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Question
The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?
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Solution
A second's pendulum means a simple pendulum having a time period T = 2 s.
For a simple pendulum, T = `2pisqrt(l/g)`
Where, `l` = Length of the pendulum
And g = Acceleration due to gravity on surface of the earth
`T_e = 2pisqrt(l_e/g_e)` .......(i)
On the surface of the moon, `T_m = 2pisqrt(K_m/g_m)` ......(ii)
∴ `T_e/T_m = (2pi)/(2pi) sqrt(l_e/g_e) xx sqrt(g_m/l_m)`
Te = Tm to maintain the second pendulum time period.
∴ 1 = `sqrt(l_e/l_m xx g_m/g_e)` ......(iii)
But the acceleration due to gravity on the moon is 1/6 of the acceleration due to gravity on earth,
i.e., `g_m = g_e/6`
Squaring equation (iii) and putting this value.
1 = `l_e/l_m xx (g_e/6)/g_e = l_e/l_m xx 1/6`
⇒ `l_e/(6l_m)` = 1
⇒ `6l_m = l_e`
⇒ `l_m = 1/6 l_e = 1/6 xx 1 = 1/6 m`
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