Question 257: Oscillations

Question

The time period of a simple pendulum inside a stationary lift is $\sqrt{3}$ second. When the lift moves upwards with an acceleration $g / 3$, the time period will be ( $\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

Given:
Time period in stationary lift, $T_1 = \sqrt{3}$ s
Acceleration due to gravity = $g$
Lift moves upwards with acceleration $a = \frac{g}{3}$
Step 1: Time period formula for a simple pendulum
In general,
$$ T = 2\pi \sqrt{\frac{l}{g_{	ext{eff}}}} $$
where $g_{	ext{eff}}$ is the effective acceleration due to gravity.
Step 2: Effective gravity in the stationary case
When the lift is stationary, $g_{	ext{eff}} = g$
So,
$$ T_1 = 2\pi \sqrt{\frac{l}{g}} $$
Step 3: Effective gravity when lift moves upward
When lift accelerates upwards with $a$,
$$ g_{	ext{eff}} = g + a = g + \frac{g}{3} = \frac{4g}{3} $$
Step 4: New time period
The new time period is
$$ T_2 = 2\pi \sqrt{\frac{l}{g_{	ext{eff}}}} = 2\pi \sqrt{\frac{l}{\frac{4g}{3}}} $$
$$ T_2 = 2\pi \sqrt{\frac{3l}{4g}} $$
Express $l$ in terms of $T_1$:
$$ T_1 = 2\pi \sqrt{\frac{l}{g}} $$
Square both sides:
$$ T_1^2 = 4\pi^2 \frac{l}{g} \implies l = \frac{g T_1^2}{4\pi^2} $$
Now, substitute $l$ into $T_2$:
$$ T_2 = 2\pi \sqrt{\frac{3}{4g} \cdot \frac{g T_1^2}{4\pi^2}} $$
$$ T_2 = 2\pi \sqrt{\frac{3 T_1^2}{16\pi^2}} $$
$$ T_2 = 2\pi \cdot \frac{T_1}{4\pi} \sqrt{3} $$
$$ T_2 = \frac{T_1}{2} \sqrt{3} $$
Step 5: Substitute $T_1 = \sqrt{3}$ s
$$ T_2 = \frac{\sqrt{3}}{2} 	imes \sqrt{3} = \frac{3}{2} = 1.5\ 	ext{seconds} $$
Final Answer:
Option
A) 1.5 s

Answer

1.5 s

Answer

2 s

Answer

$\sqrt{3} \mathrm{~s}$

Answer

3 s

Oscillations

The time period of a simple pendulum inside a stationary lift is \sqrt{3} second. When the lift moves upwards with an acceleration g / 3, the time period will be ( \mathrm{g}= acceleration due to gravity)

Choose the correct answer: