Question

If a body is executing simple harmonic motion and its current displacements 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. $\sqrt{3}:1$
• D. $3:1$

Answer 1

Answer: (D)

For displacement
$y=\frac{\sqrt{3}A}{2}$
$PE=\frac{1}{2}m{\omega }^2{y}^2$
$\Rightarrow \frac{1}{2}m{\omega }^2{\left[\frac{\sqrt{3}A}{2}\right]}^2$
$KE=\frac{1}{2}m{\omega }^2{\left(A^2-y^2\right)}^2$
$\Rightarrow \frac{1}{2}m{\omega }^2\left(A^2-{\left[\frac{\sqrt{3}A}{2}\right]}^2\right)$
$\frac{PE}{KE}=\frac{\frac{1}{2}m{\omega }^2\times \frac{3A^2}{4}}{\frac{1}{2}m{\omega }^2\frac{A^2}{4}}$
$=\frac{3}{1}$