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The amplitude of a particle executing SHM is 4 cm. At the mean position, the speed of the particle is 16 cm/s. The distance of the particle from the mean position at which the speed of the particle becomes 8 $\sqrt{3}$ cm/s 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 mean position, velocity is maximum

$i . e ., v_{\max} = ωa \Rightarrow ω = \frac{v_{\max}}{a} = \frac{16}{4} = 4$

$∴ v = ω \sqrt{a^{2} - y^{2}} \Rightarrow 8 \sqrt{3} = 4 \sqrt{4^{2} - y^{2}}$

$\Rightarrow 192 = 16 (16 - y^{2}) \Rightarrow 12 = 16 - y^{2} \Rightarrow y = 2 c m$

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