Simple Harmonic Motion on a Uniform Meter Stick

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Homework Help Overview

The problem involves a uniform meter stick pivoted at one end and held horizontal by a spring. The inquiry focuses on determining the frequency of oscillation when the stick is displaced slightly. The subject area pertains to simple harmonic motion and rotational dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty in starting the problem and seeks initial guidance. Some participants suggest drawing a force diagram to analyze the forces involved in the vertical displacement. Others discuss setting up torque equations related to the equilibrium position and the effects of displacement.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some have provided partial setups for the equations of motion, while others are questioning the next steps in their reasoning. There is no explicit consensus yet, but various approaches are being considered.

Contextual Notes

Participants are working under the assumption that small angle approximations apply, and there is a focus on the relationship between torque, force, and angular acceleration. The original poster has referenced a textbook answer, which may influence the direction of the discussion.

NathanLeduc1
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Homework Statement


A uniform meter stick of mass M is pivoted on a hinge at one end and held horizontal by a spring with spring constant k attached at the other end. If the stick oscillates up and down slightly, what is its frequency?


Homework Equations


τ=rFsinθ
f=(1/2π)√(k/m)
F=kx
x=Acos(ωt)

The Attempt at a Solution


I'm really not sure how to get started on this one. If you could just provide me with a little start, I might be able to figure it out. Thanks.

The answer, according to the textbook, is (1/2π)sqrt(3k/m)
 
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Hi NathanLeduc1! :smile:
NathanLeduc1 said:
If you could just provide me with a little start, I might be able to figure it out.

Draw a force diagram for a small vertical displacement x, and find the force as a function of x. :wink:

(assume sinx = x)
 
Ok, so I set up a force diagram and did the following work but I'm stuck again...

At equilibrium:
Ʃτ=Kxol-mg(l/2)=0

After it's been stretched:
Ʃτ=K(x+xo)-mg(l/2)=Iα

This then simplifies to:
Iα=kxol

I wrote α as the second derivative of θ with respect to time but now I'm stuck. Where should I go from here? Thanks.
 
(just got up :zzz:)
NathanLeduc1 said:
Iα=kxol

I wrote α as the second derivative of θ with respect to time but now I'm stuck. Where should I go from here? Thanks.

α = x''/l :wink:
 

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