gravitation-time period

Asked by achinvarshney | 27th Nov, 2009, 05:42: PM

Expert Answer:

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%

Regards,

Team,

TopperLearning.

Answered by  | 28th Nov, 2009, 11:34: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.