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the period of a simple pendulum in a sta...

the period of a simple pendulum in a stationary lift is 3 sec. If the lift accelerates downwards with an acceleration `(g)/(4)`, its period of oscillation will be

A

`sqrt(3)s`

B

`(1)/(2sqrt(3))s`

C

`2sqrt(3)s`

D

6s

Text Solution

Verified by Experts

The correct Answer is:
C
`(T_(1))/(T_(2))=sqrt((g)/(g')) " where " g'=g-(g)/(4)=(3g)/(4)`
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