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The amplitude of a particle executing SHM is 4cm. At the mean position the speed of the particle is $16 {cms}^{- 1}$ . The distance of the particle from the mean position at which the speed of the particle becomes $8 \sqrt{3} {cms}^{- 1}$ , will be

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$2 \sqrt{3} cm$

$\sqrt{3} cm$

1 cm

2 cm

answer is D.

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Detailed Solution

At the mean position, the speed will be maximum

$ν_{\max} = 16 {cms}^{- 1} = Aω = 4 ω$

$So, ω = \frac{16}{4} = 4 rad / s$

$and v = ω \sqrt{A^{2} - y^{2}}$

$8 \sqrt{3} = 4 \sqrt{A^{2} - y^{2}}$

$or A^{2} - y^{2} = 12$

$or y^{2} = A^{2} - 12 = (4^{2}) - 12 = 4$

$or y = 2 cm$

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