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

Formula for the Time Period of a Simple Pendulum

The time period \( T \) of a simple pendulum is given by:

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

Where:

\( L \) = Length of the pendulum

\( g \) = Acceleration due to gravity

Step 1: Calculate the Original Time Period

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

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

Step 2: Analyze the Changes

According to the problem:

• The mass of the bob is increased to three times its original mass (\( 3m \)).
• The length of the pendulum is reduced to half (\( L_0/2 \)).

Step 3: Compute the New Time Period

The time period of the new pendulum is:

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

Rewriting:

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

Comparing with \( T_0 \):

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

Step 4: Solve for \( x \)

According to the question:

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

Solving for \( x \):

$$ x = \sqrt{2} $$

Conclusion

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