Question

A man measures the period of simple pendulum in a stationary lift and finds it to be t sec.If the lift accelerates upwards

Answer 1

Qualitatively:

If the lift accelerates upwards, the effective gravity for the pendulum would increase, and time period is inversely proportional to square root of (effective) gravity, so it would decrease.

Quantitatively:

$T = 2\pi\sqrt{\frac{l}{g_{eff}}$

If lift accelerates upwards, with acceleration a,

$g_{eff} = g + a$

$T_{i} = 2\pi\sqrt{\frac{l}{g}}$ (initial)

$T_{f} = 2\pi\sqrt{\frac{l}{g + a}}$ (final)

So, $T_{f} = \sqrt{\frac{g}{g + a}} T_{i}$

Which means, the final time period is lesser than the initial one.