The amplitude of a particle executing SHM is $3 \mathrm{cm}$. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is: ______ $\mathrm{cm}$.

Let $x$ be the displacement at which its kinetic energy will be $25\%$ more than the potential energy.

Given here, $A=3 \mathrm{cm}$ and $KE=PE+\frac{PE}{4}$

So, $KE=\frac{5}{4}PE$

$\Rightarrow \frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{5}{4}\times \frac{1}{2}m{\omega }^2{x}^2$

$\Rightarrow A^2-x^2=\frac{5}{4}{x}^2$

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

$\Rightarrow x=\frac{2}{3}A$

$\therefore x=\frac{2}{3}\times 3 \mathrm{cm}$

$\Rightarrow x=2 \mathrm{cm}$