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The shortest distance travelled by a par...

The shortest distance travelled by a particle executing SHM from mean position in 2 s is equal to `(sqrt(3)//2)` times its amplitude. Determine its time period.

A

12s

B

10s

C

14s

D

15s

Text Solution

Verified by Experts

The correct Answer is:
A
`t=2 s, x=(sqrt(3))/(2) A,T =?`
Using x=A sin `omegat` , we get `(sqrt(3))/(2)A =A "sin"(2pi)/(T)xx2`
`implies "sin"(4pi)/(T)=(sqrt(3))/(2)="sin"(pi)/(3) implies (4pi)/(T)=(pi)/(3)`
`therefore T=12 s`
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