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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 isA. 4.08 sB. 1.92 sC. 3.90 sD. 2.29 s |
Answer» 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` |
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