Question

a man measures time period of a pendulum (T) in stationary lift. If the lift moves upward with acceleration g/4, then new time period will be......
(options are given below in the photo)

Answer 1

The time period of a pendulum when acceleration present is given by

$t = 2\pi \sqrt{ \frac{l}{g + a} }$

here a= g/4
therefore g+a becomes
5g/4

there for T new = 2/√5x T

we took g+a as acceleration is upwards it would be negative if acceleration is downwards.

Answer 2

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$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \large \green { \bold{ \underline { \: step \: by \: step \: explaination}}}$

$\implies\displaystyle\sf{when\: the \: lift \:is \: moving \: upward \: and\: then \: from \: Newton's \: second \: law \: of \: motion }$

$\implies\displaystyle\sf{F=ma}$

$\implies\displaystyle\sf{F=-ma}$

$\implies\displaystyle\sf{-ma=mg-R}$

$\implies\displaystyle\sf{R=mg+ma}$

$\implies\displaystyle\sf{R=m(g+a)}$

$\implies\displaystyle\sf{T=2π\sqrt{\dfrac{L}{g}} \: and \: T'=2π\sqrt{\dfrac{L}{g+a}}}$

$\implies\displaystyle\sf{when \: lift \: is \: stationary \: the \: time \: period \: is }$

$\implies\displaystyle\sf{T=2π\sqrt{\dfrac{L}{g}}}$

$\implies\displaystyle\sf{Also\: g'=g+a}$

$\implies\displaystyle\sf{g+\dfrac{g}{4}=\dfrac{5g}{4}}$

$\implies\displaystyle\sf{\dfrac{T}{T'}=\dfrac{\sqrt{5}{4g}}{g}=\dfrac{\sqrt5}{2}}$

$\implies\displaystyle\sf{ T'=\dfrac{2}{\sqrt5}T}$