A particle is performing a linear simpfe harmonic motion of amplitude 'A'. When it is midway between its mean and extreme position, the magnitudes of its velcoity and acceleration are equal. What is the periodic time of the motion ?

Options

A. `(sqrt(3))/(2pi)sec`
B. `(1)/(2pisqrt(3))sec`
C. `2 pi sqrt(3) sec`
D. `(2pi )/(sqrt(3)) sec`

Answer

D

Explanation

At `x =(A)/(2)`
The magnitude of velocity, `v= omega sqrt(A^(2)-x^(2))`
`v= omega sqrt(A^(2)-(A^(2))/(u))=(sqrt(3)Aomega)/(2)`
and the magnitude of acceleration `=omega^(2)x=omega^(2)((A)/(2))`
`therefore` It is given that
`omega^(2)((A)/(2))=(sqrt(3))/(2)A omega`
`omega=sqrt(3) therefore (2pi)/(T)=sqrt(3) therefore T=(2pi)/(sqrt(3)) s`