A particle is subjected to two simple harmonic motions.
`x_(1) = 4.0 sin (100pi t)` and `x_(2) = 3.0 sin(100pi t + (pi)/(3))`
Find
(a) the displacement at `t = 0`
(b) the maximum speed of the particle and
(c ) the maximum acceleration of the particle.

Explanation

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