Q118 / 150 · MHT CET 2021 (23 Sep Shift 1) · Physics

A body is executing S.H.M. under the action of force having maximum magnitude $50 \mathrm{~N}$. When its energy is half kinetic and half potential, the magnitude of the force acting on the particle is

  • A
    $\frac{25}{\sqrt{2}} \mathrm{~N}$
  • B
    $50 \mathrm{~N}$
  • C
    $25 \mathrm{~N}$
  • Correct answer
    D
    $25 \sqrt{2} \mathrm{~N}$

Solution

Potential energy is a half of the total energy $$ \begin{aligned} & \therefore \frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2=\frac{1}{2}\left[\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2\right] \\ & \therefore \mathrm{x}^2=\frac{\mathrm{A}^2}{2} \text { or } \mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}} \end{aligned} $$ Maximum force $F_m=m \omega^2 A$ Force at a distance $\mathrm{x}, \mathrm{F}^{\prime}=\mathrm{m} \omega^2 \mathrm{x}$ $$ \begin{aligned} & \therefore \frac{F^{\prime}}{F_m}=\frac{x}{A}=\frac{1}{\sqrt{2}} \\ & \therefore F^{\prime}=\frac{F_m}{\sqrt{2}}=\frac{50}{\sqrt{2}}=25 \sqrt{2} N \end{aligned} $$

Hence, the correct answer is option: 'D'.

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