Two parallel SHM’s represented by `X_(1) = 20 sin (8πt )cm` and `X_(2) = 10 sin (8πt + π/2 )cm `are superposed on a particle . Determine the amplitude of the resultant SHM.

Two waves represented by y_1=10 sin (2000 pi t) and y_2=10 sin (2000 pi t + pi//2) superposed at any point at a particular instant. The resultant amplitude is

If two SHMs are repersented by y_(1) = 10 sin (4 pi + pi//2) and y_(2) = 5 (sin 2 pi t +sqrt 8 cos 2 pi t) , compare their amplitudes .

x_(1) = 3 sin omega t x_(2) = 5 sin (omega t + 53^(@)) x_(3) = - 10 cos omega t Find amplitude of resultant SHM.

Two simple harmonic motions are given by y_(1)=A_(1)sinomegat andy_(2)=A_(2)sin(omegat+phi) are acting on the particles in the same direction .The resultant motion is S.H.M., its amplitude is

The equation of simple harmonic motion of two particle is x_1=A_1" "sin" "omegat and x_2=A_2" "sin(omegat+prop) . If they superimpose, then the amplitude of the resultant S.H.M. will be

x_(1)= 5 sin (omega t + 30^(@)) , x2 = 10 cos ( omega t) Find amplitude of resultant SHM.

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

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 linear SHM is represented by x=5sqrt(2) ("sin" 2pi t+ "cos" 2pi t) cm. What is the amplitude of SHM ?

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