Question

The time period of a simple oscillation is measured as T in a stationary lift. If the lift moves upwards with an acceleration of 5 g, the time period will be

• A. same
• B. increased by $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$ times
• C. decreased by $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ times
• D. none of the above.

Answer 1

The correct option is D

none of the above.

1. The time period of a simple pendulum in a stationary lift ,$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$, where l is length of the pendulum and g is the acceleration due to gravity.
2. $\Rightarrow T\alpha \sqrt{\frac{1}{g}}$
3. In the lift moving upwards with an acceleration of 5g, net acceleration,$g\text{'}=g+5g=6g$
4. In the lift moving upwards, changed time period =$T\text{'}$
5. $\frac{T\text{'}}{T}=\frac{\sqrt{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6g$}\right.}}}{\sqrt{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$g$}\right.}}}\Rightarrow \frac{T\text{'}}{T}=\frac{1}{\sqrt{6}}\Rightarrow T\text{'}=\frac{T}{\sqrt{6}}$.
6. The time period will be increased by $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{6}$}\right.$ times.

Hence, the correct option is D.