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Finding angular frewuency of the oscillator

  1. Jun 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A simple harmonic oscillator is displaced 5.00 cm from equilibrium and released at t=0s. Its position at t=1.5 s is 2.00 cm. What is the angular frequency of the oscillator?

    The book says the answer is 0.773.

    2. Relevant equations
    F= 1/T = w/ (2*pi)
    (theta,f) - (theta, i) = (1/2)(wf+wi)(t)

    3. The attempt at a solution
    My thought was to solve for w and use my answer to solve for frequency.
    (theta,f) - (theta, i) = (1/2)(wf+wi)(t)
    (.02m - .05m) = (1/2)(wf+wi)(1.5)
    w =.04 and .04/(2*pi) is not equal to 0.773

    thank you for any help
  2. jcsd
  3. Jun 15, 2008 #2


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    Homework Helper
    Gold Member

    The equation you are using is for rotational motion, and the theta variables stand for angular quantities, not distance quantities. You are going to need an equation that describes the motion of a simple harmonic oscillator. Do you remember it?

    HINT: There will be a trig function involved in the one your looking for.
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