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Simple harmonic question

  1. Jun 22, 2014 #1
    A particle of mass is connected to a spring with a force constant K. The particle undergoes simple harmonic with an amp A. What is the potential energy of the partic when the position is (1/2A)?



    2. Relevant equations
    E=1/2kA^2
    1/2kdelta^2=1/2mv^2+1/2kx^2


    3. The attempt at a solution

    U=1/2kx^2=1/2k(1/2A)^2=1/4(1/2KA^2)=1/4E= K=E-U= 3/4? Not really sure at all can somebody help i don't think i did it correctly
     
  2. jcsd
  3. Jun 22, 2014 #2

    Simon Bridge

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    Welcome to PF;
    You started strong - the formula for PE is correct.
    You substituted x=A/2 fine.

    You have been asked to find U - so stop when you have.
     
  4. Jun 23, 2014 #3
    Besides, E-U = 3/4 makes no sense.
     
  5. Jun 23, 2014 #4

    BiGyElLoWhAt

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    1/4(1/2KA^2) doesn't equal 1/4E, it equals 1/4 Umax.
    gotta be careful.
     
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