# Oscillation problems

1. Apr 27, 2005

### squib

A hoop of radius 0.130 m and mass 0.420 kg is suspended by a point on its perimeter as shown in the figure. If the hoop is allowed to oscillate side to side as a pendulum, what is the period of small oscillations?

T = 2pi (mgl/I)^.5

I assume that l = r, since center of mass is in the middle of the hoop.
I = 3mr^2/2

So T = 2pi(2g/3r)

Also tried with I = mr^2/2 in case I'm misunderstanding the problem and it is oscillating in the other direction, but this did not work either.

Anyone see my error?

Next:
A mass M is suspended from a spring and oscillates with a period of 0.860 s. Each complete oscillation results in an amplitude reduction of a factor of 0.965 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.500 of its initial value.

Tried .965^t = .5, then t = log(.5)/log(.965) but this didn't work... any other suggestions?

2. Apr 27, 2005

### Staff: Mentor

Two problems:
Check your formula for the period of a pendulum (you have terms upside down).
Check your value for the rotational inertia.

3. Apr 27, 2005

### Staff: Mentor

How does amplitude relate to total energy? (It's the energy that decreases by half, not the amplitude.)

4. Apr 27, 2005

### squib

I can't seem to find the right I... isn't 3mr^2/2 right for a ring rotated about a line tangent to the circle?

5. Apr 27, 2005

### pixelized

I tried I = (MR^2)/2 + MR^2 = (3/2)MR^2

T = 2pi * Sqrt(3/2 MR^2/MgR) = 2pi * Sqrt((3/2)r/g))

This didn't work is there something I missed?

6. Apr 27, 2005

### Staff: Mentor

Ah... here's where a picture would help. If the ring rotates through an axis tangent to the circle, then that would be correct. On the other hand, if the ring rotated on an axis perpendicular to its plane, the rotational inertia would be: I = 2mr^2. (That's how I interpreted the problem, but you are the one with the diagram!)