A particle executing linear S.H.M. has veloci | vedclass

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(A) The velocity $V$ of a particle in $S.H.M.$ at a distance $x$ from the mean position is given by the formula: $V = \omega \sqrt{A^2 - x^2}$, where

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The velocity $V$ of a particle in $S.H.M.$ at a distance $x$ from the mean position is given by the formula: $V = \omega \sqrt{A^2 - x^2}$, where $\omega$ is the angular velocity and $A$ is the amplitude.Squaring both sides, we get: $V^2 = \omega^2(A^2 - x^2) = \omega^2 A^2 - \omega^2 x^2$.For the given conditions:$V_1^2 = \omega^2 A^2 - \omega^2 x_1^2$  --- $(1)$$V_2^2 = \omega^2 A^2 - \omega^2 x_2^2$  --- $(2)$Subtracting equation $(1)$ from equation $(2)$:$V_2^2 - V_1^2 = (\omega^2 A^2 - \omega^2 x_2^2) - (\omega^2 A^2 - \omega^2 x_1^2)$$V_2^2 - V_1^2 = \omega^2 x_1^2 - \omega^2 x_2^2$$V_2^2 - V_1^2 = \omega^2(x_1^2 - x_2^2)$Solving for $\omega$:$\omega^2 = \frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}$$\omega = \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$

$A$ particle executing linear $S.H.M.$ has velocities $V_1$ and $V_2$ at distances $x_1$ and $x_2$ respectively, from the mean position. Its angular velocity is:

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