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(C) For a particle in Simple Harmonic Motion $(SHM)$,the velocity $V$ at a displacement $x$ is given by the formula: $V = \omega \sqrt{A^2 - x^2}$,whe

Vedclass · Accepted Answer

$\sqrt{\frac{V_1^2 x_2^2-V_2^2 x_1^2}{V_1^2-V_2^2}}$

Vedclass · Answer

(C) For a particle in Simple Harmonic Motion $(SHM)$,the velocity $V$ at a displacement $x$ is given by the formula: $V = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $\omega$ is the angular frequency. Squaring both sides,we get $V^2 = \omega^2(A^2 - x^2)$.Applying this to the given conditions:$V_1^2 = \omega^2(A^2 - x_1^2)$  --- $(1)$$V_2^2 = \omega^2(A^2 - x_2^2)$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{V_1^2}{V_2^2} = \frac{A^2 - x_1^2}{A^2 - x_2^2}$Cross-multiplying gives:$V_1^2(A^2 - x_2^2) = V_2^2(A^2 - x_1^2)$$V_1^2 A^2 - V_1^2 x_2^2 = V_2^2 A^2 - V_2^2 x_1^2$Rearranging terms to solve for $A^2$:$A^2(V_1^2 - V_2^2) = V_1^2 x_2^2 - V_2^2 x_1^2$$A^2 = \frac{V_1^2 x_2^2 - V_2^2 x_1^2}{V_1^2 - V_2^2}$Taking the square root,we find the amplitude:$A = \sqrt{\frac{V_1^2 x_2^2 - V_2^2 x_1^2}{V_1^2 - V_2^2}}$

$A$ particle is executing a linear simple harmonic motion. Let $V_1$ and $V_2$ be its speeds at distances $x_1$ and $x_2$ from the equilibrium position,respectively. The amplitude of oscillation is

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