Question

The total energy of the particle executing simple harmonic motion of amplitude $A$ is $100J$. At a distance of $0.707A$ from the mean position, its kinetic energy is

• A. 25 J
• B. 50 J
• C. 100 J
• D. 12.5 J
• E. 70 J

Answer 1

Answer: (B)

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