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

Question

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

Vedantu Content Team · Accepted Answer

Hint:  This problem can be solved by applying the direct formula for the time period of a simple pendulum as a function of its length and the acceleration due to gravity for both cases. The length of the pendulum for both cases will remain the same.Formula used:$T=2\pi \sqrt{\dfrac{L}{g}}$ Complete step-by-step answer:We will solve this problem by applying the direct formula for the time period of a simple pendulum. The time period $T$ of a simple pendulum of length $L$ is given by
$T=2\pi \sqrt{\dfrac{L}{g}}$ 	--(1)
where $g$ is the acceleration due to gravity.Hence, let us analyze the question.Let the length of the pendulum be $L$.
The acceleration due to gravity on the earth is $g=9.8m.{{s}^{-2}}$.
The time period of the pendulum on earth is $T=3.5s$.Therefore, using (1), we get,$T=2\pi \sqrt{\dfrac{L}{g}}$ 	--(2)On the moon, the length of the pendulum is still $L$.
The acceleration due to gravity on the moon is ${{g}_{moon}}=1.7m.{{s}^{-2}}$.
Let the time period of the simple pendulum on the moon be ${{T}_{moon}}$.Therefore, using (1), we get,${{T}_{moon}}=2\pi \sqrt{\dfrac{L}{{{g}_{moon}}}}$ 	--(3)Dividing (3) by (2), we get,$\dfrac{{{T}_{moon}}}{T}=\dfrac{2\pi \sqrt{\dfrac{L}{{{g}_{moon}}}}}{2\pi \sqrt{\dfrac{L}{g}}}=\sqrt{\dfrac{g}{{{g}_{moon}}}}$ 
$	herefore \dfrac{{{T}_{moon}}}{T}=\sqrt{\dfrac{9.8}{1.7}}=\sqrt{5.76}=2.4$ 
$	herefore {{T}_{moon}}=2.4	imes T=2.4	imes 3.5s=8.40s$ Hence, the time period of the pendulum on the moon is $8.40s$.Note: Students could solve this problem also by directly writing that the time period is inversely proportional to the square root of the acceleration due to gravity, which is, the result we arrived at anyway after writing the full formula for the time period. Students should also realize this fact and not go in for unnecessary calculations. For example, we did not try to find the length of the pendulum since it got cancelled at the end anyway. Therefore, such problems can be solved by using ratios and determining which physical quantities can be considered as constant for the problem.

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

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

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