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: $.............cm$.
JEE MAIN 2023, Medium
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$K E=P E+\frac{P E}{4}$
$K E=\frac{5}{4} P E$
$\frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{5}{4} \times \frac{1}{2} m \omega^2 x^2$ ${\left[ v =\omega \sqrt{ A ^2- x ^2}\right]}$
$A^2-x^2=\frac{5}{4} x^2$
$\frac{9 x^2}{4}=A^2$
$x=\frac{2}{3} A$
$\therefore x=\frac{2}{3} \times 3\,cm$
$x=2\,cm$
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