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Homework Help: Simple Harmonic Motion - Potential and Kinetic Energy?

  1. Mar 15, 2008 #1
    1. The problem statement, all variables and given/known data

    A 2.00 kg mass vibrates according to the equation x = 0.470 cos 8.36t, where x is in meters, and t is in seconds. Assume that x = 0.29 m.

    (a) Determine the amplitude.
    (b) Determine the frequency.
    (c) Determine the total energy.

    (d) Determine the kinetic energy.

    e) Determine the potential energy.

    I did the first two parts (a-b), but I'm stuck on c,d & e, the questions about the Energy of the spring.

    2. Relevant equations

    PE = .5kx^2


    KE = .5KA^2


    3. The attempt at a solution

    I've tried these 3 multiple times and I keep getting them wrong. One thing I'm not sure of, is is "x" the amplitude? I have been plugging in .29 for the "x" and .47 for the "A", but just curious if this is the reason I'm getting these wrong.

    I would think, that according to these formulas, KE+PE=TOTAL ENERGY, and I thought I was doing them right. I'm at my wits end.
    Last edited: Mar 15, 2008
  2. jcsd
  3. Mar 15, 2008 #2

    Doc Al

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    Staff: Mentor


    Not good. (This implies that the KE doesn't change!)

    No. "x" is displacement from equilibrium.
    These are correct.

    Absolutely. What's the total energy?

    Hint: Since total energy doesn't change, pick the easiest position to calculate PE + KE.
  4. Mar 15, 2008 #3
    K.E + P.E = total energy
    They usually give you that equation on formula sheets.
  5. Mar 15, 2008 #4
    Does Ke=.5mv^2?

    I don't know V, and I can't use w=v/r because I don't know R.


    I tried to find Total Energy when PE is maxed out and v=0, but I got it wrong. I did

    Total Energy=.5((9.8*2)/(.29))*(.29)^2
    thats .5 * k * x^2

    And Snazzy-I tried .5KA^2 as my total energy and got 2.986 and thats wrong...
  6. Mar 15, 2008 #5
    The total energy is only given by [tex]\frac{1}{2}kx^2[/tex] only if [tex]x=A[/tex]

    You find A by looking at the equation of the motion because SHM follows the equation:

    [tex]x(t)=Acos(\omega t+\phi)[/tex]

    Then you can find the potential energy at x=0.29m and then the kinetic energy at that point since you now have the total energy and the potential energy.
  7. Mar 15, 2008 #6
    But I can't find KE because I don't know V and I can't find v using W because I don't have R...






    Thank you so much for your help on this!!
    Last edited: Mar 15, 2008
  8. Mar 15, 2008 #7
    You don't need V or R or whatever to find the kinetic energy. You can use the fact that KE + PE = total energy
  9. Mar 15, 2008 #8
    Well, x is not equal to A. So I guess I can't use the fact that .5kA^2 is total, and I keep getting PE wrong. :(
  10. Mar 15, 2008 #9
    You get PE wrong because your value of k is wrong. F = kx at equilibrium, but x=0.29 is not the distance the spring has stretched at its equilibrium state. They usually give you the equation:

    [tex]\omega =\sqrt{\frac{k}{m}}[/tex]

    And you CAN use the fact that [tex]\frac{1}{2}kA^2[/tex] is the total energy because A is GIVEN TO YOU IN THE EQUATION OF THE FUNCTION. The reason you got it wrong is because your value of k is wrong.
    Last edited: Mar 15, 2008
  11. Mar 15, 2008 #10
    Okay, I tried it using x=.470 for my equation of the value of k, and I get k=41.7

    but then I still get the wrong answer for PE...=1.753
  12. Mar 15, 2008 #11
    I don't know why you're using the amplitude to find the value of k, use [tex]\omega[/tex] and the mass to find the value of k as shown in the equation above. The spring does not stretch to its amplitude at equilibrium, nor does it stretch to the value of x the question gives you.
  13. Mar 15, 2008 #12
    I GOT THEM RIGHT!!! =)

    Thanks, you were SO helpful!! I didn't realize that you can't always use the equation of F=-kx to find k!

    I'm really thankful and grateful that you stayed online and helped me through each step. Thanks again, you're a really good person. =)
  14. Mar 15, 2008 #13


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    You can always use F=-k x at two conditions: that the F you use is the force exerted by the spring and that the x you use is the displacement of the mass from the equilibrium position. I am not sure why you concluded that you can't always use that equation but I wanted to point that out.
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