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A particle executes simple harmonic motion along a straight line with maximum velocity 4 m/s and maximum acceleration $2 π m / s^{2}$ . If the particle starts its motion from mean position, then the minimum time after which its speed will be 2 m/s is

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$\frac{3}{2} \sec$

$\frac{2}{3} \sec$

$1 / 2 \sec$

$1 \sec$

answer is B.

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Detailed Solution

$ω = \frac{2 π}{T} = \frac{ω^{2} A}{ωA} = \frac{a_{\max}}{V_{\max}} = \frac{2 π}{4} \Rightarrow T = 4 \sec$

Now $V = V_{m} . cosωt$

$\Rightarrow \cos \frac{2 π}{T} t = \frac{2}{4} \Rightarrow \frac{2 πt}{T} = \frac{π}{3} \Rightarrow t = \frac{4}{6} s = \frac{2}{3} S$

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