Question

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

• A. 25 J
• B. 50 J
• C. 100 J
• D. 12.5 J

Answer 1

Answer: (B)

50 J

Answer 2

We know that,
Total energy $E=\frac{1}{2} m \omega^{2} A^{2}=100\, J$
and kinetic energy $KE$
$=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$
$=\frac{1}{2} m \omega^{2}\left[A^{2}-\left(\frac{A}{\sqrt{2}}\right)^{2}\right]$
$=\frac{1}{2} m \omega^{2}\left[\frac{2 A^{2}-A^{2}}{2}\right]$
$=\frac{1}{2} m \omega^{2} \frac{A^{2}}{2}$
$=\frac{100}{2} \,J =50\, J$