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Simple Harmonic Oscillations

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to prove that the total energy of a spring mass system is equal to (1/2)k(delta L^2 + A^2)
    The spring is in three sates, equilibrium (I proved that already), maximally stretched, and maximally compressed. The spring is at equilibrium at a height h above ground level. delta L is the amount that the string stretches after the weight is added to the free spring.


    2. Relevant equations
    EP = 1/2kx^2
    K = 1/2kx^2
    GE = mgh


    3. The attempt at a solution
    For the spring maximally compressed : I have KE = 0 becuase it is not moving. EP = (1/2)k*(A-delta L)^2. when that is extended EP = 1/2*k*(A^2-2AdeltaL+deltaL^2).
    GE = m*g*(h-a). When all of these are added together it does NOT equal (1/2)k(delta L + A^2). I can't seem to find my error.
     
  2. jcsd
  3. Apr 7, 2009 #2

    LowlyPion

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    Welcome to PF.

    What is A?
     
  4. Apr 7, 2009 #3
    A is just the distance that it moves up or down in either dirrection. For example, at its max height, h is h+A.
     
  5. Apr 7, 2009 #4

    LowlyPion

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    So A is the amplitude and ΔL is the static displacement of an added weight?
     
  6. Apr 7, 2009 #5
    Yes exactly, I just phrased it poorly.
     
  7. Apr 7, 2009 #6

    LowlyPion

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    Do you have the actual problem statement? Maybe that would mean fewer questions?
     
  8. Apr 7, 2009 #7
    Yea here it is verbatim.

    A mass "m" is attached to the free end of a light vertical spring (unstretched length l) of spring constant "k" and suspended form a ceiling. The spring stetches delta L under the load and comes to equlibrium at a height "h" above the ground level (y=0). The mass is pushed up vertically by "A" from its equilibrium position and released from rest. The mass-spring executes vertical oscillations. Assuming that gravitational potential energy of the mass "m" is zero at ground level, show that the total energy of the spring-mass system is (1/2)*k*(delta L2 + A2) + mgh.

    I already did this for equilibrium, I can't quite figure it out for when it is at its max and min positions.
     
  9. Apr 7, 2009 #8

    LowlyPion

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    That makes a big difference having the m*g*h.

    Consider the point at which you release the weight after pushing it up. Examine the sum of the Gravitational potential energy and the spring potential energy.

    PE = m*g*(h + A)

    SPE = 1/2*k*(A - ΔL)2 = 1/2*k*(A2 -2AΔL + ΔL2)

    But what is the term 1/2*k*(-2AΔL) ?

    Does that look familiar?

    Does the expression mg = kΔL hold true? The weight = the displacement times spring constant?
     
  10. Apr 7, 2009 #9
    It might by the -F of the spring, because F=-kx. If this were true, I don't see how this would help though.
    Assuming that is correct the equation would reduce to.

    (1/2)*k*(A^2 + F + ΔL^2) + mg(h+A)
     
    Last edited: Apr 7, 2009
  11. Apr 7, 2009 #10

    LowlyPion

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    Sorry, I think you missed my point.

    Total Energy = m*g*(h + A) + 1/2*k*(A2 -2AΔL + ΔL2)

    Total Energy = m*g*(h + A) + 1/2*k*(A2 + ΔL2) - 1/2*k*2AΔL

    Noting that mg = k*ΔL you get ...

    Total Energy = m*g*(h + A) + 1/2*k*(A2 + ΔL2) - A*kΔL = m*g*(h + A) + 1/2*k*(A2 + ΔL2) - A*mg

    Total Energy = m*g*h + 1/2*k*(A2 + ΔL2)
     
  12. Apr 7, 2009 #11
    ohh, it makes so much sense once I see it mapped out. Thanks for all of your help, I'll do the other state now!
     
  13. Apr 14, 2009 #12
    Why is the mgh term still there? I thought the original question was asking one to show
    Total Energy = 1/2*k*(A2 + ΔL2)
     
  14. Apr 14, 2009 #13

    LowlyPion

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    The m*g*h is there because the problem states
    ... and they do want the total energy in the system.
     
  15. Apr 15, 2009 #14
    You are absolutely right, I missed the Mathilin's verbatim restating of the original problem.
    The frame of reference is at ground level and mgh is definitely needed.
     
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