The amplitude of a particle executing S.H.M. is $3 \mathrm{~cm}$. The displacement at which its kinetic…

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
  1. $1 \mathrm{~cm}$
  2. $2 \mathrm{~cm}$
  3. $3 \mathrm{~cm}$
  4. $4 \mathrm{~cm}$

Solution

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

Asked in: MHT CET 2023 (10 May Shift 1)

Practice more Oscillations questions on Aicharya