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Question

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A
52π
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B
4π5
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C
2π3
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D
5π
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Solution

The correct option is B 4π5

Velocity of particle executing S.H.M
v=ωa2x2
Acceleration, a=ω2x
x = displacement at any time interval
a = amplitude=3 cm
ω = angular frequency =2πT
T = Time period
Now, at x=2 cm
v=|a|
ω2×2=ω3222
ω=52
2πT=52
T=4π5 s


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