What happens to time period if pendulum is set on moon (g at moon is 1/6 of g on earth)
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Time period of simple pendulum is inversely proportional to square root of acceleration due to gravity (g)
T ∝ 1/√g
T1 / T2 = √(g2 / g1)
T1 / T2 = √(g / (6g))
T1 / T2 = 1/√6
T2 = T1√6
Time period of pendulum on moon will be √6 times that of on earth.
T ∝ 1/√g
T1 / T2 = √(g2 / g1)
T1 / T2 = √(g / (6g))
T1 / T2 = 1/√6
T2 = T1√6
Time period of pendulum on moon will be √6 times that of on earth.
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