A body of mass 5, g is executing S.H.M. about | vedclass

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(C) Given: Amplitude $a = 10\, cm$,Maximum velocity $v_{\max} = 100\, cm/s$.We know that $v_{\max} = a\omega$.Therefore,$\omega = \frac{v_{\max}}{a} =

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$5\sqrt{3}\, cm$

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(C) Given: Amplitude $a = 10\, cm$,Maximum velocity $v_{\max} = 100\, cm/s$.We know that $v_{\max} = a\omega$.Therefore,$\omega = \frac{v_{\max}}{a} = \frac{100}{10} = 10\, rad/s$.The velocity $v$ at a displacement $y$ is given by the formula $v = \omega \sqrt{a^2 - y^2}$.Substituting the given values: $50 = 10 \sqrt{10^2 - y^2}$.Dividing by $10$: $5 = \sqrt{100 - y^2}$.Squaring both sides: $25 = 100 - y^2$.$y^2 = 100 - 25 = 75$.$y = \sqrt{75} = 5\sqrt{3}\, cm$.

$A$ body of mass $5\, g$ is executing $S.H.M.$ about a point with amplitude $10\, cm$. Its maximum velocity is $100\, cm/s$. Its velocity will be $50\, cm/s$ at a distance of:

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