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» Solution :`y = a SIN omegat , (SQRT(3//2))a = sin omegat, sin omegat = sqrt(3//2)= sinpi//3, t = 2 s, (2pi//T) = pi//3 , T= 12S`