Question

The displacement of a particle of mass $3gm$ executing simple harmonic motion is given by $y=3sin\left(0.2t\right)$ in SI units. The kinetic energy of the particle at a point which is at a distance equal to $\frac{1}{3}$ of its amplitude from its mean position is

• A. $12$ x ${10}^{-3}J$
• B. $25$ x ${10}^{-3}J$
• C. $0.48$ x ${10}^{-3}J$
• D. $0.24$ x ${10}^{-3}J$

Answer 1

The correct option is C $0.48$ x ${10}^{-3}J$

$y=Asin\omega t$
$y=3sin\left(0.2t\right)$
$A=3m$
$\omega =0.2rad/sec$
$m=3\times {10}^{-3}kg$
$KE=\frac{1}{2}m{\omega }^2(A^2-x^2)$
$=\frac{1}{2}3\times {10}^{-3}\left(0.2{)}^2\right(3^2-(3\times \frac{1}{3}{)}^2)$
$=\frac{1}{2}\times 3\times {10}^{-3}\times (0.2{)}^2\times 8$
$=4.8\times {10}^{-3}$