
#1
Dec2512, 01:30 AM

P: 658

1. The problem statement, all variables and given/known data
A pendulum clock that keeps correct time on the earth is taken to the moon. It will run a) at correct rate b)6 times faster c)√6 times faster d)√6 times slower 2. Relevant equations 3. The attempt at a solution [itex]T_{earth} = 2\pi \sqrt{\dfrac{L}{g}} \\ T_{moon} = 2\pi \sqrt{\dfrac{L}{g/6}} [/itex] Dividing i) by ii) [itex]\dfrac{T_{earth}}{T_{moon}} = \frac{1}{√6} \\ T_{moon} = √6T_{earth} [/itex] This implies option c) is correct but my book says it is option d). 



#2
Dec2512, 01:38 AM

PF Gold
P: 1,054

The period is longer so the frequency must be...




#3
Dec2512, 01:41 AM

P: 700

T [itex]\alpha[/itex] 1/√g As √g reduces by √6 on moon , this implies time period on moon will be √6 times that of earth , as you got. You interpreted your answer wrongly. If the time period increases , pendulum will oscillate slower or faster for a given displacement of the bob ? 



#4
Dec2512, 09:40 AM

P: 658

Pendulum clock when taken to moon 


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