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A particle executes SHM of period 12s. T...

A particle executes SHM of period 12s. Two sec after it passes through the centre of oscillation, the velocity is found to be 3.142 cm `s^(-1)` find the amplitude and the length of the path.

A

6 cm , 12 cm

B

3 cm , 6 cm

C

24 cm , 48 cm

D

12 cm , 24 cm

Text Solution

Verified by Experts

The correct Answer is:
D
Let `y=Asinomegat`
then `v=(dy)/(dt)=omegaAcosomegat`
`=(2pi)/(T)A"cos"(2pi)/(T)t" "(becauseomega=(2pi)/(T))`
`therefore3.142=(2xx3.142)/(12)xxA"cos"(2pi)/(12)xx2`
`thereforeA=12cm`
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