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Questions  

A particle is executing simple harmonic motion along X-axis with its equilibrium position at the origin. Variation of its velocity with displacement is as shown in the graph. Then maximum acceleration of the particle is

a
103m/s2
b
102m/s2
c
10m/s2
d
1m/s2

detailed solution

Correct option is A

From the v-x graph , amplitude A = 0.1mMaximum velocity Vmax=Aω=10m/s⇒ω=100.1rad/s∴Maximum accelaration amax=Aω2=0.1×1002m/s2=103m/s2

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