A partical is subjucted to two simple harmonic motions
`x_(1) = A_(1) sin omega t`
and `x_(2) = A_(2) sin (omega t + pi//3)`
Find(a) the displacement at`t = 0`, (b) the maximum speed of the partical and ( c) the maximum acceleration of the partical.

Explanation

a. At `t = 0, x_(1) = A_(1) sin omega t = 0`
and `x_(2) = A_(2) sin (omega t + pi//3)`
`= A_(2) sin ((pi)/(3)) = (A_(2) sqrt3)/(2)`
Thus, the resultant displacement at `t = 0` is
`x = x_(1) + x_(2) A_(2) (sqrt3)/(2)`
b. The resultant of the two motion is a simple harmonic motion of the resultant motion is
`A = sqrt(A_(1)^(2) + A_(2)^(2) + 2 A_(1)A_(2) (cos pi//3)) = sqrt(A_(1)^(2) + A_(2)^(2) + A_(1)A_(2))`
The maximum speed is
`u_(max) = A omega = omega sqrt(A_(1)^(2) + A_(2)^(2) + 2 A_(1)A_(2))`
c.The maximum acceleration is
`a_(max) = A omega^(2) = omega^(2) sqrt(A_(1)^(2) + A_(2)^(2) + 2 A_(1)A_(2))`