Question

The kinetic energy of a particle executing simple harmonic motion is $1/6^{th}$ of its maximum value at a distance of $5cm$ 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 $x=5cm$,
$K=\frac{1}{2}m{\omega }^2\left(A^2-{\left(5cm\right)}^2\right)$
But $K=\frac{1}{6}{K}_0$(given)
$\therefore \frac{1}{2}m{\omega }^2\left(A^2-{\left(5cm\right)}^2\right)=\frac{1}{6}\left(\frac{1}{2}m{\omega }^2{A}^2\right)$
or $A^2-{\left(5cm\right)}^2=\frac{A^2}{6}$
or $A^2-\frac{A^2}{6}={\left(5cm\right)}^2$
or $\frac{5}{6}{A}^2={\left(5cm\right)}^2$
or $A=\sqrt{\frac{6}{5}}\left(5cm\right)$
$=\sqrt{30}cm$