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A seconds pendulum is suspended from the...

A seconds pendulum is suspended from the roof of a lift. If the lift is moving up with an acceleration `9.8 m//s^(2)`, its time period is

A

1 s

B

`sqrt(2)` s

C

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

D

`2 sqrt(2) s`

Text Solution

Verified by Experts

The correct Answer is:
B
`T^(1) = 2pi sqrt((l)/(g+a)) = 2pi sqrt((l)/(g+g))=(T)/(sqrt(2)) = (2)/(sqrt(2))=sqrt(2)`
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NARAYNA-OSCILLATIONS-EXERCISE - I (C.W)
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