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Question regarding angular frequency of a SHM

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data

    image.jpg
    image.jpg


    2. Relevant equations
    KE=½m(ωa)2

    3. The attempt at a solution
    So first I did this:
    2.4x10-3= ½ mω2(1.5x10-2)2
    To find mω2=21.33
    And substitute that into the KE eqn to find the new amplitude, which is 1.30x10-2
    But I only did that because that was the only way I could think of.
    My question is, why is ω constant despite changes in the max KE?
     
  2. jcsd
  3. Apr 7, 2016 #2

    ehild

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    How does the max KE depend on ω and the amplitude?
     
  4. Apr 7, 2016 #3
    Linear velocity will change during the oscillation (x changes)
    upload_2016-4-7_10-57-15.png

    so the KE

    upload_2016-4-7_10-59-52.png

    will too:
    upload_2016-4-7_10-58-50.png
     

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  5. Apr 7, 2016 #4
    Max KE=½m(ωa)2
     
  6. Apr 7, 2016 #5
    @Physics-Tutor Thanks but I'm still confused. How do we know that the angular frequency, ω, is constant?
     
  7. Apr 7, 2016 #6

    ehild

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    Changing KE means changing the amplitude if it is the same oscillator. You know that ω=√(k/m). If the spring is the same and the mass of the load is the same, why should the frequency change?
     
  8. Apr 7, 2016 #7

    ehild

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    We use the same spring and loading mass.
     
  9. Apr 7, 2016 #8
    ehild is correct:
    upload_2016-4-7_14-11-33.png

    k is the spring constant, m the mass. These two are constant, so the angular velocity is constant.
     

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  10. Apr 7, 2016 #9
    Oh, I got it. Thanks a lot !
     
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