Question 360: Oscillations

Question

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be $T$. If the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$, the time period of the pendulum will be

Accepted Answer

Time period of simple pendulum inside a stationary lift,
$$T=2 \pi \sqrt{\frac{l}{g}}\quad	ext{.... (i)}$$
where, $I=$ length of string
and $g=$ gravitational acceleration
Acceleration of lift in upward direction,
$$a=\frac{g}{3}$$
$	herefore$ Time period of pendulum,
$$\begin{aligned}
	ext { d of pendulum, } \
\begin{aligned}
T^{\prime} & =2 \pi \sqrt{\frac{l}{g+a}}=2 \pi \sqrt{\frac{l}{g+\frac{g}{3}}} \
& =2 \pi \sqrt{\frac{3l}{4 g}}=2 \pi \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{\frac{l}{g}} \
& =\frac{\sqrt{3}}{2} \cdot 2 \pi \sqrt{\frac{l}{g}}=\frac{\sqrt{3}}{2} T 	ext { [from Eq. (i)] }
\end{aligned}
\end{aligned}$$

Answer

$\frac{T}{\sqrt{3}}$

Answer

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

Answer

$\sqrt{3} T$

Answer

$\frac{T}{3}$

Oscillations

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be T. If the lift starts accelerating upwards with an acceleration \left(\frac{g}{3}\right), the time period of the pendulum will be

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