Question

A simple pendulum attached to the ceiling of a stationary lift has a time period T. The distance y covered by the lift moving upwards varies with time t as $y=t^2$ where y is in metre and t in second. If $g=10m{s}^{-2}$, the time period of the pendulum will be

• A. $\sqrt{\frac{4}{5}}T$
• B. $\sqrt{\frac{5}{6}}T$
• C. $\sqrt{\frac{5}{4}}T$
• D. $\sqrt{\frac{6}{5}}T$

Answer 1

The correct option is B

$\sqrt{\frac{5}{6}}T$

Here, $y=t^2\Rightarrow v=2t\Rightarrow a=2m/s^2$
So, effective acceleration, $g^{\prime }=g+a=12m/s^2$
Now, $\frac{T^{\prime }}{T}=\frac{2\pi \sqrt{\frac{l}{g^{\prime }}}}{2\pi \sqrt{\frac{l}{g}}}=\sqrt{\frac{g}{g^{\prime }}}=\sqrt{\frac{5}{6}}$
Hence, $T^{\prime }=\sqrt{\frac{5}{6}}T$