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Mass Spring System

  1. Sep 24, 2005 #1
    In a vertical mass spring system, spring with spring constant 250 N/m vibrates with an amplitude of 12cm when 0.38kg hangs from it.
    What is the equation describing this motion as a function of time? ( assume the mass passes through the equilibrium point, towards the positive x(upward), at t = 0.110s)

    I do it in this way:
    Hooke`s Law: F = kx
    (0.38)(9.8)= 250 N/m (x)
    x = 0.015(m)

    [tex]\omega[/tex] = [tex]\sqrt{k/m}[/tex]
    [tex]\omega[/tex] = 25.65

    x = Acos([tex]\omega[/tex]t + [tex]\phi[/tex])
    0.015 = 0.12cos(25.65t + [tex]\phi[/tex])
    [tex]\phi[/tex] = -1.38

    So, the equation : x = 0.12cos(25.65t -1.38)

    Is it correct?
  2. jcsd
  3. Sep 24, 2005 #2


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    1. Here the force is not mg, Actually mg is balanced by initial streach of the spring when it come in equilibrium position after the mass is attached.

    2. F represents the restoring force trying to bring the mass back in the equilibrium position.
  4. Sep 25, 2005 #3
    The spring extent because of mg. Then why not F = mg?

    How should I find the phi?
  5. Sep 25, 2005 #4


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    What is the resultant force when the syatem is in equilibrium?

    Actually the force F in S.H.M. is the restoring force on the mass when it has displacement x from equilibrium position.

    your equation
    [tex] x = A cos (\omega t + \phi ) [/tex]
    is correct.

    When x = 0 ; t = 0.110s
    put these values to get [tex] \phi [/tex]
  6. Sep 25, 2005 #5
    Ya, I did it in this way before also but the value I get is [tex]\phi[/tex] equal to -1.25

    But it is actually should be about +1.89 ( from given answer)
  7. Sep 25, 2005 #6


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    [tex] cos \theta = cos \alpha [/tex] has general solution
    [tex] 2n \pi \pm \alpha [/tex]

    which sign is to be taken?
    where was the mass at t = 0 ?
  8. Sep 25, 2005 #7
    when t = 0
    x = A

    Any more clues?
  9. Sep 26, 2005 #8


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    The same solution can be written as 3pi/2 - wt = 4.712 - 2.821 = 1.891 this is given in your text book.
    Last edited: Sep 26, 2005
  10. Sep 26, 2005 #9


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    The solutions of the equation are
    [tex] \pi /2 - 2.821[/tex] and [tex] 3 \pi /2 - 2.821 [/tex]
    = -1.25 and 1.891 radians respectively.
    Now think, when t=0.11s; x=0 means at t = 0 the particle is yet to reach the equilibrium position [Where the phase angle (wt + phi) is pi/2] and x is - ve because the time period is 0.225s and 0.11s this is a bit less then half time period. For that the phase angle should be more then pi which is given by the value of phi = 1.891.
  11. Sep 26, 2005 #10
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