# Does a ball bearing oscillating on a curved track have a constant time

1. Apr 6, 2013

### mrcotton

1. The problem statement, all variables and given/known data

A ball bearing is placed on a track curved in the vertical plane. The ball bearing is released from a position on the track above the lowest point and it then oscillates with SHM. As the amplitude diminishes does the time period for each oscillation stay the same?

2. Relevant equations
For SHM of a simple pendulum the time period stays constant as the amplitude diminishes.
T=2∏√(l/g)

3. The attempt at a solution
Is this true for a ball on a curved track.
Is there a formula for this type of SHM?

2. Apr 6, 2013

### technician

how does it compare with a simple pendulum?
Do you know what that does?

3. Apr 6, 2013

### mrcotton

Thanks for responding technician
I know that for SHM the acceleration produced by the restoring force is proportional to the negative of the displacement. I am familiar with the derivation of the simple pendulum using the horizontal and vertical components of the force arrive at the formula for T.

Would I have to consider the components of the force at a normal to the track and parallel with the track to find the acceleration.
I have just read a derivation for a curved ball on a track
http://www.chaostoy.com/cd/html/pendul_e.htm
and it seems that because the ball is rotating then the derivation is using
Torque = moment of inertial * angular acceleration
It ends up with a time period formula

T=2∏√(2rR/5g)

So if the only variables in the formula are radius of ball (r), radius of track (R) and acceleration due to gravity I assume that means that the displacement from equilibrium has no effect on the time period and as the amplitude diminishes the time period remains constant.

So time period is constant in this oscillatory system?
Thanks
Dave

4. Apr 6, 2013

### haruspex

You are not told the shape of the curve. You are told the only thing that matters: the motion is SHM. That allows you to write a generic equation (no need to derive it from consideration of forces), and that equation implies the period is independent of the amplitude.

5. Apr 7, 2013

### mrcotton

Thanks Haruspex,
So by definition a body oscilating under SHM will have a constant time period

The differential equations are a little beyond me, so can I tell from the solution that the time period is always independent of the amplitude?

6. Apr 7, 2013

### haruspex

Yes. From the A cos (ωt-θ) solution, the values must repeat as t increases by multiples of 2π/ω.
Or, looking at the DE, if x=x(t) is a solution then so is Ax(t). Therefore the amplitude can change without affecting the function in any other way.