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SHM question (Is my method correct?)

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  1. Feb 13, 2017 #1
    1. The problem statement, all variables and given/known data
    A particle P is performing simple harmonic motion with amplitude 0.25m. During each complete
    oscillation, P moves with a speed that is less than or equal to half of its maximum speed for 4/3 seconds.
    Find the angular frequency of P

    3. The attempt at a solution
    First I split 4/3 into two portions, the right and left side. Since both are symmetrical, information about left side will be same as information for right side.

    Now 2/3s is the time the particle spend with velocity < 1/2 wx in one side of motion.
    So
    Time Spent = Time required to achieve max - time required to achieve 1/2 of max
    $$ \frac{2}{3} = \omega^{-1} (\sin^{-1}(\frac{Max~V}{Max~V})-\sin^{-1}(\frac{1/2~of~Max~V}{Max~V})) $$
    $$ \frac{2}{3} = \omega^{-1} (\sin^{-1} 1 - \sin^{-1} 0.5) $$
    $$ \omega = \frac{\pi}{2} $$
    Note:- The answer is correct

    Now the problem I realized it, even though my answer is coming correct but in my opinion, my method is not correct. The reason is I have calculated the time required by the particle to go from centre to the 1/2 speed points which makes no sense.
    So can you tell me whether my method is correct and if not what is the correct method?
     
  2. jcsd
  3. Feb 13, 2017 #2

    BvU

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    Hi,

    If you draw a sine and check what fraction of the period the absolute value is < 0.5 the answer is 1/3 .
    So 1/3 of a period corresponds to 4/3 s. Thereby : a full period is 4 s, hence ##\omega = \pi/2## as you 'found'.
    Don't agree: make a drawing.
     
  4. Feb 13, 2017 #3
    How did you work out the value 1/3?
     
  5. Feb 13, 2017 #4

    BvU

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    I looked at the drawing I made :smile:
     
  6. Feb 13, 2017 #5
    Î made a drawing but how did you infer the value fo 1/3?
     
  7. Feb 13, 2017 #6

    BvU

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    ## [0,\pi/6] + [5\pi/6, 7\pi/6] + [11\pi/6, 2\pi] ## totals ##4\pi/6##. And ##4\pi/6 /(2\pi) = 1/3##

    upload_2017-2-13_17-26-40.png
     
  8. Feb 13, 2017 #7
    Oh okay thanks
     
  9. Feb 13, 2017 #8

    BvU

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    You're welcome
     
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