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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 (in $sec$) on the surface of earth is $3.5\; s ?( g$ on the surface of earth is $9.8\; ms ^{-2} )$
$8.4$
$4.6$
$10.6$
$6.2$
Solution
Acceleration due to gravity on the surface of moon, $g^{\prime}=1.7\, m s ^{-2}$
Acceleration due to gravity on the surface of earth, $g=9.8\, m s ^{-2}$
Time period of a simple pendulum on earth, $T=3.5\, s$ $T=2 \pi \sqrt{\frac{l}{g}}$
Where, $l$ is the length of the pendulum
$\therefore l=\frac{T^{2}}{(2 \pi)^{2}} \times g$
$=\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8 \,m$
The length of the pendulum remains constant.
On moon's surface, time period, $T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}}$ $=2 \pi \sqrt{\frac{\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8}{1.7}}=8.4\, s$
Hence, the time period of the simple pendulum on the surface of moon is $8.4\, s$
