# Thread: Finding the period (T) of a pendulum on Mars

1. ## Finding the period (T) of a pendulum on Mars

Word Problem: The first astronauts to visit Mars are each allowed to take along some personal items to remind them of home. One astronaut takes along a grandfather clock, which, on earth, has a pendulum that takes 1 second per swing, each swing corresponding to one tick of the clock. When the clock is set up on Mars, will it run fast or slow?

The answer shows you that the clock on Mars will have a slower period on Mars (T = 2.45 sec.) than on Earth (T = 1 sec.). There's a section called "My Work Version #1", where I tried to solve the problem. The formula is pretty basic. Plug in acceleration of gravity (9.8 m/s^2) and the length of the string (1 m). With that approach I got T = 3.26.

On the section called "My Work Version #2", I pretty much set T earth and T mars and tried to solve for T mars. At the end the value that I got under the square root is 0.166/1, but the answer key that they had is 1/0.166.

So the bottom line is, I have no idea how they got the value of 0.166 how they solved for the period, T, of Mars.

2. ## Re: Finding the period (T) of a pendulum on Mars

I don't understand the step "Set each equal to each other and solve for T_mars." If a=b and c=d that does not mean that ab =cd. I think what you were thinking of is that if a=b and c = d then ad = bc. Hence what you should have is:

$T_{mars} 2 \pi \sqrt {L/g} = T_{earth} 2 \pi \sqrt { L/(0.166g)}$

3. ## Re: Finding the period (T) of a pendulum on Mars

Originally Posted by ChipB
I don't understand the step "Set each equal to each other and solve for T_mars." If a=b and c=d that does not mean that ab =cd. I think what you were thinking of is that if a=b and c = d then ad = bc. Hence what you should have is:

$T_{mars} 2 \pi \sqrt {L/g} = T_{earth} 2 \pi \sqrt { L/(0.166g)}$
I don't understand where 0.166 came from? How did they come up with that number?

4. ## Re: Finding the period (T) of a pendulum on Mars

Not sure. If I had to guess, it’s a mistake. I bet whoever came up with the answer using g_mars = 0.166 times g_earth was confused and used the gravitational acceleration of the moon by mistake, instead of Mars. It turns out that g_moon really is 0.166 times g_earth.