Question

The time period of a simple pendulum inside astationary lift is $\sqrt{5}$ . What will be the time period when the lift moves upwards with an acceleration $\frac{g}{4}?$

• A. $\sqrt{5}s$
• B. 2$\sqrt{5}s$
• C. $(2+\sqrt{5})s$
• D. 2$s$

Answer 1

The correct option is D 2$s$

We know ,

Time period $\left(T\right)=2\pi \sqrt{\frac{l}{g}}=\sqrt{5}sec$

When, lift moves upwards, $geff=g+\frac{g}{4}=\frac{5g}{4}$

Time period $\left(T^1\right)=2\pi \sqrt{\frac{l}{geff}}=2\pi \sqrt{\frac{4l}{5g}}=2\pi \sqrt{\frac{l}{g}}\sqrt{\frac{4}{5}}$

$T^1=\left(2\pi \sqrt{\frac{l}{g}}\right)\frac{2}{\sqrt{5}}=\sqrt{5}\times \frac{2}{\sqrt{5}}=2sec$