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(D) The velocity $V$ of a particle in $S.H.M.$ at displacement $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.Maximum speed $V_{\max}$ occurs at the me

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$\frac{A\sqrt{15}}{4}$

Vedclass · Answer

(D) The velocity $V$ of a particle in $S.H.M.$ at displacement $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.Maximum speed $V_{\max}$ occurs at the mean position and is given by $V_{\max} = A\omega$.According to the problem,the speed $V$ is one-fourth of the maximum speed:$V = \frac{1}{4} V_{\max}$$\omega \sqrt{A^2 - x^2} = \frac{1}{4} (A\omega)$$\sqrt{A^2 - x^2} = \frac{A}{4}$Squaring both sides:$A^2 - x^2 = \frac{A^2}{16}$$x^2 = A^2 - \frac{A^2}{16} = \frac{15A^2}{16}$$x = \frac{A\sqrt{15}}{4}$

$A$ particle executing $S.H.M.$ starts from the mean position. Its amplitude is $A$ and time period is $T$. At what displacement is its speed one-fourth of the maximum speed?

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