Question

A pendulum clock is brought to Mars. How much does the bob move on Mars as compared to the Earth?
A. It keeps time just as on Earth
B. The effect depends on the amplitude of the swing
C. It runs slower
D. It runs faster

**not sure! thank you :)

Answer 1

HMM 'how much does the bob move' is not really a well defined question.

It is very poorly phrased but looking at the answers it means 'what is th eperiod of the pendulum compared to the period of the same pendulum on earth.

the gravity on mares is LESS than that on earth (for this question you do NOT need to know any more about gravity)

The period (T) of a pendulum is given by
\[T= 2\pi \sqrt{\frac{ l }{ g}}\]

where l is the length (the same on mars as earth) and g is the accelertaiton of gravity (which is LESS on Mars than Earth)

You can derive the answer by reading the info I have given, and looking at the equation

Answer 2

http://blogs.discovermagazine.com/crux/2014/09/08/where-build-off-world-colonies/#.VLRMb9LF-4k
i hope this helps

Answer 3

okay! so would it be like this?
2pi*sq.rt. l/9.8 ?

is that what g would be? and not sure about l :/

Answer 4

as I said - you do NOT need to know the value of g on mars - just that it is LESS than that on earth

Answer 5

ohh okay so what should happen next?

Answer 6

if the value for g is less in the equation - what happens to the value of T?

Answer 7

it will be less than the period of the earth?

Answer 8

why do you say that?

the period varies with sqrt(1/g) so if g is less what happens ot T?

Answer 9

i am not sure ;(

Answer 10

so let's re-arrange my equation:

\[\frac{ T ^{2} }{2\pi}=\frac{ l }{ g }\]

What happens to the value on the right hand side as g becomes smaller?

Answer 11

oops

\[(\frac{ T }{ 2\pi })^2= \frac{ l }{ g }\]
What happens to the value on the right hand side as g becomes smaller?

Answer 12

the left side becomes larger?

Answer 13

you have a question mark in your answer
are you sure that your answer is correct?

Answer 14

not very sure :/

Answer 15

we are looking at an 'inverse relationship'

the period is related to 1/g

you MUST know what happens to 1/x as x gets smaller .....