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The displacement of a particle of mass 3 gm executing SHM is given by y = 3 sin (0.2t) in S.I units. The K.E of a particle at a point which is at a distance equal to 1/3 of its amplitude from its mean position is:

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$12 \times 10^{3} J$

$0.24 \times 10^{- 3} J$

$25 \times 10^{- 3} J$

$0.48 \times 10^{- 3} J$

answer is D.

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Detailed Solution

Here, $y = 3 \sin (0.2 t)$ , then amplitude A is 3 m and $ω$ = 0.2

given displacement, x = $\frac{A}{3}$ = 1

kinetic energy at displacement x,

K.E. = $\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{2} 3 \times 10^{- 3} {(0.2)}^{2} (3^{2} - 1^{2})$ = 0.48 x $10^{- 3}$ J

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