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Q.

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 seconds) is

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a

4π3

b

8π3

c

7π3

d

3π8

answer is B.

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Detailed Solution

At this time (when x=4 cm) velocity and acceleration have same magnitudeGiven A=5 cm and  at displacement x=4cm, velocity = acceleration ⇒ωA2-x2=−ω2x ⇒52-42=ω(4) ⇒ω=34=2πT ⇒T=8π3
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