Question

The frequency of oscillation is $\left(\frac{10}{\pi}\right)$ (in $Hz$ ) of a particle of mass $0.1 \,kg$ which executes SHM along $x$ -axis. The kinetic energy is $0.3\, J$ and potential energy is $0.2\, J$ at position $x=0.02 \,m$. The potential energy is zero at mean position. Find the amplitude of oscillations (in meters)

• A. $\frac{1}{2 \sqrt{10}}$
• B. $\frac{1}{\sqrt{10}}$
• C. $\sqrt{10}$
• D. $2 \sqrt{10}$

Answer 1

Answer: (A)

$\omega=2 \pi f=\left(\frac{K}{m}\right)^{1 / 2}$
$K=(2 \pi f)^{2} m$
$\frac{1}{2} K A^{2}=$ total energy of oscillation $=0.5 \,J$
$A=\sqrt{\frac{1.0}{K}}=\frac{1}{2 \pi f} \sqrt{\frac{1.0}{0.1}}=\frac{1}{2 \sqrt{10}}$