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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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$12 \times 10^{- 3} J$

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

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

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

answer is C.

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Detailed Solution

$large y = ,3sin left( {0.2t} right);,m, = ,3 times {10^{ - 3}}kg$

A = 3m; $large omega , = ,0.2ra{d^{ - 1}}s$

$large y, = ,frac{A}{3}, = ,1m;,,KE, = ,frac{1}{2}m{omega ^2}left( {{A^2} - {y^2}} right)$

$large = ,frac{1}{2} times 3 times {10^{ - 3}} times {left( {0.2} right)^2} times left( {{3^2} - {1^2}} right)$

$large = ,frac{1}{2} times 3 times 4 times 4 times 2 times {10^{ - 5}}, = ,0.48 times {10^{ - 3}}J$

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