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S.H.M for mass spring system

  1. Jun 12, 2005 #1
    I've been trying this question for quite some time and have given up.

    Prove that a mass,M , suspended from a fixed point by a helical spring, which obeys hooke's law, undergoes simple harmonic motion when it is displaced vertically from its equilibrium position.

    Say I displace the mass spring upwards, e metres. Then I can say:

    Mg + ke = Ma ?

    When the mass is at far bottom then

    kx - Mg = -Ma ?

    When I equate these two I get kx=-ke, which proves nothing. :yuck:

    Thanks
     
  2. jcsd
  3. Jun 12, 2005 #2

    HallsofIvy

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    What do you have to use? You certainly can't do it by looking at kinetic energy not at a few positions. I would do it by setting up the differential equation governing the motion and showing that the solution is harmonic.
     
  4. Jun 12, 2005 #3
    The book gives the answer very vaguely which confuses me more. It does not have any calculus involved..

    k(e + x) - Mg = -Ma

    I have no Idea how they've derived this

    Then they say 'in the equilibrium position, the pull of the spring ke is equal to the weight'

    ke = Mg

    But when the mass is at the equilibrium position the force on the mass by the spring is 0

    Subsittuting above..

    Mg + kx - Mg = -Ma
    kx=-Ma

    a=-(k/M)x

    Compare with a=-(omega)^2x

    Doing some math it also shows that T = 2(pi)(M/k)^(1/2)

    Thanks for your help
     
  5. Jun 12, 2005 #4
    The spring still has a tension, even in equilibrium. It's the sum of the forces on the mass that is 0, which is the same as that first statement.
     
  6. Jun 12, 2005 #5
    Thanks, it's just occured to me that I've been looking at the situation the wrong way. :zzz: . problem solved
     
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