2/(half-life period)......

    The decay constant is defined as inverse of time taken to decay to 1/e times the original.... so the your answer should be (half-life)/ln2 ....

    I'm not sure ... i think you better wait for some more replies :rolleyes:
  6. Mar 5, 2006 #5


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    physicsprasanna is right, you use the same process as nuclear physics, but instead you use energy instead of number of radioactive isotopes.
    [tex]T_{\frac{1}{2}} = \frac{\ln 2}{k}[/tex]
    Then you can work out [itex]k[/itex] which allows you to calculate [itex]E_0[/itex] and form an equation. Well thats how I understand in anyway.
  7. Mar 5, 2006 #6
    The energy in a damped oscillator goes as:
    [tex]E(t)=E_0e^{-t/ \tau}[/tex]
    Where tau is the "decay constant." This is the time it takes for the energy to be reduced by a factor of 1/e, as stated above. So we know that
    [tex]E(4)=(1/2)E_0=E_0e^{-4/ \tau}[/tex].
    From this we can find a value for tau.

    Next, the "quality factor," or "Q factor," is given by:
    [tex]Q=\omega_0 \tau[/tex]
    Where [tex]\omega_0[/tex] is the initial angular frequency (NOT frequency!) of the motion.

    The fractional energy loss per cycle is defined as:
    [tex] \left ( \frac{\Delta E}{E_0} \right )_{cycle} = \frac{E(T)-E_0}{E_0}[/tex]
    It turns out this is inversely proportional to Q, but as you can calculate it without the Q value, I leave finding that relation to you.

  8. Mar 5, 2006 #7
    Many thanks.
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