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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 - Physics

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? (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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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 14 Oscillations
Q 15 | Page 360
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