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Simple oscillation problem - help needed

  1. Jun 26, 2004 #1
    "An object is moving along the x-axis in simple harmonic motion. It starts from its equilibrium position which is at the origin at t=0 and is moving to the right. The amplitude of its motion is 2 m and its frequency is 3 hz. (1) Determine the expression for the objects displacement. (2) Where is the object located at t=1.0 s? (3) What is the maximum value of acceleration?

    For 1, using the the general SHM expression, I got x=(2 m) cos (6pi(t) + pi/2)

    But for (2), plugging in 1 sec., i get 2 m. After 1 sec, shouldn't the object be back at the origin?

    For (3), i would really appreciate a hint.

    regards
     
  2. jcsd
  3. Jun 26, 2004 #2

    HallsofIvy

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    The amplitude of motion is 2 so the 2 multiplying cosine is correct. The intial position is t=0, x= 0 so the pi/2 gives x(0)= 2*cos(pi/2)= 0 is correct (I would have been inclined to use sin() but probably the formula you were given just uses cos()). The frequency is 3 Hz= 3 cycles per second = 6pi radians per second: Okay, (6pi)t gives you that.

    Now, HOW did you get 2 m. by plugging in t= 1? 6pi(1)= pi/2= 13pi/2 and
    cos(13pi/2)= 0. Did you forget to add that pi/2?

    Since, as you correctly have, x= 2cos(6pi t+ pi/2), v= dx/dt= -12pi sin(6pi t+ pi/2) and a= dv/dt= -72pi2cos(6pi t+ pi/2). The maximum value of that is its amplitude: 72 pi2.
     
  4. Jun 26, 2004 #3
    Rereading my question, i noticed that i typed the wrong sign in the displacement equation. I meant to say x=(2 m) cos (6pi(t) - pi/2) ["minus" before pi/2]. This now gives 2 m with t = 0. Since the object is first moving to the right (positive), shouldn't the phase angle be -p/2?
     
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