A particle executing S.H.M. has velocities ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' at distances ' $x_1$ '…

A particle executing S.H.M. has velocities ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' at distances ' $x_1$ ' and ' $x_2$ ' respectively, from the mean position. Its frequency is
  1. $\frac{1}{2 \pi} \sqrt{\frac{V_1^2-V_2^2}{x_1^2-x_2^2}}$
  2. $2 \pi \sqrt{\frac{x_1^2-x_2^2}{V_1^2-V_2^2}}$
  3. $\frac{1}{2 \pi} \sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}$
  4. $2 \pi \sqrt{\frac{x_1^2-x_2^2}{V_2^2-V_1^2}}$

Solution

Particle velocities are $\begin{aligned} & V_1^2=\omega^2\left(A^2-x_1^2\right) ...(i)\\ & V_2^2=\omega^2\left(A^2-x_2^2\right)...(ii) \end{aligned}$
Subtracting (i) from (ii), $\begin{aligned} & V_2^2-V_1^2=\omega^2\left(x_1^2-x_2^2\right) \\ & \omega=\sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}} \end{aligned}$
As $\omega=2 \pi f$ we get, $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{~V}_2^2-\mathrm{V}_1^2}{\mathrm{x}_1^2-\mathrm{x}_2^2}}$ .

Asked in: MHT CET 2024 (16 May Shift 1)

Practice more Oscillations questions on Aicharya