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# The acceleration due to gravity on the surface of moon is 1.7 m s^ -2 . What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ?

Q. 14.15 The acceleration due to gravity on the surface of moon is $\inline 1.7m\; s^{-2}$ What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is $\inline 3.5\; s$ ? (g on the surface of earth is 9.8 m s–2)

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The time period of a simple pendulum of length l executing S.H.M is given by

$T=2\pi \sqrt{\frac{l}{g}}$

g= 9.8 m s-2

gm = 1.7 m s-2

The time period of the pendulum on the surface of Earth is Te = 3.5 s

The time period of the pendulum on the surface of the moon is Tm

$\\\frac{T_{m}}{T_{e}}=\sqrt{\frac{g_{e}}{g_{m}}}\\ T_{m}=T_{e}\times \sqrt{\frac{g_{e}}{g_{m}}}\\ T_{m}=3.5\times \sqrt{\frac{9.8}{1.7}} \\T_{m}=8.4s$

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