Pendulum homework problem

1. Nov 21, 2005

A certain pendulum clock that works perfectly on Earth is taken to the moon, where g=1.63m/s^2. The clock is started at 12:00:00AM and runs for one Earth day (24 hours). What will be the reading for the hours? Answer in units of h.

I'm stumped. I know that the period for the pendulum on earth is equal to 2π√(L/9.8) and on the moon it is 2π√(L/1.63).

2. Nov 22, 2005

Ouabache

This is a nice question. It gets to think about the situation and how the concepts apply. Here is something to get you started.
Would you expect the pendulum clock to run faster, slower or the same rate on the moon? (Hint: how do their periods compare?)

3. Nov 22, 2005

The period on the moon will be longer so the clock will run slower. The time will be behind compared to the clock on earth.

4. Nov 22, 2005

mathphys

Suppose you choose the lenght of the pendulum so that its period on earth is one second (or one minute, or one hour, whatever you want).

What would be its period on the moon? When the clock on earth reads one hour after 12:00 how much it will read in the moon if they started synchonized at 12:00?

5. Nov 23, 2005

Ouabache

Well the period is the time it takes for each swing (out and back). If you apply the idea like mathphys recommends; let the period on earth = 1 sec, then using the formula you gave, solve for L
L/g (earth)(m/s^2) = 1 sec^2
L = g(earth)(m/s^2) x 1 sec^2
L = 9.8 m
You know that L is the same on moon, so calculate its corresponding period.. T(moon) = ? You are correct it is longer than T(earth).

Using the same L we chose above, in 24hrs how many periods will occur?
24hrs / T(earth) =?

In 24 earth hours how many moon hours will pass?
24 hrs/ T(earth) = A / T(moon) [equation (i)]
(don't forget to keep your units consistent,e.g. if T is in seconds, need to
convert 24 hours to seconds before dividing)

Now, just solve for A, the time passed on the moon.
(if you changed units to seconds, don't forget to change them back to hours for your final answer).

As a double check of your math, instead of having T(earth) = 1sec, choose 1 minute (60sec). Solve for L as above. Plug L into equation for a period on the moon, an so on as we just did above...If you did your math correctly (and kept your units consistent), you will come out with the same number of hours for time on the moon. (I did)..

If you get stuck.. just show us what you tried and we'll get you through it..

{Hint: if you take equation (i) and solve for A directly
A = [T(moon)/T(earth)] x 24hrs , and substitute the formula you have, for the period of each.. You should see that L cancels out (meaning L does not matter). Again (keeping your units consistent), you will come out to the same number of hours passed on the moon as the two cases above. }

Last edited: Nov 23, 2005