A particle is subjected to two `SHMs x_(1) = A_(1) sin omegat` and `x_(2) = A_(2)sin (omegat +(pi)/(4))`. The resultant `SHM` will have an amplitude of

A particle is subjected to two S H Ms x_(1) = A_(1) sin wt and x_(2) = A_(2) sin (w t + pi//4) . The resultant S H M will have an amplitude of

A particle is subjected to SHM as given by equations x_1 = A_1 sin omegat and x_2 = A_2 sin (omega t + pi//3) .The maximum acceleration and amplitude of the resultant motion are it a_("max") and A ,respectively , then

A particle is subjected to two SHMs simultaneously X_(1) = a_(1) sinomegat and X_(2) = a_(2)sin(omegat + phi) Where a_(1) = 3.0 cm, a_(2) = 4.0 cm Find resultant amplitude if the phase difference phi has value 30^(0)

A particle is subjected to SHM as given by equation x_1=Asinomegat and x_2=A_2sin(omegat+(pi)/(3)) . The maximum acceleration and amplitude of the resultant motion are a_(max) and A , respectively Then.

A particle executes two types of SHM. x_(1) = A_(1) sin omega t " and "x_(2) = A_(2) sin [omega t+(pi)/(3)] , then find the maximum acceleration of the particle.