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Homework Help: Kinematics: simple harmonic motion

  1. Aug 18, 2010 #1
    1. The problem statement, all variables and given/known data
    The motion of a particle is given by x=(6.0m) cos(0.586t +0.72)
    a) find amplitude
    b) find period
    c) find the first time for t>0 when v=0
    d) find the max acceleration
    e) find the phase at time t = 1.38s

    2. Relevant equations
    T = 2pie/omega

    v = -A omega sin omega*t

    a = -A omega^2 cos omega*t

    3. The attempt at a solution
    I got the first 2.
    amplitude = 6.0m

    period = 2pie/omega = 10.7s

    i'm not sure about the next 3:
    for c) do i use the second equation above and make v = 0 and just solve for t?

    for d) do i use the third equation? but what do i use for t?

    e) Phase is looking for the angle, right?

    Can someone please help me with the last 3?

  2. jcsd
  3. Aug 18, 2010 #2


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    c)To find time when v = 0, find dx/dt and equate it to zero.

    d) max. acceleration = aω

    e) find x when t = 1.38 s. Then phase φ = 2πx/λ
  4. Aug 20, 2010 #3
    to find the phase at t = 1.38s, can't we just just multiply omega and time? Theta = omega * t.

  5. Aug 20, 2010 #4
    I'm stuck. Do I have to put into account the phase constant of 0.72??

    For part c i have the equation, v = -3.516 sin(0.586t + 0.72). I set this equal to 0, but i can't figure out how to solve for t??
  6. Aug 20, 2010 #5
    About the phase: yes, you certainly have to take into account the initial phase of 0.72 when t=0.

    A hint for part c: when is sin(x)=0?
  7. Aug 20, 2010 #6


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    In one period (T) change of phase is 2π. Ιn t seconds the change of phase is
    φ = 2πt/T + φo
  8. Aug 20, 2010 #7
    how did you get that equation?
  9. Aug 20, 2010 #8
    for part c, can i do it this way?

    v = -A*omega*sin omega*t + delta
    0 = -A*omega*sin omega*t + delta
    -delta/ -A*omega = sin omega*t

    but i still can't get t...:confused:
  10. Aug 20, 2010 #9
    Ok... sin(x) = 0 when x=0 or some multiple of pi.

    If v = -3.516 sin(0.586t + 0.72) = 0 then
    (0.586t + 0.72) = pi

    Easy to solve now, I hope.
  11. Aug 20, 2010 #10
    for part e)

    in the text it says, theta(phase) = omega * t

    so could i use this and solve for the phase in radians? what i did:
    theta = omega*t
    = 2 pie f *t
    = 2 pie (1/T) * t

    where T is found in part b and t is given, t = 1.38s.

    can i do that??
  12. Aug 21, 2010 #11
    sorry, this might be a stupid question...why pi?
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