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A simple pendulum is suspended from the ...

A simple pendulum is suspended from the ceiling of a car and its period of oscillation is T when the car is at rest. The car starts moving on a horizontal road with a constant acceleration g (equal to the acceleration due to gravity, in magnitude) in the forward direction. To keep the time period same, the length of th pendulum

A

will have to be increased by `sqrt(2)l`

B

will have to be increased by `(sqrt(2)-1)l`

C

will have to be decreased by `sqrt(2)l`

D

will have to be decreased by `(sqrt(2)-1)l`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) To keep time period same , `l'=sqrt(2)l`
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