Finding angular frewuency of the oscillator

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To find the angular frequency of a simple harmonic oscillator displaced 5.00 cm and observed at 2.00 cm after 1.5 seconds, the correct approach involves using the appropriate equations for harmonic motion rather than rotational motion. The initial attempt incorrectly applied rotational equations, leading to an incorrect calculation of angular frequency. The discussion emphasizes the need to incorporate a trigonometric function to accurately describe the oscillator's motion. The correct angular frequency, as stated in the book, is 0.773. Understanding the distinction between the equations for rotational and harmonic motion is crucial for solving such problems.
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Homework Statement


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.


Homework Equations


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

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
 
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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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