Question

The time period of a simple pendulum measured inside a
stationary lift is found to br T. If the lift states accelerating
upwards with an acceleration g/3, the time period is-
(1) T13
(2) T/3/2
(3) T/13
(4) T/3​

Answer 1

Explanation:

we know that

$t = 2\pi \sqrt{ \frac{l}{g} } \\ where \: g \: is \: acceleration$

let the new time period be t1.

$t1 = 2\pi \sqrt{ \frac{l}{ \frac{g}{3} } } = 2\pi \sqrt{ \frac{3l}{g} } \\ = \sqrt{3} \times 2\pi \sqrt{ \frac{l}{g} } \\ t1 = \sqrt{3} \times t$