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

Options

A. `sqrt(5)/(2pi)`
B. `(4pi)/(sqrt(5))`
C. `(2pi)/(sqrt(3))`
D. `(sqrt(5))/(pi)`

Answer

B

Explanation

Given, `A=3cm, x=2cm`
The velocity of a particle in simple harmonic motion is given as
`v=omegasqrt(A^(2)-x^(2))` and magnitude of its acceleration is
`a=omega^(2)x`
Given, `|v|=|a|`
`therefore omegasqrt(A^(2)-x^(2))=omega^(2)x`
`omegax=sqrt(A^(2)-x^(2))=oromega^(2)x^(2)=A^(2)-x^(2)`
`omega^(2)=(A^(2)-x^(2))/(x^(2))=(9-4)/(4)=(5)/(4)`
`omega=sqrt(5)/(2)`
Time period, `T=(2pi)/(omega)=2pi.(2)/(sqrt(5))=(4pi)/(sqrt(5))s`