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)

$\left( \frac{\sqrt{3}}{2} \right)T$

Answer 2

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 ${{T}_{2}}=\left( \sqrt{\frac{3}{2}} \right){{T}_{1}}=\left( \sqrt{\frac{3}{2}} \right)T$