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A particle executes linear simple harmon...

A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 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

`(1)/(2pisqrt(3))`

B

`2pisqrt(3)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
(c) Velocity , `v=omegasqrt(A^(2)-x^(2))` and acceleration `=omega^(2)x`
Now given, `omega^(2)x=omegasqrt(A^(2)-x^(2))`
`impliesomega^(2).1=omegasqrt(2^(2)-1^(2))`
`implies omega=sqrt(3)`
`therefore T=(2pi)/(omega)=(2pi)/(sqrt(3))`
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