(C) At the mean position, the velocity is maximum, given by $v_{\max} = A\omega$.Given $A = 4 \,cm$ and $v_{\max} = 12 \,cm/s$.$\omega = \frac{v_{\max

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(C) At the mean position, the velocity is maximum, given by $v_{\max} = A\omega$.Given $A = 4 \,cm$ and $v_{\max} = 12 \,cm/s$.$\omega = \frac{v_{\max}}{A} = \frac{12}{4} = 3 \,rad/s$.The velocity $v$ at any displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.Squaring both sides: $v^2 = \omega^2 (A^2 - x^2)$.Rearranging for $x$: $x^2 = A^2 - \frac{v^2}{\omega^2}$.Substituting the values $A = 4$, $v = 6$, and $\omega = 3$:$x^2 = 4^2 - \frac{6^2}{3^2} = 16 - \frac{36}{9} = 16 - 4 = 12$.$x = \sqrt{12} = 2\sqrt{3} \,cm$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

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$A$ particle is performing $S.H.M.$ with an amplitude $4 \,cm$. At the mean position, the velocity of the particle is $12 \,cm/s$. When the speed of the particle becomes $6 \,cm/s$, the distance of the particle from the mean position is:

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A
$\sqrt{3} \,cm$
B
$\sqrt{6} \,cm$
C
$2 \sqrt{3} \,cm$
D
$2 \sqrt{6} \,cm$

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Velocity of Simple Harmonic Motion

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$A$ particle is performing $S.H.M.$ with an amplitude $4 \,cm$. At the mean position, the velocity of the particle is $12 \,cm/s$. When the speed of the particle becomes $6 \,cm/s$, the distance of the particle from the mean position is:

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