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A simple pendulum 50cm long is suspended...

A simple pendulum `50cm` long is suspended from the roof of a acceleration in the horizontal direction with constant acceleration `sqrt(3) g m//s^(-1)`. The period of small oscillations of the pendulum about its equilibrium position is `(g = pi^(2) m//s^(2))` ltbRgt

A

`1.0sec`

B

`sqrt(2)sec`

C

`1.53 sec`

D

`1.68 sec `

Text Solution

Verified by Experts

The correct Answer is:
A
With respect to the cart , equilibrium position of the pendulum is shown.
if displaced by small angle `theta` from this position, then it will execute `SHM` about his equilibrium position position, time period of whichis given by `:`


`T=2pisqrt((L)/(g_(ef f))), g_(ef f)=sqrt(g^(2)+sqrt(3g)^(2))`
`rArr g_(ef f)=2g`
`rArrT=1.0` seconds
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