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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)/(2x)`

C

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

D

`(2pi)/(sqrt3)`

Text Solution

Verified by Experts

The correct Answer is:
C
A =3 cm,
Given, velocity is equal to acceleration at x = 2 cm
`v = omegasqrt(A^(2) - x^(2)), & a = omega^(2)x`
`omega^(2)x = omegasqrt(9-4)`
`omegax = sqrt(5) rArr (2pi)/(T)(2) = sqrt(5)`
`T = (4pi)/(sqrt(5))`
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