Asked by achinvarshney | 27th Nov, 2009, 05:42: PM
The time period of simple pendulum,
t = 2π(L/g) = 2.01L =
The acceleration due to gravity at the either poles is not affected by the rotation of the earth, and at the equator it's maximall affected.
gequator = g - Rw2 = 9.8 - (6400x103)(4x3.14x3.14/(24x3600)2) = 9.8 - 0.033 = 9.767 m/s2
Tequator = 2π(L/gequator) = 2.02L =2.02x2.01xL/2.01 = 1.005t.
The time period at the equator will increase by 0.5%
Answered by | 28th Nov, 2009, 11:34: AM
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