Four harmonics of equal frequencies and equal intensities I₀ have phase angles 0, 3 , 2 3 , . When they are superimposed, the intensity of the resultant wave is " nI₀ ". The value of n is (all phase angles are in radian and (60)^ = 1 2 )

Step-by-step solution

Let the amplitude of each harmonic wave be A₀ . The intensity of each wave is I₀ = k A₀^2 . The waves can be represented as vectors (phasors) with magnitudes A₀ and angles 0, 3 , 2 3 , . The x-component of the resultant amplitude is: A_x = A₀ (0) + A₀ ( 3 ) + A₀ ( 2 3 ) + A₀ ( ) A_x = A₀ ( 1 + 1 2 - 1 2 - 1 ) = 0 The y-component of the resultant amplitude is: A_y = A₀ (0) + A₀ ( 3 ) + A₀ ( 2 3 ) + A₀ ( ) A_y = A₀ ( 0 + 3 2 + 3 2 + 0 ) = 3 A₀ The resultant amplitude A_R is given by: A_R = A_x^2 + A_y^2 = 0 + ( 3 A₀)