For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at certain displacement…

For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at certain displacement ' $\mathrm{x}$ ' from the mean position. If ' $\mathrm{A}$ ' is the amplitude of S.H.M the value of ' $x$ ' is
  1. $\frac{\mathrm{A} \sqrt{2}}{3}$
  2. $\mathrm{A} \sqrt{3}$
  3. $\frac{2 \sqrt{2} \mathrm{~A}}{3}$
  4. $\frac{\mathrm{A}}{\sqrt{2}}$

Solution

Potential energy: $U=\frac{1}{2} m \omega^2 x^2$ and Kinetic energy: $K=\frac{1}{2} m \omega^2\left(A^2-x^2\right)$ Given: Potential energy $=8 \times$ Kinetic energy $\begin{aligned} & \mathrm{U}=8 \mathrm{~K} \\ & \frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2=8 \times \frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right) \\ & \mathrm{x}^2=8 \mathrm{~A}^2-8 \mathrm{x}^2 \\ & 9 \mathrm{x}^2=8 \mathrm{~A}^2 \\ & \mathrm{x}^2=\frac{8 \mathrm{~A}^2}{9} \\ & \mathrm{x}=\frac{2 \sqrt{2} \mathrm{~A}}{3} \end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 1)

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