Question 200: Oscillations

Question

A simple pendulum is suspended from ceiling of a lift when lift is at rest its period is ' T '. With what acceleration ' $a$ ' should lift be accelerated upward in order to reduce the period to ' $T/2$ '? (take ' g ' as acceleration due to gravity)

Accepted Answer

For a simple pendulum, time period is
$$ T = 2\pi \sqrt{\frac{l}{g}} $$
When the lift accelerates upward with acceleration $a$, the effective acceleration due to gravity becomes
$$ g' = g + a $$
So the new time period becomes
$$ T' = 2\pi \sqrt{\frac{l}{g+a}} $$
Given that the new period is reduced to half:
$$ T' = \frac{T}{2} $$
Now substitute:
$$ 2\pi \sqrt{\frac{l}{g+a}} = \frac{1}{2}\left(2\pi \sqrt{\frac{l}{g}}\right) $$
Cancel $2\pi$:
$$ \sqrt{\frac{l}{g+a}} = \frac{1}{2}\sqrt{\frac{l}{g}} $$
Squaring both sides:
$$ \frac{l}{g+a} = \frac{1}{4}\cdot \frac{l}{g} $$
Cancel $l$:
$$ \frac{1}{g+a} = \frac{1}{4g} $$
So,
$$ g+a = 4g $$
$$ a = 3g $$
Hence, the correct answer is:
$$ \boxed{3g} $$
Option B

Answer

2 g

Answer

3 g

Answer

4 g

Answer

g

Oscillations

A simple pendulum is suspended from ceiling of a lift when lift is at rest its period is ' T '. With what acceleration ' a ' should lift be accelerated upward in order to reduce the period to ' T/2 '? (take ' g ' as acceleration due to gravity)

Choose the correct answer: