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(D) The displacement of a particle in $SHM$ is given by $x(t) = A \sin(\omega t)$,where $\omega = \sqrt{K/m}$.The velocity of the particle is $v(t) =

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$\frac{K^{3/2} A^2}{2\sqrt{m}}$

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(D) The displacement of a particle in $SHM$ is given by $x(t) = A \sin(\omega t)$,where $\omega = \sqrt{K/m}$.The velocity of the particle is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t)$.The restoring force is $F(t) = -Kx = -KA \sin(\omega t)$.The power supplied by the restoring force is $P = F \cdot v = (-KA \sin(\omega t)) \cdot (A\omega \cos(\omega t))$.$P = -KA^2 \omega \sin(\omega t) \cos(\omega t) = -\frac{1}{2} KA^2 \omega \sin(2\omega t)$.The magnitude of power is $|P| = \frac{1}{2} KA^2 \omega |\sin(2\omega t)|$.The maximum power occurs when $|\sin(2\omega t)| = 1$,so $P_{max} = \frac{1}{2} KA^2 \omega$.Substituting $\omega = \sqrt{K/m}$,we get $P_{max} = \frac{1}{2} KA^2 \sqrt{\frac{K}{m}} = \frac{K^{3/2} A^2}{2\sqrt{m}}$.

$A$ particle of mass $m$ is performing linear simple harmonic motion. Its equilibrium position is at $x = 0$,the force constant is $K$,and the amplitude of $SHM$ is $A$. The maximum power supplied by the restoring force to the particle during $SHM$ will be:

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