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.

Given amplitude $$A = 3\text{ cm}$$.

In simple harmonic motion, total energy $$E$$ is constant and given by the formula:
$$E = \frac12\,k\,A^2 \quad -(1)$$

At a displacement $$x$$, potential energy $$PE$$ is given by:
$$PE = \frac12\,k\,x^2 \quad -(2)$$

Kinetic energy $$KE$$ at displacement $$x$$ is:
$$KE = E - PE \quad -(3)$$

The problem states that kinetic energy is 25% more than potential energy. This means:
$$KE = 1.25\,PE = \frac54\,PE \quad -(4)$$

Substitute (3) into (4):
$$E - PE = \frac54\,PE$$
Substitute (1) and (2) into this equation:
$$\frac12\,k\,A^2 - \frac12\,k\,x^2 = \frac54\;\Bigl(\frac12\,k\,x^2\Bigr)$$

Cancel common factor $$\frac12\,k$$ from both sides:
$$A^2 - x^2 = \frac54\,x^2 \quad -(5)$$

Rearrange equation (5):
$$A^2 = x^2 + \frac54\,x^2 = \frac94\,x^2$$

Therefore
$$x^2 = \frac49\,A^2 \quad\Longrightarrow\quad x = \frac23\,A$$

Substitute $$A = 3\text{ cm}$$:
$$x = \frac23 \times 3\text{ cm} = 2\text{ cm}$$

Final answer: The displacement at which kinetic energy is 25% more than potential energy is $$2\text{ cm}$$.