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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). |
Answer» On the surface of the earth, g = 9.38 ms-2, T = 3.5 s. Using T = 2π√{l/g} ⇒ l = T2g/4π2 or, l = {(3.5)2 x 9.8}/{4 x 9.87} = 3.04 m On the surface of the moon, g = 1.7 ms-2, l = 3.04 m Hence T = 2π√{l/g} = 2 x 3.14 x √{3.04/1.7} = 8.4 sec. |
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