Theacceleration 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 moonif its time period on the surface of earth is 3.5 s? (gon the surface of earth is 9.8 ms^–2)

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Solution

Acceleration due togravity on the surface of moon, =1.7 m s^–2
Acceleration due togravity on the surface of earth, g = 9.8 m s^–2
Time period of a simplependulum on earth, T = 3.5 s
Where,
l is the lengthof the pendulum
The length of thependulum remains constant.
On moon’ssurface, time period,

Hence, the time periodof the simple pendulum on the surface of moon is 8.4 s.

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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 on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms^–2)

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(1) If the weight of a body on the surface of the moon is 100N, what is its mass? (61.34 kg)

(2) If the acceleration due to gravity one the surface of the earth is 9.8 ms² what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth? (4.9 m/s²).

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Theacceleration 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 moonif its time period on the surface of earth is 3.5 s? (gon the surface of earth is 9.8 ms–2)

Theacceleration 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 moonif its time period on the surface of earth is 3.5 s? (gon the surface of earth is 9.8 ms^–2)