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Simple Harmonic Motion

  1. Nov 9, 2011 #1
    There is a horizontal spring-mass system lie on top of a frictionless table.
    let m=mass
    k=spring constant
    x=displacement from equilibrium position
    A=amplitude
    T=period

    By using conventional way, time taken to complete a quarter of cycle=T/4=(∏/2)√(m/k)

    consider another way of doing, F=-kx
    a=-(k/m)x
    then we can consider the average acceleration of first quarter of motion a=(1/2)(k/m)A
    since s=ut+1/2at^2
    then, A=(1/2)*(1/2)(k/m)A*t^2
    finally, we get t=2√(m/k)

    why is the ans from both way diffrent...Both ways seem to be equivalent...
     
  2. jcsd
  3. Nov 9, 2011 #2

    SammyS

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    Are you simply assuming that's the average acceleration? ... or can you show it?

    The acceleration due to the spring, is not uniform.
     
  4. Nov 9, 2011 #3
    If u consider a body moving with constant acceleration(constant rate of change of velocity with time), then we know that average velocity across any time interval=(V2-V1)/2
    Thus, the motion can be regarded as constant velocity with the average velocity

    In this system, rate of change of acceleration with respect to x is constant, so when can conclude that average acceleration across any interval of x=(a2-a1)/2
    For this case,
    a2= acceleration at maximum displacement=(k/m)A
    a1=acceleration at equilibrium position=0

    isn't this show that the motion can be regarded as a constant acceleration with the average acceleration?
     
  5. Nov 9, 2011 #4
    Its true that if velocity is linear with respect to time, then the average velocity can be treated as a constant velocity. Its also true that if acceleration is linear with respect to time, then the average acceleration can be treated as a constant acceleration. But in the case of the spring, the acceleration is not linear with respect to time, its linear with respect to position. Your method would be fine if acceleration were linearly related to time, but its not; its sinusoidaly related.
     
    Last edited: Nov 9, 2011
  6. Nov 9, 2011 #5

    SammyS

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    In fact, over one complete cycle of the harmonic oscillator, the average acceleration is zero !
     
  7. Nov 11, 2011 #6
    Can you explain or show why it so when it is linearly related to time?
     
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