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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 surface of moon if its time period on the surface of earth is 3.5 s? (*g *on the surface of earth is 9.8 ms^{–2})

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#### Solution 1

Acceleration due to gravity on the surface of moon, g' = 1.7 m s^{–2}

Acceleration due to gravity on the surface of earth, g = 9.8 m s^{–2}

Time period of a simple pendulum on earth, T = 3.5 s

`T = 2pisqrt(1/g)`

Where l is the length of the pendulum

`:.l = T^2/(2pi)^2 xx g`

`=(3.5)^2/(4xx(3.14)^2) xx 9.8 m`

The length of the pendulum remains constant.

On moon’s surface, time period, `T' = 2pi sqrt(1/g^')`

`= 2pi sqrt(((3.5)^2/(4xx3.14)^2 xx 9.8)/1.7) = 8.4 s`

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.

#### Solution 2

Here `g_m = 1.7 ms^(-2), g_e= 9.8 ms^2; T_m = ? T_e= 3.5 s^(-1)`

Since `T_e = 2pi sqrt(1/(g_e))` and `T_m = 2pisqrt(1/g_m)`

`:. T_m/T_e = sqrt(g_e/g_m) => T_m = T_e = sqrt(g_e/g_m)`

`= 3.5 sqrt(9.8/1.7) = 8.4 s`

#### Solution 3

Here `g_m = 1.7 ms^(-2), g_e= 9.8 ms^2; T_m = ? T_e= 3.5 s^(-1)`

Since `T_e = 2pi sqrt(1/(g_e))` and `T_m = 2pisqrt(1/g_m)`

`:. T_m/T_e = sqrt(g_e/g_m) => T_m = T_e = sqrt(g_e/g_m)`

`= 3.5 sqrt(9.8/1.7) = 8.4 s`

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