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Simple harmonic motion energy question

  1. Jun 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Can I get help in part (b) of this question ?
    Q1.jpg

    2. Relevant equations
    KE=1/2 m v2
    v= (2π f )√(A2 - x2)

    3. The attempt at a solution
    I substituted the second equation into first one, so i got
    KE= 1/2 m (2π f )2 (A2 - x2)

    but then couldnt complete
     
  2. jcsd
  3. Jun 5, 2015 #2

    jbriggs444

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    You are apparently concerned with the answer to part b. One approach starts with the insight that total energy is conserved. The sum of PE + KE is a constant.

    If KE is 3/4 of total energy then what fraction of total energy is PE?

    The next insight is that the force required for harmonic motion is exactly the force supplied by a perfect spring. Or, simpler yet, we could simply assume that the oscillation is produced by a mass attached to a perfect spring. Do you know the formula for the potential energy of a spring?
     
  4. Jun 5, 2015 #3
    Do you mean hooks law F=-kx ?
     
  5. Jun 5, 2015 #4

    jbriggs444

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    Yes. That's part of it. There is also a formula for energy in a spring as a function of displacement x. That formula can be derived by integrating F=-kx over distance. The result is PE = 1/2kx2
     
  6. Jun 5, 2015 #5
    Hi,
    You know that the energy is conserved right, so E is constant, why not calculating is at any point, example would be when there's no kinetic energy, Once you got E, you know that when KE = 3E/4, what is left if energy is potential energy, do this sound familier kx^2 /2 ?
     
  7. Jun 5, 2015 #6
    well , iam not used to PE = 1/2kx2 :(
     
  8. Jun 5, 2015 #7
    Kx^2/2 is the elastic potential energy of a system, where K is the spring's constant, x is the strech of the spring, and 1/2 is one half (obviously, ) and if you want a proof, W = F.x so dW = F*dx = F*x'dt = k*x*x' dt and you integrate both sided and it yields to W = 1/2kx^2 ! And now you can work with it, good luck
     
  9. Jun 5, 2015 #8

    jbriggs444

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    If you don't like using PE = 1/2 kx2 there are still a couple of other approaches to solving the original problem. Do you prefer trigonometry or differential calculus?
     
  10. Jun 5, 2015 #9
    Isnt it possible to solve it by the equation I have given ?
    it will be the same on both sides , but one will have the coefficient (3/4).
    At the same time , it say in equilibrium position , so one side will have x=0, and so max. velocity (K.E)
    but i am facing problem in which side to place them ( right or left ).
    correct me if I am wrong please .
     
  11. Jun 5, 2015 #10

    jbriggs444

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    The equation seems to be missing a factor of sin(theta).

    Edit: I take that back. That's what the ##\sqrt{A^2-x^2}## is supposed to cover.

    So if you know that the total energy E is equal to KE alone when x=0 (because this is the point where PE is presumably taken as zero), that should give you an equation.
     
    Last edited: Jun 5, 2015
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