Question

A point particle of mass $0.1 \,kg$ is executing SHM of amplitude $0.1 \,m$. When the particle passes through the mean position, its kinetic energy is $8 \times 10^{-3} J$. The equation of motion of this particle, if its initial phase of oscillation is $45^{\circ}$, is

• A. $y=0.1 \sin \left(\frac{r}{4}+\frac{\pi}{4}\right)$
• B. $y=0.1 \sin \left(\frac{t}{2}+\frac{\pi}{4}\right)$
• C. $y=0.1 \sin \left(4 t-\frac{\pi}{4}\right)$
• D. $y=0.1 \sin \left(4 t+\frac{\pi}{4}\right)$

Answer 1

Answer: (D)

Kinetic energy at mean position $=\frac{1}{2} m \omega^{2} a^{2}=8 \times 10^{-3}$
Or $\omega=\left(\frac{2 \times 8 \times 10^{-3}}{m a^{2}}\right)^{1 / 2}=\left[\frac{2 \times 8 \times 10^{-3}}{0.1 \times(0.1)^{2}}\right]^{1 / 2}=4$
Equation of SHM is,
$y=a \sin (\omega t+\theta)=0.1 \sin \left(4 t+\frac{\pi}{4}\right)$