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. `(2pi)/(sqrt(3))s`
B. `(sqrt(3))/(2pi)s`
C. `2pi sqrt(3) s`
D. `(1)/(2pi sqrt(3))`s

Answer

A

Explanation

In liner simple harmonic motion , the velocity of particle is given by
`v = omega sqrt(A^(2) - x^(2))" "……..(i)`
where `,omega` = angular frequency
A = maximum displacement of amplitude
and x = displacement form mean position.
The acceleration of a particle in simple harmonic motion , (SHM) is given by
`a = omega^(2)x" "......(ii)`
Here , `" "x =(A)/(2)`
Also , `V = a" "("give")`
`omegasqrt((A^(2) - x^(2)))=omega^(2) x` [from Eqs. (i) and (ii) we, get ]
`sqrt((A^(2) -(A^(2))/(4))) = omega xx (A)/(2) rArr (sqrt(3)A)/(2) = omega xx (A)/(2)`
`rArr" "(2pi)/(T) = sqrt(3) " "[because omega = (2pi)/(T)]`
`rArr " "T = (2pi)/(sqrt(3))s`