Question

A lift is ascending with an acceleration equal to g/3 What will be the time period of a simple pendulum suspended from its ceiling if its time period in stationary lift isT:

• A. $\frac{T}{4}$
• B. $\left(\frac{\sqrt{3}}{4}\right)T$
• C. $\left(\frac{\sqrt{3}}{2}\right)T$
• D. $\frac{T}{2}$

Answer 1

Answer: (C)

Here: Acceleration of lift $(a)=g\text{/}3$ Initial time period $T_1=T$ The effective acceleration when it is ascending $g_2=g-a=g-\frac{8}{3}=\frac{2g}{3}$ Time period of simple pendulum $T=2\pi \sqrt{\frac{l}{g}}$ or $T\propto \frac{\sqrt{1}}{g}$ Hence $\frac{T_1}{T_2}=\sqrt{\frac{g_2}{g_1}}=\sqrt{\frac{\frac{2}{3}g}{g}}$ or $T_2=\left(\sqrt{\frac{3}{2}}\right)T_1=\left(\sqrt{\frac{3}{2}}\right)T$