Torsion Pendulum Oscillation (SHM)

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Homework Help Overview

The problem involves a torsion pendulum consisting of a uniform meter stick suspended from a wire, which oscillates with a given period. After modifying the length of the stick, the question seeks to determine the new period of oscillation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the moment of inertia and the length of the stick, with some attempting to derive the new period based on changes in mass and length. Questions arise regarding how to appropriately factor these changes into the period formula.

Discussion Status

Participants are actively engaging with the problem, exploring the implications of the uniformity of the rod and how it affects mass and moment of inertia. Some guidance has been provided regarding the proportional relationships involved, and a productive direction is emerging as participants clarify their understanding of the relationships between length, mass, and period.

Contextual Notes

There is an emphasis on the uniformity of the rod and its implications for mass distribution, which is central to the discussion. The original period and the new length are key variables under consideration.

SuperCass
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Homework Statement



A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation.

With what period does it now oscillate?

Homework Equations



T=2pi(sqrt(I/K))
I=(1/12)ML^2

The Attempt at a Solution



I tried solving for m in terms of k and then plugging that into a new equation solving for T with the length being .76, but that didn't work. Any suggestions?

Thanks!
 
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Hi SuperCass! :smile:

(have a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)

Hint: by how much is I reduced? :wink:
 
(1-.76)² ?
So since the formula is T=2π√(I/k), would I multiply the original T by (1-.76)?(Thanks for your signature!)
 
No, I is proportional to ML2, and both M and L are reduced. :wink:
 
Remember that you've been told that the rod is uniform, that means that its mass is proportional to its length! If we define a mass per unit length, \mu\equiv \frac{M_0}{L_0} then the mass of a rod of length \ell made of the same material would be..?
 
.76 of the original mass?
 
SuperCass said:
.76 of the original mass?

Exactly! That makes for m \propto \ell which implies I \propto \ell^3
 
Oh!
I got it now!
So I just do the original period times the square root of .76^3!

Thank you all very very much!
 

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