Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

  2. jcsd
  3. Jun 12, 2005 #2


    User Avatar
    Science Advisor

    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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook