Question

The kinetic energy of a particle executing simple harmonic motion is $ 1/6^{th} $ of its maximum value at a distance of $ 5 \,cm $ from its equilibrium position. The amplitude of its motion is

• A. $ \sqrt{30}\, cm $
• B. $ \sqrt{10}\, cm $
• C. $ \sqrt{20}\, cm $
• D. $ \sqrt{5}\, cm $

Answer 1

Answer: (A)

The kinetic energy of a particle executing simple harmonic motion at a distance $x$ from its equilibrium position is
$K=\frac{1}{2}m\omega^{2} \left(A^{2}-x^{2}\right)$
where $m$ is the mass of the particle, $\omega$ is the angular frequency and $A$ is the amplitude of oscillation
The kinetic energy is maximum at equilibrium position $(x=0)$ and its maximum value $(K_{0})$ is
$K_{0}=\frac{1}{2} m\omega^{2}\,A^{2} \dots(i)$
At $x=5\,cm$,
$K=\frac{1}{2}m\omega^{2} \left(A^{2}-\left(5\,cm\right)^{2}\right)$
But $K=\frac{1}{6}K_{0}$(given)
$\therefore \frac{1}{2}m\omega^{2}\left(A^{2}-\left(5\,cm\right)^{2}\right)=\frac{1}{6}\left(\frac{1}{2}m\omega^{2}A^{2}\right)$
or $A^{2}-\left(5\,cm\right)^{2}=\frac{A^{2}}{6}$
or $A^{2}-\frac{A^{2}}{6}=\left(5\,cm\right)^{2}$
or $\frac{5}{6}A^{2}=\left(5 cm\right)^{2}$
or $A=\sqrt{\frac{6}{5}}\left(5\,cm\right)$
$=\sqrt{30}\,cm$