Time period of simple pendulum inside a stationary lift,
$T=2\pi \sqrt{\frac{l}{g}}$ … (i)
Where, $l$ = length of string
and g = gravitational acceleration
Acceleration of lift in upward direction,
$a=\frac{g}{3}$
$\therefore$ Time period of pendulum,
$T^{\text{'}}=2\pi \sqrt{\frac{l}{g+a}}=2\pi \sqrt{\frac{l}{g+\frac{g}{3}}}$
$=2\pi \sqrt{\frac{3l}{4g}}=2\pi .\frac{\sqrt{3}}{2}.\sqrt{\frac{l}{g}}$
$=\frac{\sqrt{3}}{2}.2\pi \sqrt{\frac{l}{g}}=\frac{\sqrt{3}}{2}T$ [from Eq. (i)]