Four harmonics of equal frequencies and equal intensities $I_{0}$ have phase angles $0, \frac{π}{3}, \frac{2 π}{3}, π$ . When they are superimposed, the intensity of the resultant wave is " $n I_{0}$ ". The value of n is
(all phase angles are in radian and $cos (60)^{\circ} = \frac{1}{2}$ )

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Let the amplitude of each harmonic wave be $A_{0}$ . The intensity of each wave is $I_{0} = k A_{0}^{2}$ .

The waves can be represented as vectors (phasors) with magnitudes $A_{0}$ and angles $0, \frac{π}{3}, \frac{2 π}{3}, π$ .

The x-component of the resultant amplitude is:
$A_{x} = A_{0} cos (0) + A_{0} cos (\frac{π}{3}) + A_{0} cos (\frac{2 π}{3}) + A_{0} cos (π)$
$A_{x} = A_{0} (1 + \frac{1}{2} - \frac{1}{2} - 1) = 0$

The y-component of the resultant amplitude is:
$A_{y} = A_{0} sin (0) + A_{0} sin (\frac{π}{3}) + A_{0} sin (\frac{2 π}{3}) + A_{0} sin (π)$
$A_{y} = A_{0} (0 + \frac{3}{2} + \frac{3}{2} + 0) = 3 A_{0}$

The resultant amplitude $A_{R}$ is given by:
$A_{R} = A_{x}^{2} + A_{y}^{2} = 0 + (3 A_{0})^{2} = 3 A_{0}$

The resultant intensity $I_{R}$ is proportional to the square of the resultant amplitude:
$I_{R} = k A_{R}^{2} = k (3 A_{0})^{2} = 3 k A_{0}^{2} = 3 I_{0}$

Comparing this with $n I_{0}$ , we get $n = 3$ .

Answer: $3$

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