The length of a second's pendulum on the surf | vedclass

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Since $T$ is constant $(2 \ s)$ for a second's pendulum on both the

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$1/6 \ m$

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Since $T$ is constant $(2 \ s)$ for a second's pendulum on both the earth and the moon,we have $T \propto \sqrt{\frac{l}{g}}$.Therefore,$\frac{T_m}{T_e} = \sqrt{\frac{l_m}{g_m} \cdot \frac{g_e}{l_e}}$.Given $T_e = T_m = 2 \ s$,we get $1 = \sqrt{\frac{l_m}{l_e} \cdot \frac{g_e}{g_m}}$.Squaring both sides,$1 = \frac{l_m}{l_e} \cdot \frac{g_e}{g_m}$.We know that $g_m = \frac{g_e}{6}$,so $\frac{g_e}{g_m} = 6$.Given $l_e = 1 \ m$,substituting these values gives $1 = \frac{l_m}{1} \cdot 6$.Thus,$l_m = \frac{1}{6} \ m$.

The length of a second's pendulum on the surface of the earth is $1 \ m$. What will be the length of a second's pendulum on the moon?

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