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)

Asked by Abhisek | 1 year ago |  65

##### Solution :-

Given

Acceleration due to gravity on the surface of moon, g’ = 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 π$$\sqrt{\dfrac{I}{g}}$$

Where,

l is the length of the pendulum

Therefore,

l = $$\dfrac{T^2}{(2 π)^2}$$ x 2

On substituting, we get,

l = $$\dfrac{3.5^2}{(4 \times (3.14))^2}$$ x 9.8 m

The length of the pendulum remains constant

On moon’s surface, time period, T = 2 π$$\sqrt{\dfrac{I}{g}}$$

= 2 π $$\sqrt{\dfrac{(3.5)^2}{\dfrac{4 \times (3.14)^2 \times 9.8}{1.7}}}$$

We get,

= 8.4 s

Therefore, the time period of the simple pendulum on the surface of the moon is 8.4 s

Answered by Pragya Singh | 1 year ago

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