The amplitude of the vibrating particle due to superposition of two `SHMs`,
`y_(1)=sin (omega t+(pi)/(3)) and y_(2)=sin omega t` is :

The resultant ampiltude of a vibrating particle by the superposition of the two waves y_(1) = a sin ( omega t + (pi)/(3)) and y_(2) = a sin omega t is :

Find the amplitude of the resultant wave produced due to interference of two waves given as, y_1 =A_1 sin (omega)t y_2 = A_2 sin [ (omega)t + (phi) ]

The ratio of amplitudes of following SHM is x_(1) = A sin omega t and x_(2) = A sin omega t + A cos omega t

The resultant amplitude due to superposition of three simple harmonic motions x_(1) = 3sin omega t , x_(2) = 5sin (omega t + 37^(@)) and x_(3) = - 15cos omega t is

The resultant amplitude due to superposition of three simple harmonic motions x_(1) = 3sin omega t , x_(2) = 5sin (omega t + 37^(@)) and x_(3) = - 15cos omega t is

Wave equations of two particles are given by y_(1)=a sin (omega t -kx), y_(2)=a sin (kx + omega t) , then

Two particle A and B execute simple harmonic motion according to the equation y_(1) = 3 sin omega t and y_(2) = 4 sin [omega t + (pi//2)] + 3 sin omega t . Find the phase difference between them.

Two wave are represented by equation y_(1) = a sin omega t and y_(2) = a cos omega t the first wave :-

Determine resultant amplitude after super position of given four waves with help of phasor diagram. y_(1) = 15 sin omega t mm,y_(2) = 9 sin (omega t -pi//2) mm,y_(3)=7 sin (omega t +pi//2)mm & y_(4) = 13 sin omega t mm .

Four simple harmonic vibrations x_(1) = 8 sin omega t, x_(2) = 6 sin (omega t + pi//2), x_(3) = 4 sin(omega t + pi) and x_(2) = 2 sin (omega t + 3pi //2) are superimposed on each other. The resulting amplitude is