The time period of a simple pendulum ins...

The time period of a simple pendulum inside a stationary lift is `sqrt(5)` s. What will be the time period when the lift moves upward with an acceleration `(g)/(4)`?

A

`sqrt(5)` s

B

`2 sqrt(5) s`

C

`(2 + sqrt(5))`s

D

`2 s`

Text Solution

Verified by Experts

The correct Answer is:

D

As, `T = 2pi sqrt((I)/(g))`
`rArr" "T' = 2 pi sqrt((I)/((g+(g)/(4))))=(2)/(sqrt(5))T = 2s`

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