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(C) The velocity of a particle in $S.H.M.$ at a displacement $x$ is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is

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$\frac{\sqrt{3}}{2}v$

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(C) The velocity of a particle in $S.H.M.$ at a displacement $x$ is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.At the mean position,$x = 0$,so the velocity $v = \omega \sqrt{a^2 - 0^2} = a\omega$.At a distance equal to half of the amplitude,$x = \frac{a}{2}$.Substituting this into the velocity formula: $V' = \omega \sqrt{a^2 - (\frac{a}{2})^2} = \omega \sqrt{a^2 - \frac{a^2}{4}} = \omega \sqrt{\frac{3a^2}{4}} = \frac{\sqrt{3}}{2}a\omega$.Since $v = a\omega$,we have $V' = \frac{\sqrt{3}}{2}v$.

The velocity at the mean position of a particle executing $S.H.M.$ is $v$. What is the velocity of the particle at a distance equal to half of the amplitude?

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