Question

The moon's gravity is one-sixth that of the earth. What is the period of a 0.25 m long pendulum on the moon?

0.32 s, 1.0 s, 2.5 s, or 3.5 s

**Not sure how to calculate?

Answer 1

Physics?

Answer 2

yes:(

Answer 3

In order to calculate it you need to know the equation for the period of a pendulum

Answer 4

it's 2pi*square root l/g ?

Answer 5

this is\[T=2\pi \sqrt{\frac{ l }{ g }}\]

Answer 6

ok so we are given what l is in the question and we are told that g of the moon is 1/6 that of earth

Answer 7

do you see how you would calculate this now?

Answer 8

2pi* sqrt(.25/?)

is it 9.8 for where i put the"?" ?

Answer 9

would it be 1.63?

Answer 10

sorry because 1/6 of 9.8 right?

Answer 11

so then 2pi*sq.rt.(.25/1.63)=2.459=2.46=2.5 s ? is that correct? :)

Answer 12

@recon14193 ?

Answer 13

it's not 1/6

because it is proportional to 1/sqrt g

so if gravity is 1/6 then th eperiod I increase by 1/sqrt 1/6

Answer 14

oh so it would be 2pi*sqrt(.25/1.667) ?

Answer 15

in fact - it doesn't ask for relative period compared to earth

so ig gmoon = 9.81/6 m/s^2 just put that into the formula to get T

Answer 16

g earth is NOT 10 - it is a poor approximation.

Unless you are told otherwise you should use 9.81m/s^2

Answer 17

ohh so how would i calculate it?

Answer 18

you have the correct formula above:

oh so it would be 2pi*sqrt(.25/1.667) ?

just put 9.81/6 instead of 1.6667 though

Answer 19

@MrNood the question states it is 1/6 while in reality it may not be 1/6 it is a close approximation same thing with 10. 10 can be an approximation he is told to use as part of his class for solving this problem

Answer 20

but yes @iheartfood you equation above would be correct so plug that in and solve for it

Answer 21

okay, so 2.196?

Answer 22

I'm getting something different when I plug it into my calculator. How are you typing it in?

Answer 23

@recon14193
there is nothing in the question to say 'take g =10' but there IS the statement that g=1/6

I don't really care what he uses - but he should be aware that 10m/s^2 is NOT the value acceptec for g

g varies by about 1% across the earth's surface but 9.806... is accepted as the mean value

Answer 24

i just did 2pi*sqrt(0.25/(9.81/6))
?
and got 2.196?

but earlier i did sq rt (.25/(1/6)) and got 2.5?

Answer 25

for what it is worth - I get 2.46, using g= 9.81

Answer 26

alright when you do (.25/(9.81/6)) what do you get?

Answer 27

2.45? so it should be 2.5 s final answer?

Answer 28

I tried it using g as 10 and I get 2.43 which is close to 2.46 but my problem is that 2.196 is not close enough for rounding error

Answer 29

ah okay, sorry i had a typo with the 2.196 earlier!

Answer 30

yes it would be 2.5 but I'm just trying to see if we can figure out how you are getting 2.19 so that a similar error doesnt happen on a test or something else

Answer 31

oh i plugged something wrong because when i tried it again, i got 2.45! :)

Answer 32

so error on my part :(

Answer 33

alright that's good to hear glad we could help!

Answer 34

thanks so much for helping me @recon14193 :)