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Simple Harmonic Motion - Acceleration and Time Period

  1. Aug 24, 2011 #1
    Hi,

    I am a bit confused regarding the relation between the acceleration and time period in simple harmonic motion.

    This questions asks what would be one change to the motion of a car's wing mirror in case the glass was switched for a heavier one. I don't understand for example, how would the period change. I can think that the same force will have to move a greater mass over the same distance but, wouldn't the amplitude decrease if the mass increases?

    Thanks.
     
  2. jcsd
  3. Aug 24, 2011 #2

    lightgrav

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    the oscillation is going to occur around the _equilibrium_ location
    (more massive glass will have a lower equilibrium);
    the oscillation Energy depends on Amplitude (squared, times ½k)
    but the frequency depends on the spring "stiffness" compared to the inertia
    (inside a square root). If you have the same Force, but a bigger mass, what can you tell about the acceleration at these turn-around points?
    (that is where it spends most of its time, since its speed is slow there).
     
  4. Aug 24, 2011 #3
    Sorry, I couldn't follow very well the beginning. But, for the same force with a bigger mass the acceleration will decrease.

    So, from the equation, either omega must decrease (frequency decreases/time period increases) or the displacement will decrease...

    How do I know which of these will happen?
     
  5. Aug 24, 2011 #4

    lightgrav

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    IF the oscillation has the same (total) Energy, it has the same maximum PE (spring),
    so it will have the same amplitude (maximum displacement from equilibrium).
    . . . . if the mass is doubled . . .
    acceleration will be halved, in x = ½at^2 , (here, t is roughly time for ¼ oscillation)
    but it has to get just as far, so t^2 is doubled . . . so t is increased by sqrt(2)
    that's why the omega formula is squareroot(k/m).
     
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