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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. Find the time period in seconds.

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-displacement relationship is given by
v=ωA2x2
Acceleration-displacement relationship is given by
|a|=ω2x
Given that,
|v|=|a|
ωA2x2=ω2x
ω2(A2x2)=ω4x2
Given that amplitude A=3 cm & required distance x=2 cm
ω2=A2x2x2=944=54
ω=52
T=4π5 seconds
(T=2πω)

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