Q. A point particle of mass 0.1 kg is executing SHM of amplitude 0.1m. When the particle passes through the mean position, its KE is $8 \times 10^{- 3}J.\ \ $The equation of motion of this particle, if its initial phase of oscillation is $45^{0}$ 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) $y = 0.1\sin\left( 4t - \frac{\pi}{4} \right)$
D) $y = 0.1\sin\left( 4t + \frac{\pi}{4} \right)$
Explanation
(d) KE at mean position$= \frac{1}{2}\ m\omega^{2}a^{2} = 8 \times 10^{- 3}$
or $\omega = \left( \frac{2 \times 8 \times 10^{- 3}}{ma^{2}} \right)^{1/2}$
$= \left\lbrack \frac{2 \times 8 \times 10^{- 3}}{0.1 \times (0.1)^{2}} \right\rbrack^{1/2} = 4$
Equation of SHM is,
$y = a\sin{(\omega t + \theta)} = 0.1\sin\left( 4t + \frac{\pi}{4} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.