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Simple Harmonic Motion question

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height 3 cm above the top of the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 10 cm. What is the oscillation frequency?

    2. Relevant equations

    Not sure if this energy equation applies. 1/2k A^2 = 1/2 k x_0 ^2 + 1/2 m v_0 ^2
    w = 2 pi / T T = 2pi root( m/ k)
    f = 1 /T
    w = root ( k/m)
    3. The attempt at a solution

    I really have no clue on how to start this problem. But i believe i can use the first energy equation to find k . But i am really stuck. Can you guys help me out.
  2. jcsd
  3. Nov 10, 2007 #2


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    You are correct about where to start. You can use the energy equation to find k. Can you do this or are you having trouble computing k? If you are, show me your work and I may be able to help you find your error or help you get past a stumbling block.

    After you find k, do you know where to go from there?
  4. Nov 10, 2007 #3
    well i don't have the mass for the energy equation, and i am not sure exactly what xo is, is it zero? and what is v_o .
  5. Nov 10, 2007 #4


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    v_o is whatever the blocks speed was when it hit the spring. Also, you can set x_o as 0 in the problem.

    Regarding the the mass of the block, would it help if you solved the equation for k/m instead of just k?
  6. Nov 10, 2007 #5
    Thanks for the help.

    This is my method, but i am off by .6 from the answer.

    i used the kinematics equation to solve for the speed.

    Vf ^ 2 = vi ^2 + 2 a s

    i get vf = 0.766 m/s

    This vf is my new Vi for the spring.

    THen i use

    1/2 k A^2 = 1/2mvi^2

    I solve for k

    and i get k = m vi^2 / A^2

    I use another formula F = 1/2pi * root ( k/m ) this formula is the frequency formula

    I sub k into that formula and i get that the freq is 1.22

    but the answer is 1.83. Can someone see if my method is right?
  7. Nov 11, 2007 #6


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    Your method seems correct to me. Are you sure your numbers are correct?
  8. Nov 11, 2007 #7


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    There are 2 mistakes I'm seeing here:

    first, you're not taking into account the gravitational potential energy of the block... the block is going to drop a certain height...

    second the energy stored in the spring is not (1/2)kA^2 when all the energy is converted to elastic potential energy...

    The amplitude is the displacement from the equilibrium position. When the mass is at 0 amplitude... there is already energy stored in the spring....

    how much has the spring compressed when it is at the equilibrium position... ie: when the mass would be hanging at rest.

    so the form of the equation you need is:

    1/2 k (A + x0)^2 = 1/2mvi^2 + mg(A + x0).

    A little better way, is to just use the energy at the very top (before it moves at all), before it drops to the top of the spring... this way you don't have the kinetic energy term.

    (1/2) k (A + x0)^2 = mg(A + x0 + 0.03).

    So using the bottom most level the mass reaches as having 0 grav. potential energy, the mass initially has mg(A + x0 + 0.03) energy. At the bottom it only has elastic potential energy in the amount of (1/2) k (A + x0)^2

    So find x0... plug it into the above equation... and try to use that equation to find w... try to manipulate the equation so that you won't need to deal with m and k... use the relationship between, w, m and k.
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