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SHM Problem

  1. Jun 6, 2005 #1

    DDS

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    A 2.12 kg mass on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.83 N/m. The mass is displaced 3.12 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the force (including direction, choose the right to be positive) acting on the mass 3.46 s after it is released?
    B) How many times does the mass oscillate in 3.46 s?

    The above is my problem. I am having a hard time linking the SHM aspects with the kinematics aspects.

    I have calculated my w, found my period and frequency but i dont know where to take it from there. Any suggestions?
     
  2. jcsd
  3. Jun 6, 2005 #2

    DDS

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    I have also taken another approach to solving this question, i am thinking of using conservation of energy and solving for v and then plugging that into the impulse forumal but i am not sure if that would be solving the problem incorrectly since the system is obeying SHM
     
  4. Jun 6, 2005 #3
    The first half can be solved via a direct application of Hooke's Law.
     
  5. Jun 6, 2005 #4

    DDS

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    I dont think it can because the block is displaced to the a particular side. Its not displaced in a plane.

    Plus i tried to solve it that way and the answer is wrong.

    Anyone elese?
     
  6. Jun 6, 2005 #5
    This doesnt make sense. It is displaced along the axis of the spring, is there something I'm missing?

    You need to find the position at t=3.46sec. It is asking for the force at this time.
     
  7. Jun 6, 2005 #6

    DDS

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    whozum, im really sorry i misread my own question. I was confusing it with another one. Really sorry
     
  8. Jun 6, 2005 #7
    Thats fine, do you need any more help?
     
  9. Jun 6, 2005 #8

    DDS

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    Now that i know what its asking,

    i found the velocity of the block and the total distance its traveled in that time span but when i plug back into hookes law i get a fairly large answer. Which results in it being wrong. How would i best go about finding its position at the given time interval?
     
  10. Jun 6, 2005 #9
    Hooke's Law requires the displacement from the equilibrium position. You probably are entering the total distance covered by the oscillating block, which is not what you need, you need the distance between the position at t=3.46s and the equilibrium position (t=0 for this case).
     
  11. Jun 6, 2005 #10

    DDS

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    Let me show you what ive done so far:
    .5kx^2=.5mv^2
    v=4.7

    d=v*t
    =16.262 this becomes my displacement from eqm.

    subbing into hookes law i get

    F=56.2 N
     
  12. Jun 7, 2005 #11
    The equation you cited above gives the kinetic energy of an object due to its PE from the spring's displacement. It is not suitable for this specific problem. Also note that in the spring, the velocity for simple harmonic motion is NOT constant. Surely you have expressed SHM problems in terms of sine and cosine functions, this application is what you need to use for this problem.

    Again, that would be the TOTAL displacement that the mass has experienced in the 3.46 seconds. This is not what you are looking for, you want to find the distance that the block is form its original starting position. Since it is in SHM it is oscillating about the equilibrium point, it just keeps going back and forth in the same location, so it is easy to see that AT MOST this value will be that of the wave's amplitude.

    If it had travelled 16.262m total, it would not be 16.262 meters away, it would have made several 'laps' around the origin and returned to a previous position within the lap (kind of like a swimmer swimming 300m in a 50m pool, he is not 300m away from where he started, instead he is exactly where he started.)
     
  13. Jun 7, 2005 #12

    DDS

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    Yes i see what you are saying:

    would i have to manipulate

    x(t)=Acos(wt +pi) in order to find my amplitude.

    I have a feeling that it takes 1/4 of oscialtiong to reach its max distance or lets say amplitude
     
  14. Jun 7, 2005 #13
    You could definitely do it that way, however the pi in the equation should be a theta, and the trig function should (doesnt have to be for correct values of theta) be a sin function as opposed to a cosine.

    If you know some calculus, you could easily find the force from the above equation knowing that acceleration is the second derivative of position..
     
    Last edited: Jun 7, 2005
  15. Jun 7, 2005 #14
    [tex]x(t) = A cos(\omega t + \phi)[/tex]

    at t = 0, x = A. Use this fact to determine the phase constant. Then proceed with Hooke's law as you planned.
     
  16. Jun 7, 2005 #15

    Chi Meson

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    There is an easier way.

    Use the equation for acceleration as a function of time. Then Newton's 2nd Law.
     
  17. Jun 7, 2005 #16
    That is another way, but why is that easier?

    In the first x = ..., F = kx,

    In the second a = ..., F = ma.
     
  18. Jun 7, 2005 #17

    Chi Meson

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    Point taken. It is equivalent to your solution. Both are much easier than the direction this thread was going.
     
  19. Jun 7, 2005 #18

    DDS

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    Thanks everyone i got the answer
     
  20. Jun 7, 2005 #19
    The thread was going to this same conclusion, if you read my last post.
     
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