Finding the period (T) of a pendulum on Mars

Nov 2012
744
2
Hawaii
[FONT=&amp]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?[/FONT]

[FONT=&amp]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.[/FONT]

[FONT=&amp]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.[/FONT]

[FONT=&amp]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.


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Attachments

ChipB

MHF Helper
Jun 2014
309
127
NJ
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:

$T_{mars} 2 \pi \sqrt {L/g} = T_{earth} 2 \pi \sqrt { L/(0.166g)}$
 
Last edited:
Nov 2012
744
2
Hawaii
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?
 

ChipB

MHF Helper
Jun 2014
309
127
NJ
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.