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Need help: occillatory motion.

  1. Oct 21, 2004 #1
    ok here's the problem:

    When a mass of 0.2kg is suspended from a spring, it stretches 0.04m. The mass is pulled down an additional distance 0.1m from its equilibrium position and released.


    how long after being released is the posiiton of the mass equal to 0.05m
  2. jcsd
  3. Oct 21, 2004 #2


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    Use the first statement to find the spring constant. Divide it by the mass and find its square root. That gives you the angular frequency and you should be able to take it from there.
    Last edited: Oct 22, 2004
  4. Oct 21, 2004 #3

    i've found all of those. there are several parts to this question. i've done them up to this question which i'm stuck on.

    the constant k=49
    period T= .4
    frequency f= 2.5
  5. Oct 22, 2004 #4


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    If you have those then you should know that, taking x to be the distance from the equilibrium position, positive above, negative below, x= -0.1 cos(2pi t/.4). To answer the question "how long after being released is the posiiton of the mass equal to 0.05m", set x= 0.05 and solve for t.

    By the way, do you understand that saying "period T= .4 frequency f= 2.5" doesn't mean anything unless you give the units?
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