Two waves are represented by `y_(1)= a sin (omega t + ( pi)/(6)) and y_(2) = a cos omega t `. What will be their resultant amplitude

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

Two light waves are represented by y_(1)=a sin_(omega)t and y_(2)= a sin(omega t+delta) . The phase of the resultant wave is

Two waves are represented by the equations y_1=a sin omega t and y_2=a cos omegat . The first wave

When the waves y_(1) = A sin omega t and y_(2) = A cos omega t are superposed, then resultant amplitude will be

Two waves represented as y_(1) = A_(1) sin omega t and y_(2) = A_(2) cos omega t superpose at a point in space. Find out the amplitude of the resultant wave at that point .

Two waves are given by y_(1) = a sin (omega t - kx) and y_(2) = a cos (omega t - kx) . The phase difference between the two waves is

Two waves are represented by the equations y_(1)=a sin (omega t+kx+0.785) and y_(2)=a cos (omega t+kx) where, x is in meter and t in second The phase difference between them and resultant amplitude due to their superposition are

Two waves are represented by the equations y_(1)=a sin (omega t+kx+0.785) and y_(2)=a cos (omega t+kx) where, x is in meter and t in second The phase difference between them and resultant amplitude due to their superposition are

When two displacement represented by y_(1) = a sin (omega t) and y_(2) = b cos (omega t) are superimposed, the motion is