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

For a particle executing SHM, magnitude of velocity of particle when it is at displacement x from mean position                 =ω A2-x2Also, magnitude of acceleration of particle in SHM =ω2xGiven, when x = 2 cm                Magnitude of velocity = Magnitude of acceleration⇒               ω A2-x2=ω2x⇒                               ω A2-x2x=9-42⇒ Angular velocity, ω=52∴Time period of motion,                                  T=2πω=4π5s