Question

A particle is performing SHM. When the displacement is one half of its amplitude, find what fraction of total energy are the kinetic and potential energies of the particle?

• A. $\frac{3}{4},\frac{1}{4}$
• B. $\frac{3}{8},\frac{5}{8}$
• C. $\frac{1}{3},\frac{2}{3}$
• D. $\frac{2}{3},\frac{1}{3}$

Answer 1

The correct option is A $\frac{3}{4},\frac{1}{4}$

For a particle executing SHM, We know that
Total energy of the particle $E=\frac{1}{2}m{\omega }^2{A}^2$
Kinetic energy of the particle $K=\frac{1}{2}m{\omega }^2(A^2-x^2)$
and Potential energy of the particle $U=\frac{1}{2}m{\omega }^2{x}^2$
From the data given in the question, $x=\frac{A}{2}$
$\therefore K=\frac{1}{2}m{\omega }^2(A^2-\frac{A^2}{4})=\frac{3}{4}\times \left(\frac{1}{2}m{\omega }^2{A}^2\right)$
$\Rightarrow \frac{K}{E}=\frac{3}{4}$
Similarly, $U=\frac{1}{2}m{\omega }^2\frac{A^2}{4}=\frac{1}{4}\times \left(\frac{1}{2}m{\omega }^2{A}^2\right)$
$\Rightarrow \frac{U}{E}=\frac{1}{4}$
Thus, option (a) is the correct answer.