a second pendulum is taken on the surface of an imaginary planet where the acceleration due to gravity is 1/4th to that of the earth.Find out new time-period.
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T = 2 π sqrt(L / g)
T ∝ 1/sqrt(g)
T1 / T2 = sqrt(g2 / g1)
T2 = T1 * sqrt(g1 / g2)
T2 = 2 * sqrt(g1 / (g1/4))
T2 = 2 * 2
T2 = 4s
The new time period is 4 seconds
T ∝ 1/sqrt(g)
T1 / T2 = sqrt(g2 / g1)
T2 = T1 * sqrt(g1 / g2)
T2 = 2 * sqrt(g1 / (g1/4))
T2 = 2 * 2
T2 = 4s
The new time period is 4 seconds
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