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Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsOscillations5 Marks
Question
The acceleration due to gravity on the surface of moon is $1.7m s^{-2}$. 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 $9.8m s^{-2}$)
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Answer

Acceleration due to gravity on the surface of moon, $g' = 1.7m s^{–2}$ Acceleration due to gravity on the surface of earth, $g = 9.8m s^{–2}$ Time period of a simple pendulum on earth, T = 3.5s $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ Where l is the length of the pendulum $\therefore\ \text{l}=\frac{\text{T}^2}{(2\pi)^2}\times\text{g}$
$=\frac{(3.5)^2}{4\times(3.14)^2}\times9.8\text{m}$ The length of the pendulum remains constant. On moon's surface, time period, $\text{T}'=2\pi\sqrt{\frac{\text{l}}{\text{g}'}}$
$=2\pi\sqrt{\frac{\frac{(3.5)^2}{(4\times3.14)^2}\times9.8}{1.7}}=8.4\text{s}$Hence, the time period of the simple pendulum on the surface of moon is 8.4s.

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