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Q.

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

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a

12π3

b

2π3

c

2π3

d

32π

answer is C.

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Detailed Solution

Velocity v=ωA2−x2  and acceleration  =ω2xNow given,  ω2x=ωA2−x2 ⇒ω2.1=ω22−12   ⇒ω=3  ∴T=2πω=2π3
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A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is