1. Tardigrade
  2. Question
  3. Physics
  4. The acceleration due to gravity on the surface of moon is 1.7 ms-2. The time period of a simple pendulum on the surface of moon if its time period on the surface of earth being 3.5 s is (g on the surface of earth is 9.8 m s-2).

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

Oscillations

Solution:

$Earth$, Time period, T = $2 \pi \sqrt{\frac{l}{g}}$ i.e. 3.5 = $2 \pi \sqrt{\frac{l}{9.8}}$
$Moon$, Here, $g = 1.7 \, m \, s^{-2}$
$T' = \sqrt{\frac{l}{1.7}}$
Dividing, $\frac{T'}{3.5} = 2 \pi \sqrt{\frac{l}{1.7}}/ 2\pi \sqrt{\frac{l}{9.8}}$
$T' = \sqrt{\frac{9.8}{1.7}} \times 3.5 = 8.4 \, s$