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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.For the earth: $T_e = 2 \pi \sqrt{\frac{l}{g_e}} = 3.5 \; s$,where

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$8.4$

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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.For the earth: $T_e = 2 \pi \sqrt{\frac{l}{g_e}} = 3.5 \; s$,where $g_e = 9.8 \; m s^{-2}$.For the moon: $T_m = 2 \pi \sqrt{\frac{l}{g_m}}$,where $g_m = 1.7 \; m s^{-2}$.Taking the ratio: $\frac{T_m}{T_e} = \sqrt{\frac{g_e}{g_m}}$.$T_m = T_e 	imes \sqrt{\frac{g_e}{g_m}} = 3.5 	imes \sqrt{\frac{9.8}{1.7}}$.$T_m = 3.5 	imes \sqrt{5.7647} \approx 3.5 	imes 2.4 = 8.4 \; s$.Thus,the time period on the moon is $8.4 \; s$.

The acceleration due to gravity on the surface of the moon is $1.7 \; m s^{-2}$. What is the time period of a simple pendulum on the surface of the moon if its time period (in $s$) on the surface of the earth is $3.5 \; s$? ($g$ on the surface of the earth is $9.8 \; m s^{-2}$)

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