A particle executes simple harmonic with an amplitude of 5 cm. When the particle is at 4 cm from the mean position the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is

Options

A. `(8pi)/(3)`
B. `(4pi)/(3)`
C. `(3)/(8pi)`
D. `(7)/(3pi)`

Answer

A

Explanation

Velocity of particle performing SHM,
`v= pm omega sqrt(A^(2)-y^(2))`
`therefore v= pm omega sqrt(25-16)`
`therefore v= pm 3 omega" ""…….."(1)`
and acceleration `a= pm omega^(2)y`
`therefore a= pm 4 omega^(2)" ""…….."(2)`
From equation (1) and (2),
`a= v`
`therefore pm 4 omega^(2) = pm 3 omega`
`therefore omega = (3)/(4)`
`therefore (2pi)/(T) = (3)/(4)`
`therefore T= (8pi)/(3)s`.