A particle is attached to a vertical spring and is pulled down a distance 0.04m below its equilibrium position and is released from rest. The initial upward acceleration of the particle is `0.30 ms^(-2)`. The period of the oscillation is
A. 4.08 s
B. 1.92 s
C. 3.90 s
D. 2.29 s
A. 4.08 s
B. 1.92 s
C. 3.90 s
D. 2.29 s
Correct Answer – D
(d)`A=0.04m,omega^(2)A=0.3 ms^(-2)`
`therefore omega=2.74 “rad”s^(-1)`
Now, `T=(2pi)/(omega)=2.29 s`