A particle executes a simple harmonic motion of amplitude $\mathrm{A}$. The distance from the mean position at which its kinetic energy is equal to its potential energy is

The kinetic energy $K$ and potential energy $U$ of a particle executing simple harmonic motion with amplitude $A$ at a distance $x$ from the mean position are given by $K = \dfrac{1}{2} k(A^2 - x^2)$ and $U = \dfrac{1}{2} k x^2$, where $k$ is the force constant.

Given that the kinetic energy is equal to the potential energy, we set $K = U$:

$\dfrac{1}{2} k(A^2 - x^2) = \dfrac{1}{2} k x^2$

Simplifying the equation:

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

$A^2 = 2x^2$

$x^2 = \dfrac{A^2}{2}$

$x = \dfrac{A}{\sqrt{2}}$

Using the value $\sqrt{2} \approx 1.414$, we calculate:

$x = \dfrac{A}{1.414} \approx 0.707 A$

Rounding to two decimal places, we get $x = 0.71 A$.

Answer: 0.71 A