Question

The time period of a simple pendulum of length L is 3 s. If the length of the pendulum is increased 4 times, the increase in time period of the simple pendulum is

• A. 6s
• B. 4s
• C. 5s
• D. 3s
• E. 2s

Answer 1

The Correct Option is D

The time period \(T\) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. Initially, the time period \(T_1\) for length \(L\) is: \[ T_1 = 2\pi \sqrt{\frac{L}{g}} = 3 \text{ s} \] Now, if the length of the pendulum is increased 4 times, the new length becomes \(4L\). The new time period \(T_2\) is: \[ T_2 = 2\pi \sqrt{\frac{4L}{g}} = 2\pi \sqrt{4} \sqrt{\frac{L}{g}} = 2 \times 2\pi \sqrt{\frac{L}{g}} = 2T_1 \] Thus, the new time period is \(T_2 = 6 \text{ s}\). The increase in time period is: \[ \Delta T = T_2 - T_1 = 6 \text{ s} - 3 \text{ s} = 3 \text{ s} \] Thus, the increase in time period is 3 s. 

Answer 2

The time period of a simple pendulum is given by:  

$$ T = 2\pi \sqrt{\frac{L}{g}} $$ 
Given: initial time period \( T_1 = 3 \, \text{s} \) 
Let the initial length be \( L \), then: $$ 3 = 2\pi \sqrt{\frac{L}{g}} \quad \text{(1)} $$ 
When the length is increased 4 times, the new length is \( 4L \), so the new time period becomes: 
$$ T_2 = 2\pi \sqrt{\frac{4L}{g}} = 2\pi \cdot 2 \sqrt{\frac{L}{g}} = 2 \cdot (2\pi \sqrt{\frac{L}{g}}) $$ 
From equation (1), we know: $$ 2\pi \sqrt{\frac{L}{g}} = 3 $$ 
Therefore: $$ T_2 = 2 \times 3 = 6 \, \text{s} $$ 
So, the increase in time period is: $$ T_2 - T_1 = 6 - 3 = \boxed{3\, \text{s}} $$