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Simple harmonic motion question

  1. Jun 30, 2007 #1
    1. The problem statement, all variables and given/known data
    A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation

    A=Ai * sin(wt+ (pi/3)) ,

    Where w = pi radians per second, t is in seconds, and Ai = 2.4m.
    What is the phase of motion at t = 9.4 seconds? Answer in units of radians.

    2. Relevant equations

    A is the amplitude.
    Ai is the initial amplitude.
    w is actually "omega" but I didn't know how to enter that. That is the given angular velocity in rad/s.
    Pi is 3.14.....

    3. The attempt at a solution
    I honesty don't know where to start. I just plugged into the equation with the given data and got

    -1.78355 meters.

    The answer wants radians. Also, it asks for the "phase of motion". The answer I got is just the final amplitude at the given time.

    Any help would be appreciated.
  2. jcsd
  3. Jun 30, 2007 #2


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    The phase is simply the argument of the sine function, namely the [itex] \omega t + \frac{\pi}{3} [/itex] That's all there is to it.
  4. Jun 30, 2007 #3


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    I think the "phase of motion" is the argument of the sine function (=ωt+φ)

    So at time t=0, the phase of motion would just be the phase constant (in your problem, π/3). And I think your answer should be between 0 and 2π, so if you compute something larger than 2π, you should subtract multiples of 2π until you are in that range.

    E.T.A.: Looks like I was too slow...and Greek letters don't work the way they used to...
  5. Jun 30, 2007 #4

    Doc Al

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    You can trace the SHM motion through [itex]2\pi[/itex] radians of "phase" as the body moves past the origin, goes to maximum + displacement, returns to the origin, goes to maximum - displacement, and then back where it started. When the body crosses the origin, consider its phase to be 0; when it reaches maximum amplitude, phase = [itex]\pi/2[/itex]; back to the origin, phase = [itex]\pi[/itex]. Etc.

    Hint: Consider the argument of the sine function.

    (Looks like nrqed and jamesrc both beat me to it!)
  6. Jun 30, 2007 #5
    Thanks for the help. I got it.
    Last edited: Jun 30, 2007
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