A particle executes linear simple harmonic motion with an amplitude of 3cm. When the particle is at 2cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

Amplitude, A = 3cm
In Harmonic Motion, x = A sinωt, where ω = 2π/T
v = Aω cosωt
Acceleration, a = -Aω² sinωt
At x = 2cm
|v| = |a|
|Aω cosωt| = |-Aω² sinωt|
Asinωt = 2cm
3sinωt = 2
sinωt = 2/3
cosωt = √1 - (2/3)² = √5/3
From eqn. (1)
|3ω√5/3| = | -ω² x 2/3 |
ω = √5/2
2π/T = √5/2
T = 4π/√5