Question

If a body is executing simple harmonic motion and its current displacement is $\frac{\sqrt{3}}{2}$ times the amplitude from its mean position, then the ratio between potential energy and kinetic energy is:

• A. $3:2$
• B. $2:3$
• C. $\sqrt{3}:2$
• D. $3:1$

Answer 1

The correct option is D $3:1$

Let the amplitude of oscillation be $A$.

Thus the displacement of body $x=\frac{\sqrt{3}}{2}A$

$\therefore$ Potential energy $P.E=\frac{1}{2}k{x}^2$ where $k=m{\omega }^2$

$⟹$ $P.E=\frac{1}{2}m{\omega }^2\times \frac{3}{4}{A}^2$

Kinetic energy $K.E=\frac{1}{2}m{\omega }^2(A^2-x^2)$

$⟹$ $K.E=\frac{1}{2}m{\omega }^2\times (A^2-\frac{3}{4}{A}^2)=\frac{1}{2}m{\omega }^2\times \frac{1}{4}{A}^2$

$\therefore$ $\frac{P.E}{K.E}=\frac{\frac{3}{4}}{\frac{1}{4}}=3$

$⟹$ $P.E:K.E=3:1$