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)

87

Answer 2

The relation for kinetic energy of S.H.M. is given by $=\frac{1}{2} m \omega^{2}\left(a^{2}-y^{2}\right)$ ...(1) Potential energy is given by $=\frac{1}{2} m \omega^{2} y^{2}$ ...(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, $y =\frac{a}{2} \sqrt{3}=0.866 a$ $\approx 87 \%$ of amplitude