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A particle vibrating simple harmonically...

A particle vibrating simple harmonically has an acceleration of `16cm*s^(-2)` when it is at a distance of 4 cm from the mean position. Its time period is

A

1 s

B

2.572 s

C

3.142 s

D

6.028 s

Text Solution

Verified by Experts

The correct Answer is:
C
Acceleration of particle executing SHM at a distance x from equilibrium position, `a=omega^(2)x`
`therefore` Angular velocity, `omega=sqrt(a/x)=sqrt(16/4)=2rad//s`
Therefore, time period, `T=(2pi)/omega=(2pi)/2=pi=3.142s`
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