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Finding time in a simple harmonic motion

  1. Jan 17, 2016 #1
    1. The problem statement, all variables and given/known data
    image.jpg
    I seem to have problems finding time in SHM.
    To find the time after projection, I know that either x=asinωt or x=acosωt needs to be used, so since R is projected away from O, it means that it is moving towards the equilibrium position, therefore I used x=acosωt to find the time.
    2. Relevant equations
    x=asinωt x=acosωt

    3. The attempt at a solution
    x=acosωt
    -L/4 = 3L/4 cos (√2g/L)t
    t= (√L/2g) cos-1(-1/3)

    But, the correct answer should be
    t=(√L/2g) sin-1(1/3)

    Why can't cosine be used? Since it is moving towards the equilibrium position?

    Also, for the equations x=asin(ωt+α) or x=acos(ωt+α)
    I am not sure when is α needed to take into consideration.
    So far, I never encountered any questions needing to find α, but I would like to know when is α needed.

    Thanks very much in advance!
     
  2. jcsd
  3. Jan 17, 2016 #2
    Drawing a picture always helps.
     
  4. Jan 18, 2016 #3
    Yeah I drew a diagram image.jpg

    But I am confused on when to use sine or cosine in the displacement formula..
     
  5. Jan 18, 2016 #4
    In choosing sin or cos, do you want the displacement to be a maximum or a minimum at t = 0?
    Either to can be converted to the other by including a phase angle of pi / 2.
    Here it appears that the displacement is decreasing at t = 0 so it may be that
    the cos might be a starting point.
     
  6. Jan 18, 2016 #5
    Sin when displacement is a minimum and cos when displacement is a maximum at t=0.

    But when t=0, the displacement is -L/4 which is not the amplitude (3L/4), and since it's moving towards the equilibrium position, why not cos?
     
  7. Jan 18, 2016 #6
    Isn't mass m being projected from x = - L / 4.
    It appears from the wording that L = constant and that x is the displacement from the new equilibrium position at 5 L / 4.
    How did you get 3 L / 4 for the amplitude?
     
  8. Jan 19, 2016 #7
    Yeah
    v22(a2-x2)
    x is the displacement down the slope from the new equilibrium position which is at 5L/4 from O,
    When x= -L/4, v=√gL
    So, gL = 2g/L (a2-(-L/4)2)
    ∴ a=3L/4

    But, if the particle is displaced at x=-L/4, then it is moving towards the equilibrium position, but it seems that cosine can't be used but sine .
    Or has the equilibrium position changed to a distance L from O? Is that why it is sine?
     
  9. Jan 19, 2016 #8
    Isn't the displacement increasing at t = 0?
    Also, it appears that a phase angle is involved since the displacement
    is not zero at t = 0.
     
  10. Jan 19, 2016 #9
    Sorry, I'm a little confused.. from which point do you define your displacement from? Is it at a distance L or 5L/4 from O?
     
  11. Jan 20, 2016 #10
    The problem states that the new equilibrium point is 5 L / 4 and the ring is
    moved to distance L so the initial displacement must be (L - 5 L / 4) = - L / 4.
    The problem also states that the general position is OR = 5 L / 4 + x which
    means that x is the displacement variable and at t = 0, x = - L / 4.
    Since the ring is projected away from O, the displacement variable is increasing towards
    positive values.
    Choosing the sine function fits this description since the sine is also increasing at t = 0.
    Note that the speed is also increasing which is described by the cosine function, and the acceleration
    in this case would be described by the negative sin function since the restoring force will be
    decreasing as the ring moves towards equilibrium at 5 L / 4 as measured from O.
    The cosine function can be increasing at t = 0 if the phase angle is negative.
    Examining the initial conditions will help in choosing sin or cos to describe the resulting motion.
     
  12. Jan 22, 2016 #11
    Thanks, I've got it
     
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