calculate the time period of simple pendulum of length 1.44 m on the surface of moon the acceleration due to gravity on the surface of moon is 1/6 the acceteration due to gravity on earth .
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The period T of a pendulum at the surface of the Moon is given (approximately) by
T=2πLg−−√T=2πLg
where L is the length of the pendulum and g= 1.62 m/s22 is the acceleration due to gravity on the Moon.
A “seconds pendulum” is one that takes one second to swing each way; thus a “seconds pendulum” on the Moon (or anywhere else) has a period of two seconds. You could use the above formula to calculate the length of a seconds pendulum on the Moon
T=2πLg−−√T=2πLg
where L is the length of the pendulum and g= 1.62 m/s22 is the acceleration due to gravity on the Moon.
A “seconds pendulum” is one that takes one second to swing each way; thus a “seconds pendulum” on the Moon (or anywhere else) has a period of two seconds. You could use the above formula to calculate the length of a seconds pendulum on the Moon
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Answer:
(Ans-5.90s)
Explanation:
Solve and get the answer
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