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Pendulums and SHM

  1. Nov 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Two identical simple pendulums of length 0.25m and period T, are mounted side-by-side. One is released from rest with an initial angular displacement of [itex]\displaystyle\frac{\pi}{12}rad[/itex], and the other is started with an initial angular velocity of 0.1 rad/s at θ=0. They are both started in motion simultaneously. Which are true or false?

    A. After a time of T, they have both reversed their initial angular velocities.
    B. They both have the same phase.
    C. After 0.1s they both have the same phase.
    D. The equation of motion for both oscillators reads: [itex]x=A\cos(\omega t+\delta)[/itex]

    2. Relevant equations

    -

    3. The attempt at a solution

    I understand that the first pendulum should not have any phase as the initial displacement would be already the amplitude of the oscillation of the pendulum (thus wouldn't B and D be false?). On the other hand, after a time of T, they would both have the same velocities (what does reversed mean in this context?). I found that T = 1.00s. I'm not so sure on how C works (what is the exact definition of "phase"? Does it refer to the initial conditions or the angle at a time t?)
     
    Last edited: Nov 25, 2013
  2. jcsd
  3. Nov 25, 2013 #2
    The "phase" is the therm in the parenthesis (in the equation given in part D).
    Phase = ω t + δ. So the phase is a function of time.

    The initial phase is the phase at t=0 so it will be δ in the above equation. The initial phase is a constant, for a given system.

    Then the first pendulum has an initial phase of zero, assuming that the equation in D is used.
     
  4. Nov 25, 2013 #3
    So then [itex]δ=\displaystyle\frac{\pi}{2}[/itex] or is it 0? Then shouldn't the second pendulum as well have an initial phase of zero as it starts from an angle 0? Or do we have to consider the initial velocity as well? I have that:

    [itex]0.1=-Aω\sin(wt+\phi)[/itex] (then this would mean that the initial phase is not zero?)

    I'm completely lost right now.
     
  5. Nov 25, 2013 #4
    OK, so assume we use the equation with cosine, as in part D.
    Then at t=0, the first pendulum has displacement x=A and the second has x=0. (the displacement is measured from equilibrium position).
    Put t=0 in the equation x=Acos(ωt+δ) and calculate δ so that you have x=A for the first pendulum and x=0 for the second pendulum.

    No need to use the equation for speed here.
     
  6. Nov 25, 2013 #5
    So i found the phase of the second pendulum is [itex]-\displaystyle\frac{\pi}{2}rad[/itex]. Then they would not have the same phase after 0.1 seconds, and they would not have the same initial phase. After a time T, they would have reverted to their initial angular velocities (so i'm not sure what they mean with prompt A). On the other hand, we would have that the equation of motion reads [itex]x=A\cos(\omega t+\delta)[/itex] (for the exception that the amplitude of both oscilators is different, and the phase is different as well). I tried with putting all of them false, but that is not the case. I as well tried with T F F F and F F T T. Now I'm thinking that maybe it could be T F F T. Is my reasoning correct? The question is kind of vague.
     
  7. Nov 25, 2013 #6
    In think that the last one is true. O course they may have different values for A and δ but the equation is the same.

    A says that the their velocities will be reversed not that they revert to initial velocities.
    And this is not true. They will have the same velocities as in the beginning, after T.
     
  8. Nov 25, 2013 #7
    You were right, I got it with F F F T. Thank you very much for your patience!
     
  9. Nov 30, 2013 #8
    Hmm... I have almost the same question yet when I got the correct answer, the last statement was false. (i.e. False: The equation of motion for both oscillators reads: x = Acos(ωt+δ) ) This is because although the motion of each oscillator separately is described by that equation, together they are not or something like that. Not too sure myself, guessed the answer via elimination.
     
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