Question

The acceleration due to gravity on the surface of moon is 1.7 m s${}^{-2}$. 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 m s${}^{-2}$)

Answer 1

Acceleration due togravity on the surface of moon, $g^{\prime }=1.7m{s}^{-2}$

Acceleration due togravity on the surface of earth, $g=9.8m{s}^{-2}$
Time period ofa simple pendulum on earth, $T=3.5s$

$T=2\pi \sqrt{\frac{l}{g}}$

where,

$l$ is the lengthof the pendulum

$\therefore l=\frac{T^2}{(2\pi {)}^2}\times g$

$=\frac{(3.5{)}^2}{4\times (3.14{)}^2}\times 9.8m$

The length ofpendulum remains constant

On moon's surface,time period, $T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}$

$=2\pi \sqrt{\frac{\frac{(3.5{)}^2}{4\times (3.14{)}^2}\times 9.8}{1.7}}=8.4s$

Hence, the timeperiod of the simple pendulum on the surface of moon is 8.4 s.