# Torsion Pendulum Oscillation (SHM)

## 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?

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!

tiny-tim
Homework Helper
Hi SuperCass!

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

Hint: by how much is I reduced?

(1-.76)² ?
So since the formula is T=2π√(I/k), would I multiply the original T by (1-.76)?

tiny-tim
Homework Helper
No, I is proportional to ML2, and both M and L are reduced.

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?

.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!!