The ends of a stretched wire of length `L` are fixed at `x = 0 and x = L`. In one experiment, the displacement of the wire is `y_(1) = A sin(pi//L) sin omegat` and energy is `E_(1)` and in another experiment its displacement is `y_(2) = A sin (2pix//L ) sin 2omegat` and energy is `E_(2)`. Then

The ends of a stretched wire of length L are fixed at x = 0 and x = L, In one experiment, the displacement of wire is y_(1) = A sin (pi x//L) sin omegat and energy is E_(1) and in another experiment its displacement is y_(2)= A sin (2pix//L) sin 2 omegat and energy is E_(2) . Then

The ends of a stretched wire of length 'L' are fixed at x = 0 and x = L. In one experiment, the displacement of the wire is y_(1) = A "sin"(pi x)/(L) sin wt and the energy is E_(1) and in another experiment its displacement is y_(2)=A "sin" (2pi x)/(L) sin 2 wt and the energy E_(2) then

The ends of a stretched wire of length l are fixed at x = 0 and x = l. in one experiment, the displacement of the wire is y_(1) = a sin ((pi x)/(l)) (sin omega t) and energy E_(1) and in another experiment, its displacement is y_(2) = a sin ((3pi x)/(l)) (sin 3 omega t) and energy is E_(2) then

The displacement of the interfering light waves are y_1 = 4 sin omegat and y_2 = 3 sin (omegat + pi//2) . The amplitude of the resultant wave is

When two displacement represented by y_1 = a sin t (omegat) and y_2 = b cos t (omegat) are superimposed the motion is

Displacement of two light waves are e_(1)=4 sin omega t and e_(2)= 3 sin (omegat + pi/2) . so amplitude of resultant wave = .....