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 $