Question

When the displacement of a particle executing SHM is one - fourth of its amplitude, what fraction of the total energy is the kinetic energy?

• A. $\frac{16}{15}$
• B. $\frac{15}{16}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

Answer: (B)

$\frac{15}{16}$

$x=\frac{A}{4}$

$v=\omega \sqrt{A^2-x^2}$

$v=\omega \sqrt{A^2-\frac{A^2}{16}}$

$v=\omega \sqrt{\frac{15A^2}{16}}$

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2\left(\frac{15A^2}{16}\right)=\frac{15}{16}\left(TE\right)$ $[TE=\frac{1}{2}m{\omega }^2{A}^2]$