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Question

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

A
12π3
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B
2π3
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C
2π3
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D
32π
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Solution

The correct option is B 2π3
Given,
A=2cm
x=1cm
The relation of velocity and displacement is given by
v=ωA2x2
Acceleration, a=ω2x
According to question,
ω2x=ωA2x2
ωx=A2x2
2πT×1=2212=41
T=2π3
The correct option is C.

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