The amplitude of a particle executing S.H.M. is \(3 \mathrm{~cm}\). The displacement at which its kinetic energy will be \(25 \%\) more than the potential energy is

\(\text {Given : K.E }=\text {P.E }+\frac{25}{100} . P . E \)
\( \mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}+\frac{1}{4} \mathrm{P} . \mathrm{E} \)
\( \text {We know } \mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right) \text { and } \mathrm{P} \cdot \mathrm{E}=\) \(\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2 \)
\( \therefore \quad \mathrm{K} . \mathrm{E}=\frac{5}{4} \text {P.E } \)
\( \frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{5}{4}\left(\frac{1}{2} m \omega^2 x^2\right) \)
\( \mathrm{A}^2-\mathrm{x}^2=\frac{5}{4} \mathrm{x}^2 \)
\( \mathrm{~A}^2=\frac{9}{4} \mathrm{x}^2 \)
\( \therefore \quad \mathrm{A}=\frac{3}{2} \mathrm{x} \)
\( \therefore \quad \mathrm{x}=\mathrm{A} \times \frac{2}{3}=2 \mathrm{~cm}\)