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
Velocity of particle executing S.H.M
v=ω√a2−x2
Acceleration, a=−ω2x
x = displacement at any time interval
a = amplitude=3 cm
ω = angular frequency =2πT
T = Time period
Now, at x=2 cm
v=|a|
⇒ω2×2=ω√32−22
⇒ω=√52
⇒2πT=√52
⇒T=4π√5 s