Question

Hi,

I'm stuck with the following question: The acceleration due to gravity on Mars is only 3.75m/s^2. You would like to modify an antique grandfather clock so that it would still keep accurate time on Mars. If the length of the pendulum on Earth is l, what should the length of the pendulum be on Mars so that the pendulum oscillates with the same frequency that it does on Earth. [Use g=10m/s]


So far, I assumed the frequency of the clock to be 1 Hz. Therefore, it's length on earth would be 0.25m. I calculated the new length on Mars to be 0.094m. However, that answer is incorrect, so, obviously, I have made a mistake somewhere.

Any help is greatly appreciated.
Thank you.

Answer 1

period=2PI sqrt(L/g)

so the ratio of L/g has to be the same.

Earth:
1sec=2PI sqrt( L/g)
L=g/(4PI^2)=.253
Mars:
L=g'/4PI^2)= 0.095m