The amplitude of a particle executing SHM is 3cm . The displacement...

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

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

So, $KE = \dfrac{5}{4}PE$

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

$⇒ {A}^{2} - {x}^{2} = \dfrac{5}{4}{x}^{2}$

$⇒ \dfrac{9{x}^{2}}{4} = {A}^{2}$

$⇒ x = \dfrac{2}{3}A$

$∴ x = \dfrac{2}{3} \times 3cm$

$⇒ x = 2cm$