Q. The displacement of a particle of mass 3 g executing simple harmonic motion is given by $Y = 3\ sin(0.2t)$ in SI units. The KE of the particle at a point which is at a distance equal to 1/3 of its amplitude from its mean position is
A) $12 \times 10^{- 3}\ J$
B) $25 \times 10^{- 3}\ J$
C) $0.48 \times 10^{- 3}\ J$
D) $0.24 \times 10^{- 3}\ J$
Explanation
(c) Displacement of particle in the case of SHM
$y = A\sin{(\omega t + \varnothing)}$ …(i)
$y = 3\sin{(0.2\ t)}$ …(ii)(given)
Comparing Eqs. (i) and (ii), we get
$A = 3,\ \omega = 0.2$
Now, particle distance $x = \frac{A}{3} = 1$
Kinetic energy in SHM
= $\frac{1}{2}m\omega^{2}(A^{2} - x^{2})$
= $\frac{1}{2} \times 3 \times 10^{- 3}(0.2)^{2}\lbrack 3^{2} - 1^{2}\rbrack$
$= 0.48 \times 10^{- 3}$ J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.