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(C) The maximum velocity of a particle in simple harmonic motion is given by $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angul

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$2\sqrt{3} \, cm$

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(C) The maximum velocity of a particle in simple harmonic motion is given by $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.Given $v_{\max} = 10 \, cm/s$ and $a = 4 \, cm$,we have $\omega = \frac{v_{\max}}{a} = \frac{10}{4} = 2.5 \, rad/s$.The velocity $v$ at a displacement $y$ from the mean position is given by $v = \omega \sqrt{a^2 - y^2}$.Squaring both sides,we get $v^2 = \omega^2(a^2 - y^2)$.Rearranging for $y$,we get $y^2 = a^2 - \frac{v^2}{\omega^2}$,so $y = \sqrt{a^2 - \frac{v^2}{\omega^2}}$.Substituting the values $a = 4 \, cm$,$v = 5 \, cm/s$,and $\omega = 2.5 \, rad/s$:$y = \sqrt{4^2 - \frac{5^2}{(2.5)^2}} = \sqrt{16 - \frac{25}{6.25}} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \, cm$.

$A$ particle executes simple harmonic motion with an amplitude of $4 \, cm$. At the mean position,the velocity of the particle is $10 \, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \, cm/s$ is

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