Question 309: Oscillations

Question

In a stationary lift, time period of a simple pendulum is '$$\mathrm{T}$$'. The lift starts accelerating downwards with acceleration $$\left(\frac{\mathrm{g}}{4}\right)$$, then the time period of the pendulum will be

Accepted Answer

Time period of pendulum: $$T=2 \pi \sqrt{\frac{L}{g}}$$
When lift is accelerated downward with acceleration $$\frac{\mathrm{g}}{4}$$,
$$\begin{aligned}
& g=g-\frac{g}{4} \
& g=\frac{3 g}{4}
\end{aligned}$$
$$	herefore\quad$$ New Time period will be
$$\begin{aligned}
& T_1=2 \pi \sqrt{\frac{4 L}{3 g}} \
& T_1=2 \pi \frac{2}{\sqrt{3}} \sqrt{\frac{L}{g}} \
& T_1=\frac{2}{\sqrt{3}} T
\end{aligned}$$

Answer

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

Answer

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

Answer

$$\frac{3}{4} \mathrm{~T}$$

Answer

$$\frac{4}{3} \mathrm{~T}$$

Oscillations

In a stationary lift, time period of a simple pendulum is '$\mathrm{T}'. The lift starts accelerating downwards with acceleration \left(\frac{\mathrm{g}}{4}\right)$, then the time period of the pendulum will be

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