Physics - SHM Question with Solution | TestHub

When the lift is at rest, the time period of the simple pendulum is $T=2\pi \sqrt{\frac{l}{g}}$ Now, if the lift is moving up with an acceleration $\frac{g}{4}$ then the effective acceleration due to gravity $$\begin{array}{rl}& g^{\prime }=g+\frac{g}{4}=\frac{5g}{4}\\ & \therefore \text{ Time period }{T}^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}\\ & =2\pi \sqrt{\frac{l}{5g/4}}\\ & =\frac{2}{\sqrt{5}}\times 2\pi \sqrt{\frac{l}{g}}=\frac{2}{\sqrt{5}}T\end{array}$$