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.

The shortest distance travelled by a particle executing SHM from mean

1.	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.
Answer» `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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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.

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