 18
 0
1. Homework Statement
A simple pendulum and a massspring system have the same oscillation frequency f at the surface of the Earth. The pendulum and the massspring system are taken down a mine where the acceleration due to gravity is less than at the surface. What is the change in the frequency of the simple pendulum and the change in the frequency of the mass spring system
Simple Pendulum  MassSpring
A f increases  f decreases
B f decreases  f decreases
C f increases  f stays unchanged
D f decreases  f stays unchanged
2. Homework Equations
Potential Energy = mgh
Kinetic Energy = 1/2mv^2
3. The Attempt at a Solution
I thought that because the acceleration due to gravity was now lower that the new maximum potential energy would now be lower as it is dependent on the value of g which in now lower, and a lower maximum potential energy means a lower maximum kinetic energy. Therefore if the pendulum has to travel the same path but at a lower maximum speed then it should take longer so the frequency should be less. For the mass spring I assumed that the frequency would be unchanged as the lower speed it will gain travelling downwards is counteracted by the fact it loses speed slower on the way up so I would have given D as my answer. However the mark scheme gives C as the answer. Can someone explain why the frequency increases when the value of g is lowered?
Thanks
A simple pendulum and a massspring system have the same oscillation frequency f at the surface of the Earth. The pendulum and the massspring system are taken down a mine where the acceleration due to gravity is less than at the surface. What is the change in the frequency of the simple pendulum and the change in the frequency of the mass spring system
Simple Pendulum  MassSpring
A f increases  f decreases
B f decreases  f decreases
C f increases  f stays unchanged
D f decreases  f stays unchanged
2. Homework Equations
Potential Energy = mgh
Kinetic Energy = 1/2mv^2
3. The Attempt at a Solution
I thought that because the acceleration due to gravity was now lower that the new maximum potential energy would now be lower as it is dependent on the value of g which in now lower, and a lower maximum potential energy means a lower maximum kinetic energy. Therefore if the pendulum has to travel the same path but at a lower maximum speed then it should take longer so the frequency should be less. For the mass spring I assumed that the frequency would be unchanged as the lower speed it will gain travelling downwards is counteracted by the fact it loses speed slower on the way up so I would have given D as my answer. However the mark scheme gives C as the answer. Can someone explain why the frequency increases when the value of g is lowered?
Thanks
Attachments

14.7 KB Views: 1,510
Last edited: