Question 223: 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 ' $T$ '? (take ' g ' as acceleration due to gravity)

Accepted Answer

Step 1: Understand how period depends on gravity
The formula for the time period of a simple pendulum is: $ T = 2\pi\sqrt{\frac{L}{g}} $. This means the time period $ T $ gets smaller when gravity $ g $ is bigger. We can write: $ T \propto \frac{1}{\sqrt{g}} $.
Step 2: Set up the ratio for periods and gravity
We compare the old period $ T $ to the new period $ T' $ when gravity changes:

$$ \frac{T}{T'} = \sqrt{\frac{g'}{g}} $$

where $ g' $ is the new effective gravity.
Step 3: When the lift moves upward
If the lift goes up with acceleration $ a $, the new gravity felt (effective gravity) becomes $ g' = g + a $. We are told we want the new period to be half as long: $ T' = T/2 $.
Step 4: Put the values into the ratio
Substitute these values into our ratio formula:

$$ \frac{T}{T / 2} = \sqrt{\frac{g + a}{g}} $$

This simplifies to $ 2 = \sqrt{1 + \frac{a}{g}} $.
Step 5: Solve for acceleration $ a $
Square both sides: $ 2^2 = 1 + \frac{a}{g} $, so $ 4 = 1 + \frac{a}{g} $.
Solve for $ a $: $ 4 - 1 = \frac{a}{g} $, so $ \frac{a}{g} = 3 $.
Therefore, $ a = 3g $.

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 '? (take ' g ' as acceleration due to gravity)

Choose the correct answer: