The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is: _____ cm.

We have amplitude $$A = 3$$ cm and need to find the displacement where KE is 25% more than PE.

For a particle in SHM, the total energy, kinetic energy, and potential energy are:

$$E = \frac{1}{2}kA^2, \quad KE = \frac{1}{2}k(A^2 - x^2), \quad PE = \frac{1}{2}kx^2$$

The condition is KE = 1.25 × PE:

$$\frac{1}{2}k(A^2 - x^2) = 1.25 \times \frac{1}{2}kx^2$$

$$A^2 - x^2 = 1.25x^2$$

$$A^2 = 2.25x^2$$

$$x^2 = \frac{A^2}{2.25} = \frac{9}{2.25} = 4$$

$$x = 2 \text{ cm}$$

So, the answer is $$2$$ cm.