Finding angular frewuency of the oscillator

[crazyog]

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

 

[G01]

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.

 

[Moetasim]

I would first check the units being used in the problem. The given displacement and time are in centimeters and seconds, respectively, while the angular frequency is typically measured in radians per second. This could explain the discrepancy between the calculated value and the given answer of 0.773.

To solve for the angular frequency, we can use the equation for simple harmonic motion: x(t) = A*cos(wt + phi), where A is the amplitude, w is the angular frequency, and phi is the phase angle. In this case, we are given the initial displacement (A=5cm) and the position at t=1.5s (x=2cm). Plugging in these values, we get:

2cm = 5cm*cos(w*1.5s + phi)

Solving for w, we get w = 1.71 rad/s. This is equivalent to 0.272 Hz, which is not the same as the given answer of 0.773.

To find the correct value of angular frequency, we can use the equation F= 1/T = w/ (2*pi) and solve for w:

w = 2*pi/T = 2*pi/1.5s = 4.19 rad/s

This is equivalent to 0.666 Hz, which is much closer to the given answer of 0.773. It is possible that the given answer was rounded or approximated. In any case, it is always important to check units and use the appropriate equations when solving problems in science.