Superposition of Four Harmonic Waves with Phase Differences

To determine the intensity of the resultant wave, we can use the method of phasor addition (or vector addition of amplitudes).

Let the amplitude of each harmonic wave be $A_0$. Since the intensity is directly proportional to the square of the amplitude ($I \propto A^2$), the intensity of an individual wave is: $I_0 \propto A_0^2$

The phase angles of the four waves are: $\phi_1 = 0, \quad \phi_2 = \frac{\pi}{3}, \quad \phi_3 = \frac{2\pi}{3}, \quad \phi_4 = \pi$

Representing the waves as phasors, the horizontal (real) and vertical (imaginary) components of the resultant amplitude $A_R$ are given by:

1. Horizontal Component ($A_x$): $A_x = A_0 \left[ \cos(0) + \cos\left(\frac{\pi}{3}\right) + \cos\left(\frac{2\pi}{3}\right) + \cos(\pi) \right]$ $A_x = A_0 \left[ 1 + \frac{1}{2} - \frac{1}{2} - 1 \right] = 0$

2. Vertical Component ($A_y$): $A_y = A_0 \left[ \sin(0) + \sin\left(\frac{\pi}{3}\right) + \sin\left(\frac{2\pi}{3}\right) + \sin(\pi) \right]$ $A_y = A_0 \left[ 0 + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + 0 \right] = \sqrt{3} A_0$

The magnitude of the resultant amplitude $A_R$ is: $A_R = \sqrt{A_x^2 + A_y^2} = \sqrt{0^2 + (\sqrt{3} A_0)^2} = \sqrt{3} A_0$

The resultant intensity $I_{\text{resultant}}$ is proportional to the square of the resultant amplitude: $I_{\text{resultant}} \propto A_R^2 = (\sqrt{3} A_0)^2 = 3 A_0^2$ $I_{\text{resultant}} = 3 I_0$

Comparing this with $I_{\text{resultant}} = n I_0$, we obtain: $n = 3$