# Oscillating piece of wire

1. Jul 12, 2009

1. The problem statement, all variables and given/known data
A uniform piece of wire is bent into an upside down V shape with angle $\theta$ between two legs of length L. The wire is placed over a pivot. Show that the angular frequency of small-amplitude oscillations about the equilibrium is
$$\omega=\sqrt{\frac{3g\cos(\frac{\theta}{2})}{2L}}$$

2. Relevant equations
maybe centre of mass of a rod comes into it
maybe $\omega=\sqrt{\frac{k}{I}}$

3. The attempt at a solution
Don't know how to start this problem

2. Jul 12, 2009

### Staff: Mentor

Hint: Treat it as a physical pendulum. (Which is what it is.)

3. Jul 12, 2009

the shape of the thing makes it a bit tricky because i think i have to find out the centre of mass
perhaps the centre of mass could be found by
$$\cos(\frac{\theta}{2})}=\frac{x}{\frac{L}{2}}$$
where x is the vertical distance as measured from the pivot point
so
$$x=\frac{L}{2}\cos(\frac{\theta}{2})}$$
and let the mass of the wire be m
then
$$I\alpha=mgx\sin\phi$$
$$I=mx^2$$
where
$$\alpha=\frac{d^2\phi}{dt^2}$$
$$mx^2\frac{d^2\phi}{dt^2}=mgx\sin\phi$$
$$x\frac{d^2\phi}{dt^2}=g\sin\phi$$
$$\frac{L}{2}\cos(\frac{\theta}{2})}\frac{d^2\phi}{dt^2}\approx g\phi$$
not sure where a 3 will pop in but thats my best effort

4. Jul 12, 2009

### diazona

It's not true for this case that $I = mx^2$, because it's not a point mass, it's a rod. (Actually it's like two rods)

5. Jul 13, 2009

$$I=\frac{2}{3}mL^2$$
$$I\alpha\approx 2mgx\phi$$
$$\frac{1}{3}L^2\alpha\approx gx\phi$$