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A particle executes linear simple harmon...

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

`(sqrt(5))/(pi)`

B

`(sqrt(5))/(2pi)`

C

`(4pi)/(sqrt(5))`

D

`(2pi)/(sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
C
`V=omega sqrt(A^(2)-x^(2))`
`a=x omega^(2)`
`v=a`
`omega sqrt(A^(2)-x^(2))=x omega^(2)`
`sqrt((3)^(2)-(2)^(2))=2((2pi)/(T))`
`sqrt(5)=(4pi)/(T)`
`T=(4pi)/(sqrt(5))`
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