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 performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $$K_2$$

We are given a pendulum executing simple harmonic motion with maximum kinetic energy $$K_1$$. When the length is doubled and it performs SHM with the same amplitude, the maximum kinetic energy is $$K_2$$. We need to find the relation between $$K_1$$ and $$K_2$$.

For a simple pendulum with length $$L$$ and angular amplitude $$\theta_0$$ (small angle), the maximum kinetic energy equals the maximum potential energy (by conservation of energy).

The height raised by the bob at maximum displacement is:

$$h = L(1 - \cos\theta_0)$$

By conservation of energy, the maximum kinetic energy is:

$$K_{\max} = mgh = mgL(1 - \cos\theta_0)$$

Alternatively, using the SHM approach: $$K_{\max} = \frac{1}{2}m\omega^2 A^2$$, where $$\omega = \sqrt{\frac{g}{L}}$$ and the linear amplitude $$A = L\theta_0$$.

$$K_{\max} = \frac{1}{2}m \cdot \frac{g}{L} \cdot L^2\theta_0^2 = \frac{1}{2}mgL\theta_0^2$$

This is consistent with the energy conservation result (using $$1 - \cos\theta_0 \approx \frac{\theta_0^2}{2}$$ for small angles).

For the first pendulum with length $$L$$:

$$K_1 = \frac{1}{2}mgL\theta_0^2$$ $$-(1)$$

When the length is doubled to $$2L$$, and the pendulum performs SHM with the same amplitude (same angular amplitude $$\theta_0$$):

$$K_2 = \frac{1}{2}mg(2L)\theta_0^2 = mgL\theta_0^2$$ $$-(2)$$

Dividing equation $$(2)$$ by equation $$(1)$$:

$$\frac{K_2}{K_1} = \frac{mgL\theta_0^2}{\frac{1}{2}mgL\theta_0^2} = 2$$

Therefore, $$K_2 = 2K_1$$.

The correct answer is Option A: $$K_2 = 2K_1$$.