Question

The time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $\frac{g}{2}$, the time period of the pendulum will be:

• A. $\begin{array}{l}\sqrt{\frac{3}{2}}T\end{array}$
• B. $\begin{array}{l}\frac{T}{\sqrt{3}}\end{array}$
• C. $\begin{array}{l}\sqrt{\frac{2}{3}}T\end{array}$
• D. $\begin{array}{l}\sqrt{3}T\end{array}$

Answer 1

The correct option is C

$\begin{array}{l}\sqrt{\frac{2}{3}}T\end{array}$

Step 1: Given Data

The time period of the simple pendulum $=T$

Acceleration of the lift $=\frac{g}{2}$

Step 2: Formula Used

The time period of a simple pendulum,

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

where, $L$ is the length of the pendulum.

Step 3: Calculate the effective acceleration

When the lift is stationary, the time period can be given as,

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

A pseudo force will act downwards when the lift is accelerating upwards.

A pseudo force is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.

Therefore,

$\begin{array}{l}{g}_{eff}=g+\frac{g}{2}=\frac{3g}{2}\end{array}$

Step 4: Calculate the new time period

Therefore, the new time period,

$$\begin{gathered} \begin{array}{l}T\text{'}=2\pi \sqrt{\frac{L}{g_{eff}}}\end{array} \\ \Rightarrow \begin{array}{l}T\text{'}=2\pi \sqrt{\frac{2L}{3g}}\end{array} \\ \Rightarrow \begin{array}{l}T\text{'}=\sqrt{\frac{2}{3}}T\end{array} \end{gathered}$$

Hence, the correct answer is option (C).