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Vikas TU Grade: 12th Pass
        At moon the weight of things become 1/6th of weight of earth . What is the ratio of time period of simple

pendulum at earth to that on the moon.
6 years ago

Answers : (3)

vikas askiitian expert
510 Points
										

W = mg


Wmoon = Wearth/6


gmoon = gearth/6


T(time period) = 2pi(L/g)1/2


 Tearth/Tmoon  = (gmoon/gearth)1/2


g(moon) = g(earth/6) so


Tearth/Tmoon = (1/6)1/2

6 years ago
Sudheesh Singanamalla
114 Points
										

Consider a simple pendulum of mass 'm' and length 'L' .


time period of a simple pendulum = 2*pi* sqrt( L/g ) ----------------------------- (on the earth) -------- (i)


time period of the simple pendulum = 2*pi* sqrt (6L / g ) ------------------------ (on the moon) -------- (ii)


 


ratio of time period of simple pendulum of earth to that on moon = (i) / (ii)


                                                                                           = 2*pi* sqrt( L/g) / 2*pi* sqrt ( 6L/g)


                                                                                           = sqrt (L/g ) / sqrt ( 6L/g )


                                                                                           = (L/g)/(6L/g)


                                                                                           = 1/6


 


therefore the ratio of timeperiod of the pendulum on the earth to that of the moon is 1:6 .


Please Approve !!

6 years ago
Ritvik Gautam
85 Points
										

We know, time period of a pendulum,


T=2π√l/g


 


Hence, substituting the values and using formula,


we get,


Tmoon = √6 x Tearth


 


 


Please Approve!!

6 years ago
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