What should be the length of a simple pendulum on the surface of the moon if its period does not differ from that on the surface of the earth? Mass of earth is 80 times that of the moon and radius of the earth is 4 times that of the moon.
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Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.
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Answer:
Acceleration due to gravity on the surface of moon, g
′
=1.7ms
−2
Acceleration due to gravity on the surface of earth, g=9.8ms
−2
Time period of a simple pendulum on earth, T=3.5s
T=2π
g
l
where,
l is the length of the pendulum
∴l=
(2π)
2
T
2
×2
=
4×(3.14)
2
(3.5)
2
×9.8m
The length of pendulum remains constant
On moon's surface, time period, T
′
=2π
g
′
l
=2π
1.7
4×(3.14)
2
(3.5)
2
×9.8
=8.4 s
Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.
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