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A particle is executing simple harmonic motion along a straight line. Acceleration of the particle is ‘a’ and its distance from the equilibrium position is ‘x’. Variation of a with x is shown in the graph. Then maximum velocity of the particle is

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8 cm/s

2 cm/s

4 cm/s

16 cm/s

answer is C.

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Detailed Solution

$a = - ω^{2} x ∴ ω^{2} = \frac{8}{2} \Rightarrow ω = 2 rad / \sec$

From the graph it is evident that amplitude of oscillation is A = 2 cm

$∴ V_{\max} = ωA = 2 \times 2 cm / s = 4 cm / \sec$

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