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

Answer 1

The correct option is B $50J$

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(A^2-y^2)$
Now, position of particle from its mean position $y=0.707A=\frac{A}{\sqrt{2}}$
$KE=\frac{1}{2}m{\omega }^2[A^2-{\left(\frac{A}{\sqrt{2}}\right)}^2]$
$KE=\frac{1}{2}m{\omega }^2\left[\frac{2A^2-A^2}{2}\right]$
$KE=\frac{1}{2}m{\omega }^2\frac{A^2}{2}$
$KE=\frac{100}{2}J=50J$