Question 315: Oscillations

Question

The time period of a simple pendulum inside a stationary lift is '$$T$$'. When the lift starts accelerating upwards with an acceleration $$\left(\frac{\mathrm{g}}{3}\right)$$, the time period of the pendulum will be

Accepted Answer

Time period of a simple pendulum is $$\mathrm{T}=2 \pi \sqrt{\left(\frac{l}{\mathrm{a}}ight)}$$
In stationary lift, the value of acceleration is $$\mathrm{a}=\mathrm{g}$$
$$\mathrm{T}=2 \pi \sqrt{\left(\frac{l}{\mathrm{~g}}ight)}$$
When the lift is accelerating in an upward direction, there is a pseudo force acting in a downward direction.
$$\begin{aligned}
& 	herefore \quad \mathrm{ma}=\mathrm{mg}+\frac{\mathrm{mg}}{3} \
& 	herefore \quad \mathrm{a}=\mathrm{g}+\frac{\mathrm{g}}{3} \
& 	herefore \quad \mathrm{a}=\frac{4 \mathrm{~g}}{3}
\end{aligned}$$
$$	herefore \quad$$ Period for a pendulum in accelerating lift is
$$\begin{aligned}
\quad \mathrm{T}^{\prime} & =2 \pi \sqrt{\frac{l}{\mathrm{a}}}=2 \pi \sqrt{\frac{3 l}{4 \mathrm{~g}}} \
	herefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2}\left(2 \pi \sqrt{\frac{l}{\mathrm{~g}}}ight) \
	herefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2} \mathrm{~T}
\end{aligned}$$

Answer

$$\frac{\sqrt{5}}{2} \mathrm{~T}$$

Answer

$$\frac{\sqrt{3}}{2} \mathrm{~T}$$

Answer

$$\frac{2 \mathrm{~T}}{\sqrt{3}}$$

Answer

$$\frac{2 \mathrm{~T}}{\sqrt{5}}$$

Oscillations

The time period of a simple pendulum inside a stationary lift is '$T'. When the lift starts accelerating upwards with an acceleration \left(\frac{\mathrm{g}}{3}\right)$, the time period of the pendulum will be

Choose the correct answer: