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Homework Help: Finding the phase angle for simple harmonic motion

  1. Dec 22, 2011 #1
    1. The problem statement, all variables and given/known data
    A .25 kg block oscillates on the end of a spring with a force constant of 200 N/m. If the oscillation started by elongating the spring 0.15 m and giving the block a speed of 3 m/s, (a) what is the amplitude of the oscillation, and (b) If the clock is started when the block is at the right hand extreme of its motion, how long does it take to reach a point where the kinetic energy of the block equals the elastic potential energy stored in the spring?

    2. Relevant equations
    x(t) = Acos (ωt + phase angle)
    v(t) = -ωAsin(ωt + phase angle).

    3. The attempt at a solution

    I know that the first part of this question can be solved using energy relationships, but I tried to use the x = Acos (ωt + phase angle) and v(t) = -ωAsin(ωt + phase angle).

    I divided v(t) by x(t) and got

    3/0.15 = -[(rad)(k/m)][tan(phase angle)]
    20 = -28.28tan(phase angle)
    phase angle = -.616 radians

    I then went back to the the x(t) equation and plugged this value in:

    .15 = Acos[28.28(0) - .616]
    A = 0.18 m

    My first question is, is this method valid? The correct answer is 0.18, but just making sure this reasoning is sound.

    If so, I don't understand why I can't carry this information into part (b) of the question.

    My reasoning for the 2nd part was that 1/2kx^2 = 1/2kA^2

    Therefore, x = 0.13.

    I then plugged this into the x(t) equation from above:

    .13 = .184cos(28.28t - .616)

    This gave a t of .0496 seconds.

    However, my professor did not include a phase angle of -.616 in his answer; his phase angle is 0. What am I doing wrong?!?!
  2. jcsd
  3. Dec 22, 2011 #2

    Doc Al

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    Staff: Mentor

    Looks good to me.

    Sure you can.

    I assume you had a factor of 1/2 in there.

    For one thing, note that they want the time measured from the right hand extreme--not from the starting point from part a. You could use your original equation, but then you'd have to find two times. Or you can just start with a new equation, starting with t = 0 at the extreme (no phase factor needed)--I assume that's what your professor did.
  4. Dec 22, 2011 #3
    Hi Doc Al,

    Thanks a million for replying! I have my final in a few hours! Quick followup -- can you explain why you don't need a phase angle if you begin at the extreme? I thought you didn't need a phase angle if you started at the equilibrium position.

  5. Dec 22, 2011 #4

    Doc Al

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    It just depends on whether you represent the position using sine or cosine, which is arbitrary.

    If you use x = Acos(ωt + θ), then θ = 0 if you start the clock at the maximum position x = A. (But you'll need a phase factor if you start the clock at equilibrium, x = 0.)

    Just the opposite if you use x = Asin(ωt + θ).
  6. Dec 22, 2011 #5
    After I realized that the phase angle would not be the same as the initial phase angle, I tried researching phase angles and found a website that said, "When the system is displaced and given no initial velocity, then the system starts out its initial position equal to its amplitude. In this case the phase angle has to be +p/2 or -p/2." However, this also doesn't equal a phase angle of 0, which both you and the prof agree is the true phase angle. It's a head scratcher for me.

    That website is http://faculty.wwu.edu/vawter/PhysicsNet/Topics/SHM/PhaseAngle.html [Broken].
    Last edited by a moderator: May 5, 2017
  7. Dec 22, 2011 #6
    Oh, ok. I think that website was using sin to express the position. I think it's starting to make sense.
    So if you begin at t=0 at the equilibrium position and want to express the location using the cosine function, then the phase angle would be pi/2, but it it's 0 when you begin at the amplitude.
  8. Dec 22, 2011 #7

    Doc Al

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    See my previous post (we posted simultaneously). That site uses the x = Asin(ωt + θ) version; your prof uses the cosine version.
  9. Dec 22, 2011 #8

    Doc Al

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