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Simple harmonic motion for block of mass

  1. May 8, 2005 #1
    the problem gives
    x(t)= A cos( (k/m)^(1/2) * t)

    A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. View Figure . Assume that the +x direction is to the right.

    The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero initial velocity.

    and asks

    At what time t_1 does the block come back to its original position for the first time?
    Express your answer in terms of k and m.

    So when t=0
    cos( (k/m)^(1/2) * 0) = 1

    so at t_1
    cos( (k/m)^(1/2) * t_1) = 1

    (k/m)^(1/2) * t_1 = 2pi

    so t_1 = 2pi(m/k)^(1/2)

    but when i submit it it says it's off by a multiplicative factor!!!!!

    what's wrong with that?

  2. jcsd
  3. May 8, 2005 #2


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    Homework Helper

    That's funky. [itex]T=2\pi\sqrt{\frac{m}{k}}[/itex] is correct as you can easily check.
  4. May 8, 2005 #3
    sigh. apparently the original position it was refering to was the equalibrium position, not the position it was initially pulled to.
    So cos has to be solved to equal 0.

    cos( (k/m)^(1/2) * t) = 0

    t_1 = (pi/2)(m/k)^(1/2)

    I think the wording is poorly chosen.
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