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A body of mass $1 \mathrm{~kg}$ is executing simple harmonic motion. Its displacement $y(\mathrm{~cm})$ at $t$ seconds is given by $y=6 \sin (100 t+\pi / 4)$. Its maximum kinetic energy is
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The correct answer is:
$18 \mathrm{~J}$
So $a=6 \mathrm{~cm} \omega=100 \mathrm{rad} / \mathrm{sec}$
$K_{\max }=\frac{1}{2} m \omega^2 a^2=\frac{1}{2} \times 1 \times(100)^2 \times\left(6 \times 10^{-2}\right)^2=18 \mathrm{~J}$
$K_{\max }=\frac{1}{2} m \omega^2 a^2=\frac{1}{2} \times 1 \times(100)^2 \times\left(6 \times 10^{-2}\right)^2=18 \mathrm{~J}$
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