Question

In a SHM, when the displacement is one half the amplitude, what fraction of the total energy is kinetic?

• A. Zero
• B. 3/4
• C. 1/4
• D. 1/2

Answer 1

Answer: (B)

In case of SHM

$KE=\frac{1}{2}m{\omega }^2(a^2-x^2)$

and at $x=a/2$

$KE=\frac{1}{2}m{\omega }^2[a^2-(a/2{)}^2]$

$=\frac{3}{4}\left[\frac{1}{2}m{\omega }^2{a}^2\right]$

Total energy = $\frac{1}{2}m{\omega }^2{a}^2$

$\therefore$ Fraction of total energy

$=\frac{KE}{Totalenergy}=\frac{\frac{3}{4}\left[\frac{1}{2}m{\omega }^2{a}^2\right]}{\frac{1}{2}m\omega a^2}=\frac{3}{4}$