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Doubling the energy of an oscillating mass on a spring

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    From my calculations, option A is correct. What am I doing wrong? Also, I can't find a relationship between amplitude and energy.

    [tex]{ KE }_{ max }=\frac { 1 }{ 2 } m{ { v }_{ max } }^{ 2 }\\ =\frac { 1 }{ 2 } m{ (rω) }^{ 2 }\\ =\frac { 1 }{ 2 } mr^{ 2 }{ ω }^{ 2 }\\ { { KE }'_{ max } }=\frac { 1 }{ 2 } mr^{ 2 }{ (\sqrt { 2 } ω })^{ 2 }\\ =m{ r }^{ 2 }{ ω }^{ 2 }\\ =2{ KE }_{ max }[/tex]

    Attached Files:

  2. jcsd
  3. Apr 16, 2013 #2


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    By conservation of mechanical energy during SHM,

    KE + PE = constant.

    Total energy = constant = ½kA2

    Where A is the amplitude, so option A should be correct.
  4. Apr 16, 2013 #3
    Option A doesn't involve amplitudes though. You mean option B?
  5. Apr 16, 2013 #4
    Okay, I understand now how I increasing the amplitude by √2 doubles total energy. However, it seems that increasing angular frequency by √2 also doubles total energy, so are both option A and option B correct?
  6. Apr 16, 2013 #5


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    yep. Nice work!

    Edit: is the mark scheme saying that only B is correct? as you said, it looks like A and B are both true.
  7. Apr 16, 2013 #6
    Arghh... I'm not sure how my physics professor missed that. Now I'm a little nervous that she might mark my correct answers as wrong for the final tomorrow.

    Thanks for the help! :)
  8. Apr 16, 2013 #7


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    I meant option B, sorry.

    In your equations, you used v=rω which is for circular motion. Your motion happens to be a mass oscillation.
  9. Apr 16, 2013 #8
    Oh, I see. This is my fault for plugging and chugging. I have to understand what the equations mean.

    So I can only use v=rω for circular motion (i.e. two-dimensional oscillations) only? It doesn't apply to one-dimensional oscillations?
  10. Apr 16, 2013 #9


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    still, even for mass oscillation, the 'angular frequency' is defined by the equation:
    [tex]\omega = \sqrt{\frac{k}{m}} [/tex]
    Even though it is not related to a circular motion. So I think A and B should both be correct.

    Edit: but yeah, as rockfreak says, it is not circular motion, so the calculation should be a bit different.
  11. Apr 16, 2013 #10


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    yeah. After all, the velocity in one-dimensional SHM should be varying, right?
  12. Apr 16, 2013 #11
    Yes, it does vary. But what does that have to do with it?

    30 seconds of thinking...

    Oh wow! Brilliant!

    So the equation v=rω makes use of the fact that the magnitude of velocity in circular motion stays constant!
  13. Apr 16, 2013 #12


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    yes, for uniform circular motion.

    edit: I mean yes, for uniform circular motion the magnitude of velocity is constant, so the angular frequency is constant. (i.e. a parameter of the system).
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