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Q.
The total energy of the particle executing simple harmonic motion of amplitude $A$ is $100 \,J$. At a distance of $0.707\, A$ from the mean position, its kinetic energy is
We know that,
Total energy $E=\frac{1}{2} m \omega^{2} A^{2}=100\, J$
and kinetic energy $KE$
$=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$
$=\frac{1}{2} m \omega^{2}\left[A^{2}-\left(\frac{A}{\sqrt{2}}\right)^{2}\right]$
$=\frac{1}{2} m \omega^{2}\left[\frac{2 A^{2}-A^{2}}{2}\right]$
$=\frac{1}{2} m \omega^{2} \frac{A^{2}}{2}$
$=\frac{100}{2} \,J =50\, J$