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A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in second is

(a) 8π/3

(b) 4π/3

(c) 3π/8

(d) 7π/3

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The correct option is answer a) 8π/3 

EXplaination ::

v=\(w\sqrt{A^2-x^2}\)   --{speed}

Acceleeation a=\(-\omega ^2x\)  

As \(|v|=|a|\) 

\(ω ={3\over4}⇒T = 2{π\over ω }= 8{π\over3}\)

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