the length of a seconds pendulum on the surface of earth is 100cm. Find the length of the seconds pendulum on the surface of the moon. Take gM=1/6 gE.
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Dear student,
The time period of a simple pendulum on earth is given as
T = 2π(l/g)1/2
now, for moon g' = g/6
this would mean that the gravitational force exerted by moon would be six times less compared to that of earth.
so, the time period would be
T' = 2π(l / g/6)1/2
or
T' = √6T
T' > T
thus, time period of simple pendulum would increase on moon.
Regards
====================
Dear student,
The time period of a simple pendulum on earth is given as
T = 2π(l/g)1/2
now, for moon g' = g/6
this would mean that the gravitational force exerted by moon would be six times less compared to that of earth.
so, the time period would be
T' = 2π(l / g/6)1/2
or
T' = √6T
T' > T
thus, time period of simple pendulum would increase on moon.
Regards
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