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Question

A simple pendulum is suspended in a lift which is going up with an acceleration of 5 m/s2. An electric field of magnitude 5 N/C and directed vertically upward is also present in the lift. The charge of the bob is 1 μC and mass is 1 mg. Taking g=π2 and length of the simple pendulum 1 m, find the time period of the simple pendulum (in sec).

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Solution


We know that time period of a simple pendulum is given by
T=2πlg
But in this case g is not the same. So we have to take g effective (geff)
T=2πlgeff
Given that acceleration of the lift is 5 m/s2
As acceleration of the lift and acceleration due to electric field is in same direction. So we consider relative acceleration as
=qEM5

geff=g(qEM5)=gqEM+5geff=π2+51×5×1061×106
geff=π2

Now,
T=2πlπ2=2 sec

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