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)
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`
