A particle is executing a linear simple harmonic motion. Let ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' are…

A particle is executing a linear simple harmonic motion. Let ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' are its speed at distance ' $x_1$ ' and ' $x_2$ ' from the equilibrium position. The amplitude of oscillation is
  1. $\sqrt{\frac{V_1^2 x_2^2-V_2^2 x_2^2}{V_1^2-V_2^2}}$
  2. $\sqrt{\frac{V_1^2-V_2^2}{V_1^2 x_2^2-V_2^2 x_1^2}}$
  3. $\sqrt{\frac{V_1^2 x_2^2-V_2^2 x_1^2}{V_1^2-V_2^2}}$
  4. $\sqrt{\frac{V_1^2 x_1^2-V_2^2 x_2^2}{V_1^2-V_2^2}}$

Solution

For S.H.M, velocity is given by, $\begin{aligned} & \quad V=\omega \sqrt{A^2-x^2} \Rightarrow V^2=\omega^2\left(A^2-x^2\right)...(i) \\ & \therefore \quad V_1^2=\omega^2\left(A^2-x_1^2\right) \\ & \text { and } V_2^2=\omega^2\left(A^2-x_2^2\right) ...[From(i)]\\ & \frac{V_1^2}{V_2^2}=\frac{\omega^2\left(A^2-x_1^2\right)}{\omega^2\left(A^2-x_2^2\right)} \\ & V_1^2\left(A^2-x_2^2\right)=V_2^2\left(A^2-x_1^2\right) \\ & A=\sqrt{\frac{V_1^2 x_2^2-V_2^2 x_1^2}{V_1^2-V_2^2}} \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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