Energy quantization of oscillator

[w3390]

Homework Statement

A simple pendulum has a length equal to 0.6 m and has a bob that has a mass equal to 0.5 kg. The energy of this oscillator is quantized, and the allowed values of energy are given by En = (n + 1/2)hf0, where n is an integer and f0 is the frequency of the pendulum. Find n if the angular amplitude is 1.0°.

Homework Equations

En = (n + 1/2)hf0
f0=w(omega)/(2pi)

The Attempt at a Solution

I'm stuck at the beginning of finding out the total energy of the pendulum. I think I might have to use the equation U(x)=(1/2)m(w0)^2x^2 to find the energy but I get stuck with unknown values. My biggest question is what does the angular amplitude of 1 degree get used for. I don't think it can just be plugged into the equation as is for an amplitude. Any help on how to approach the problem would be much appreciated.

 

[kuruman]

Express the potential energy in terms of θ then expand it for small values of θ. The total energy of the oscillator is the potential energy at maximum displacement. That's where the 1 degree comes in.

 

[w3390]

Okay, so I'm using the equation U=(1/2)lw^2(theta)^2. I found this equation and I think the l in it stands for the length of the pendulum. So after doing the calculations I get the potential energy to be .0855 but my units are confusing me. Should theta be in radians or degrees and is this in fact what l stands for?

 

[kuruman]

Yes, on both accounts. The angle must be expressed in radians and l is the length of the pendulum in meters. With these units, the energy should come out in Joules.

 

[w3390]

When I do the calculations, I get U=.00149 m/s^2. I do not know why I am missing units.

 

[kuruman]

Show me the numbers you put in and your answer. The correct expression is

U = mglθ²/2.

 

[w3390]

I'm confused where you got mgl from, but using your formula I was able to get the correct answer

 

[kuruman]

Derive (or look up) the potential energy for a pendulum referenced to the lowest point of the motion (equilibrium position) in terms of the angle θ. Then use the small angle approximation for the cosine

$$cos\theta\approx1-\frac{\theta^{2}}{2}.$$