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

A particle executes simple harmonic with an amplitude of 5 cm. When th

1.	A particle executes simple harmonic with an amplitude of 5 cm. When the particle is at 4 cm from the meanposition the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is
Answer» `(8pi)/(3)` `(4pi)/(3)` `(3)/(8pi)` `(7)/(3pi)` SOLUTION :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`.