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 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 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 values (a) 0^(0) , (b) 60^(0) , (c) 90^(0)

x_(1) = 5 sin (omegat + 30^(@)) x_(2) = 10 cos (omegat) Find amplitude of resultant SHM .

x_(1)=A sin (omegat-0.1x) and x_(2)=A sin (omegat-0.1 x-(phi)/2) Resultant amplitude of combined wave is

(a) A particle is subjected to two simple harmonic motions x_1=A_1sin omegat and x_2=A_2 sin (omegat+pi//3) . Find (i) the displacement at t=0 , (ii) the maximum speed of the particle and (iii) the maximum acceleration of the particle. (b) A particle is subjected to two simple harmonic motions, one along the x-axis and the other on a line making an angle of 45^@ with the x-axis. The two motions are given by xy=x_0sinomegat and s=s_0sin omegat Find the amplitude of the resultant motion.

Two waves y_(1) =A_(1) sin (omega t - beta _(1)), y_(2)=A_(2) sin (omega t - beta_(2) Superimpose to form a resultant wave whose amplitude is

x_(1) = 3 sin omega t ,x_(2) = 4 cos omega t Find (i) amplitude of resultant SHM, (ii) equation of the resultant SHM.

Equations y_(1) A sinomegat and y_(2) = A/2 sin omegat + A/2 cos omega t represent S.H.M. The ratio of the amplitudes of the two motions is

x_(1) = 3 "sin" omega t , x_(2) = 4 "cos' omega t Find (i) amplitude of resultant SHM. (ii) equation of the resultant SHM.