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Simple harmonic motion of particles

  1. Mar 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean position is separated by distance X0 (X0 > A). If the maximum separation between them is (X0 + A), the phase difference between their motion is

    My
    answer is π/2 but the answer given is π/3.

    2. Relevant equations
    x = Asin(ωt + φ)

    3. The attempt at a solution
    To me it looks like this: When the first particle is at it's mean position, the second particle is at it's extreme position, so that the distance between both of them is X0 + A. So, the phase difference should be π/2

    In terms of coordinates, let mean pos of P1 be x=0, mean pos of P2 be X0. Let P1 go from x= -A to x = +A and P2 go from x = X0 - A to x = X0 + A. The situation here represents P1 at x=0 and P2 at x = X0 + A.
     
  2. jcsd
  3. Mar 28, 2016 #2

    ehild

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    You can check the validity of your logic, by finding the maximum separation if the phase difference is pi/2. Is it really xo+A?
     
  4. Mar 28, 2016 #3

    BvU

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    In that case A isn't the maximum difference ! Work out ##A \sin \omega t (t + \phi) - A \sin(\omega t)## to get an expression in ##\phi## and see where you can go from there ....

    [edit] ah! Elisabeth was faster. I leave you in her hands...
     
  5. Mar 29, 2016 #4
    x1 = Asin(ωt + Φ1)
    x2 = X0 + Asin(ωt + Φ2)
    x2 - x1 = X0 + Asin(ωt + Φ2) - Asin(ωt + Φ1)
    = X0 + 2A##cos\dfrac{2ωt + φ_1 + φ_2}{2}sin\dfrac{φ_2 - φ_1}{2} ##
    X0 + A = X0 + 2A##cos\dfrac{2ωt + φ_1 + φ_2}{2}sin\dfrac{φ_2 - φ_1}{2} ##

    ##\dfrac{1}{2} = cos\dfrac{2ωt + φ_1 + φ_2}{2}sin\dfrac{φ_2 - φ_1}{2} ##

    This can be visualised as SHM with amplitude ##sin\dfrac{φ_2 - φ_1}{2}##

    ##\dfrac{1}{2} = sin\dfrac{φ_2 - φ_1}{2}##

    ##φ_2 - φ_1 = \dfrac{π}{3} ##

    While I have gotten the answer, I'm not sure I understand why it's true at all.
    I think I'm very confused here. I don't know how to.
     
    Last edited: Mar 29, 2016
  6. Mar 29, 2016 #5

    ehild

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    You can choose one of the phase constant zero.
    x1 = Asin(ωt )
    x2 = X0 + Asin(ωt + φ)
    x2 - x1 = X0 + Asin(ωt + φ) - Asin(ωt )
    ##cos\dfrac{2ωt + φ}{2}sin\dfrac{φ}{2}##
    At what time is the last expression maximum and what is the maximum value?
     
  7. Mar 29, 2016 #6
    Oh, I got that, I edited my post.
     
  8. Mar 29, 2016 #7

    ehild

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    It is confusing to the reader that you edited the first post. This makes to look all helpers were stupid.
     
  9. Mar 29, 2016 #8
    I didn't edit my first post. I edited my answer post before you replied.
     
  10. Mar 29, 2016 #9

    ehild

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    Yes, I see. So I was stupid only, answering a solved problem.
    Anyway. If you do correct Maths and you got a result it is correct.
     
  11. Mar 29, 2016 #10
    I'm really sorry if I shouldn't have.
     
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