The velocity at the mean position of a partic | vedclass

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(C) The velocity of a particle in $S.H.M.$ at the mean position is given by $v = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequ

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

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(C) The velocity of a particle in $S.H.M.$ at the mean position is given by $v = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.At a distance $y = \frac{a}{2}$,the velocity $v'$ is given by the formula $v' = \omega \sqrt{a^2 - y^2}$.Substituting $y = \frac{a}{2}$ into the formula:$v' = \omega \sqrt{a^2 - (\frac{a}{2})^2}$$v' = \omega \sqrt{a^2 - \frac{a^2}{4}}$$v' = \omega \sqrt{\frac{3a^2}{4}}$$v' = \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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