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Homework Help: Simple harmonic motion finding velocity trouble

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data

    A block with mass m =6.6 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.26 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.4 m/s. The block oscillates on the spring without friction.

    at time 0.39 s what is the speed of the block
    2. Relevant equations

    v= -Awsin(wt)

    3. The attempt at a solution

    i sub A= 0.716681695
    w= 6.139406135
    t= 0.39 into the equation and get 2.990262698

    my calculator is on rads and i really dont understand what is going wrong in this calculation i found the other variables using the given values
  2. jcsd
  3. Dec 2, 2011 #2
    It looks to me that at t = 0s, the velocity is 4.4m/s.

    But the equation that you are using i.e. v= -Awsin(wt)
    is inconsistent with this because it gives v = 0 at t = 0,
  4. Dec 2, 2011 #3
    which equation did you use?
  5. Dec 2, 2011 #4
    ok so you use cos instead but how do you know which one to use? the sin equation or the cos equation?
  6. Dec 2, 2011 #5
    An equation consistent with the conditions of the problem is v = Awcos(wt)
  7. Dec 2, 2011 #6
    It all depends at which point the time is taken as 0.
    The displacement equation is x = Asinwt if the stopwatch is started at the equilibrium i.e.t = 0 at x = 0.

    But the displacement is x = Acoswt if the stopwatch is started at the positive extreme end of the oscillation i.e. x = A when t = 0.

    Then the corresponding equation for the velocity is used.
  8. Dec 2, 2011 #7
    yes but how would i know which one to use sin or cos? is there a general rule for telling which one should be used? or telling when there should be a phase shift etc?

    thankyou for the help so far!
  9. Dec 2, 2011 #8
    Sometimes the information given in the problem gives at which point in the shm the time t = 0. For example in the problem quoted, the mass is given a speed from the equilibrium position. And then one is asked to find something later on after that. So one can take t = 0 when the particle was at equilibrium i.e when displacement x = 0.

    Otherwise one must make an assumption about at which point t = 0 and must remain consistent with that assumption.
  10. Dec 2, 2011 #9
    ok thankyou so much! although i am confused a little bit because when the box is pushed on the timer starts, is the block not at the equilibrium when it is pushed on in this problem?
  11. Dec 2, 2011 #10
    It WAS at equilibrium BEFORE t = 0 but then AT t = 0 it was given a velocity.
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