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Q.

The displacement of a particle of mass 3gm executing S.H.M 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

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a

12 × 10³J

b

25 × 10^–3J

c

0.48 × 10^–3J

d

0.24 × 10^–3 J

answer is C.

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Detailed Solution

$\begin{array}{l} y = 3 \sin (0 .2 t); m = 3 \times 10^{- 3} kg \\ A = 3 m; ω = 0 .2 {rad}^{- 1} s \\ y = \frac{A}{3} = 1 m; KE = \frac{1}{2} {mω}^{2} (A^{2} - y^{2}) \\ = \frac{1}{2} \times 3 \times 10^{- 3} \times (0 .2)^{2} \times (3^{2} - 1^{2}) \\ = \frac{1}{2} \times 3 \times 4 \times 4 \times 2 \times 10^{- 5} = 0 .48 \times 10^{- 3} J \end{array}$

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