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Simple Harmonic motion

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  1. Apr 10, 2017 #1
    1. The problem statement, all variables and given/known data
    The question is uploaded.
    SHM.jpg
    3. The attempt at a solution
    I have completed the whole question, however, stuck on the last part.
    How to find the value about which ## \rm \small \theta## now oscillates?
    A source stated that ## \rm \small \alpha## is the value about which ## \rm \small \theta## is oscillating. Can anybody tell me if that is correct? If yes, then why?
     
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  3. Apr 10, 2017 #2

    TSny

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    What value of ##\phi## does the pendulum oscillate about?
     
  4. Apr 10, 2017 #3
    I am not sure about the answer to your question. All I know is that the value of ##\theta## is technically the "amplitude" of the motion.
     
  5. Apr 10, 2017 #4

    TSny

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    What value of ##\phi## corresponds to the equilibrium position of the pendulum. See equation [2].
     
  6. Apr 10, 2017 #5
    ##\theta##?
    If yes? Can you tell why?
     
  7. Apr 10, 2017 #6
    Maybe I got this.
    Before the acceleration, the ##7/24 cos\theta## term wasn't present. Thus if I consider, the before acceleration state as equilibrium then ##\theta## is the value that corresponds to equilibrium. But after applying the acceleration, the ##7/24 cos\theta## term emerged which gave rise to the ##-\alpha## value. As a result, the ##-\alpha## value must be the maximum amplitude or the value about which ##\phi## oscillates.
    Rearranging the equation, we get that ##\theta ## oscillates about the value of ##\alpha##.
    Correct?
     
  8. Apr 10, 2017 #7

    gneill

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    Hint: Usually when you want to arrive at a result of the form trig(a - b) where "trig" is one of the trig functions, you are looking to employ one of the trig angle sum or difference identities in some fashion...
     
  9. Apr 10, 2017 #8
    Apologies but I can't seem to relate your hint with my current confusion.
     
  10. Apr 10, 2017 #9

    gneill

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    Start with the expression:

    ##g~\left( sin(θ) - 7~cos(θ) \right)##

    You need to get it into a form where the sin(a - b) identity can be applied. That will mean massaging the expression until the coefficients of the existing sin(θ) and cos(θ) are the cosine and sine of some new angle, ##\alpha##.
     
  11. Apr 10, 2017 #10
    Yes, I did all of that and also obtained the correct expression ## \rm \small -g\sin(\theta - 16.3) ##
    I am just stuck at "the value at which theta oscillates"? I mean why the value of alpha is the value at which theta oscillates?
    Is this explanation correct?
    Before the acceleration, the ##7/24 cos\theta## term wasn't present. Thus if I consider, the before acceleration state as equilibrium then ##\theta## is the value that corresponds to equilibrium. But after applying the acceleration, the ##7/24 cos\theta## term emerged which gave rise to the ##-\alpha## value. As a result, the ##-\alpha## value must be the maximum amplitude or the value about which ##\phi## oscillates.
    Rearranging the equation, we get that ##\theta ## oscillates about the value of ##\alpha##.
     
  12. Apr 10, 2017 #11

    TSny

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    Try to be precise in how you state things. ##\theta## is a variable, so it cannot correspond to equilibrium. There is a specific value of ##\theta## that corresponds to equilibrium.

    First, find the value of ##\phi## that corresponds to equilibrium. If the pendulum hangs at rest in the equilibrium position, what is the value of ##\frac{d^2 \phi}{dt^2}##?

    Then, what does equation [2] imply for the value of ##\phi## at equilibrium?
     
  13. Apr 10, 2017 #12
    Well if the pendulum is in equilibrium and hangs in free fall, then ##\phi=0## and then ## \small \frac{d^2\phi}{dt^2}## should be equal to 0.
    I get the meaning of your second question but unfortunately, I don't know the answer. Maybe ##-\alpha## because ##\theta## keeps changing.
     
    Last edited: Apr 10, 2017
  14. Apr 10, 2017 #13

    haruspex

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    Did you mean φ=0?
     
  15. Apr 10, 2017 #14
    Oh yes.
     
  16. Apr 10, 2017 #15

    haruspex

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    Your equation is ##\ddot \phi=k\sin(\phi)##. For what value of φ will ##\ddot \phi## be zero?
     
  17. Apr 10, 2017 #16
    ## \phi = 2k\pi ## but since in this question ## \phi ## is a small angle so ## \phi = 0 ##
     
  18. Apr 10, 2017 #17

    haruspex

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    Right. (I see you edited your post #14 after my post #15.)
    So what does that give for θ at equilibrium?
     
  19. Apr 10, 2017 #18
    Yes it gives ##\theta = \alpha##
     
    Last edited: Apr 10, 2017
  20. Apr 10, 2017 #19

    haruspex

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    Ok. All done?
     
  21. Apr 11, 2017 #20
    Is this explanation correct?
    WhatsApp Image 2017-04-11 at 4.43.09 AM.jpeg
     
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