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

Vedclass · Accepted Answer

$\frac{1}{2 \pi} \sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}$

Vedclass · Answer

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

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

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