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The acceleration due to gravity on the surface of moon is `1.7 ms^(-2)` . What is the time period of a simple pendulum on the moon if its time period on the earth is 3.5 s ? (g on earth = `9.8 ms^(-2)`)

Text Solution

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Here, `g_m = 1.7 ms^(-2) , g_e = 9.8 ms^(-2) , T_e = 3.5 s^(-1)`
As `T_e = 2pi sqrt(l/(g_e)) and T_m = 2pi sqrt(l/g_m)`
`therefore (T_m)/(T_e) =sqrt((g_e)/(g_m)) ` (or) `T_m = T_e sqrt((g_e)/(g_m)) = 3.5 sqrt((9.8)/(1.7)) =8.4 s`
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