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