A pendulum is executing simple harmonic motion and its maximum kinetic energy is `K_(1)`. If the length of the pendulum is doubled and it perfoms simple harmonuc motion with the same amplitude as in the first case, its maximum kinetic energy is `K_(2)` Then:

Options

A. `K_(2) =2K_(1)`
B. `K_(2) = (K_(1))/(2)`
C. `K_(2) =(K_(1))/(2)`
D. `K_(2) =K_(1)`

Answer

To solve the problem, we need to analyze the kinetic energy of a pendulum executing simple harmonic motion (SHM) under two different conditions.

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy in SHM**:
The maximum kinetic energy (K) of a pendulum in simple harmonic motion can be expressed as:
\[ K = \frac{1}{2} m \omega^2 A^2 \]
where:
- \( m \) is the mass of the pendulum bob,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of the motion.

2. **Finding Angular Frequency**:
The angular frequency \( \omega \) for a simple pendulum is given by:
\[ \omega = \sqrt{\frac{g}{L}} \]
where \( g \) is the acceleration due to gravity and \( L \) is the length of the pendulum.

3. **Calculating Maximum Kinetic Energy (K1)**:
For the initial pendulum with length \( L \):
\[ K_1 = \frac{1}{2} m \left(\sqrt{\frac{g}{L}}\right)^2 A^2 = \frac{1}{2} m \frac{g}{L} A^2 \]

4. **Doubling the Length of the Pendulum**:
When the length of the pendulum is doubled, \( L \) becomes \( 2L \). The new angular frequency \( \omega' \) is:
\[ \omega' = \sqrt{\frac{g}{2L}} = \frac{1}{\sqrt{2}} \sqrt{\frac{g}{L}} \]

5. **Calculating Maximum Kinetic Energy (K2)**:
The maximum kinetic energy for the new pendulum with length \( 2L \) is:
\[ K_2 = \frac{1}{2} m \left(\frac{1}{\sqrt{2}} \sqrt{\frac{g}{L}}\right)^2 A^2 = \frac{1}{2} m \frac{g}{2L} A^2 = \frac{1}{4} \left(\frac{1}{2} m \frac{g}{L} A^2\right) = \frac{1}{4} K_1 \]

6. **Conclusion**:
Therefore, the relationship between the maximum kinetic energies is:
\[ K_2 = \frac{1}{4} K_1 \]

### Final Answer:
\[ K_2 = \frac{1}{4} K_1 \]