Question

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is \(\frac{x}{2}\) times its original time period. Then the value of x is :

• A. \(\sqrt3\)
• B. \(\sqrt2\)
• C. \(2\sqrt3\)
• D. 4

Answer 1

The Correct Option is B

Simple Pendulum Time Period Formula

The time period \( T \) of a simple pendulum is defined as:

$$ T = 2\pi \sqrt{\frac{L}{g}} $$

Where:

\( L \) denotes the length of the pendulum.

\( g \) represents the acceleration due to gravity.

Step 1: Determine the Initial Time Period

For a pendulum with an initial mass \( m \) and length \( L_0 \), the time period is given by:

$$ T_0 = 2\pi \sqrt{\frac{L_0}{g}} $$

Step 2: Evaluate the Modifications

As per the problem statement:

• The mass of the bob is tripled, becoming \( 3m \).
• The pendulum's length is halved, becoming \( L_0/2 \).

Step 3: Calculate the Revised Time Period

The time period for the modified pendulum is:

$$ T_{\text{new}} = 2\pi \sqrt{\frac{L_0/2}{g}} $$

This can be rewritten as:

$$ T_{\text{new}} = 2\pi \sqrt{\frac{L_0}{2g}} $$

In comparison to \( T_0 \):

$$ T_{\text{new}} = T_0 \frac{1}{\sqrt{2}} $$

Step 4: Ascertain the Value of \( x \)

According to the problem's conditions:

$$ \frac{x}{2} = \frac{1}{\sqrt{2}} $$

Solving for \( x \):

$$ x = \sqrt{2} $$

Final Result

The determined value of \( x \) is \(\sqrt{2}\).