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SHM Question

  1. Nov 8, 2003 #1
    A particle vibrates in SHM with the period of 2.4 s and the amplitude of .1 m.

    a) How long is required for the particle to move from one end of its path to the nearest point .05 m from the equilibrium position?
    b) What is the particle's speed at this position?

    for a) this is what I did
    x=Acos(ωt), 5=10cos(ωt), ω=2pi/T=2pi/2.4=2.62
    t=arccos(.5)/2.62= 0.4 s

    I'm a little confused on part b.

    Please see if I did a correctly, and show me how to answer b.

  2. jcsd
  3. Nov 8, 2003 #2
    Part (a) looks fine. For part (b), you have to take the time derivative of position to get the velocity,

    v = dx/dt = -Aω sin(ωt)
  4. Nov 8, 2003 #3
    so v=-(.1)(2.62)(sin(2.62*.4))= .00479 m/s

    Is that correct?
  5. Nov 8, 2003 #4
    Almost; you have to switch your calculator to radians to take the sine.
  6. Nov 8, 2003 #5
    Okay, so it should be .227 m/s?
  7. Nov 8, 2003 #6
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