# 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)

#### 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

Concept: Some Systems Executing Simple Harmonic Motion
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#### APPEARS IN

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