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 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 rms value of the emf given by E = 8 sin omegat + 6 sin 2omegat .

The displacement of the interfaring light waves are y_1 =4 sin omega t and y_2=3sin (omegat +(pi)/( 2)) What is the amplitude of the resultant wave?

The displacement of two interfering light waves are given by y_(1)=3 sinomegat,y_(2)=4 sin(omegat+pi//2) . The amplitude of the resultant wave is

Two particle are executing simple harmonic motion. At an instant of time t their displacement are y_(1)=a "cos"(omegat) and y_(2)=a "sin" (omegat) Then the phase difference between y_(1) and y_(2) is

The length of a steel wire is l_(1) when the stretching force is T_(1) and l_(2) when the stretching force is T_(2) . The natural length of the wire is

A x and y co-ordinates of a particle are x=A sin (omega t) and y = A sin(omegat + pi//2) . Then, the motion of the particle is