# Simple Harmonic Motion Question

• zstraught

#### zstraught

Homework Statement
A copper rod (length=2.0 m, radius=3.0×10−3 m) hangs down from the ceiling. A 9.0-kg object is attached to the lower end of the rod. The rod acts as a “spring,” and the object oscillates vertically with a small amplitude. Ignoring the rod's mass, find the frequency f of the simple harmonic motion.
Relevant Equations
F=-kx
w = sqrt(k/m)
T = 2pi x r sqrt(m/k)
First I use young's modulus to solve for delta y. I get 5.67x10 -5.
I am not sure what to do after this, but this is my attempt.

Next I do T = 2delta y sqrt(m/k) (I am not sure if I am supposed to put 2 delta y)

Solving for f, i get f = 1/(2delta y sqrt(m/k))

F = kx, mg = kx, m = kx/g

Substituting in the equation above, I get f = 1/(2deltay sqrt(x/g))

Plugging in the numbers, I get a very large number around 3 million, which is totally wrong.

First I use young's modulus to solve for delta y. I get 5.67x10 -5.
That just gives you the equilibrium position. Consider oscillations around there.