Question

When the kinetic energy of a body executing S.H.M. is $1/3$ of the potential energy. The displacement of the body is $x$ percent of the amplitude, where $x$ is :

• A. 33
• B. 87
• C. 67
• D. 50

Answer 1

Answer: (B)

The relation for kinetic energy of S.H.M. is given by
$=\frac{1}{2}m{\omega }^2\left(a^2-y^2\right)$
Potential energy is given by
$=\frac{1}{2}m{\omega }^2{y}^2$
Now for the condition of question and from eqs. (1) and (2)
$\frac{1}{2}m{\omega }^2\left(a^2-y^2\right)=\frac{1}{3}\times \frac{1}{2}m{\omega }^2{y}^2$
or $\frac{4}{6}m{\omega }^2{y}^2=\frac{1}{2}m{\omega }^2{a}^2$
or $y^2=\frac{3}{4}{a}^2$
So, $y=\frac{a}{2}\sqrt{3}=0.866a$
$\approx 87\%$ of amplitude