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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.
8 years ago

Answers : (3)

vikas askiitian expert
509 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

8 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 !!

8 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!!

8 years ago
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