Q.

A particle executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.5 m/s2. The maximum velocity of the particle is (in m/s)

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a

0.01

b

0.05

c

0.5

d

0.25

answer is C.

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Detailed Solution

Acceleration A=ω2y ⇒ω=Ay=0.50.02=5Maximum velocity  vmax=aω=0.1×5=0.5
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A particle executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.5 m/s2. The maximum velocity of the particle is (in m/s)