Question

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

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Add the four equal amplitudes as phasors with the given phase angles; intensity is proportional to amplitude squared.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Waves of equal frequency add as phasors, each of amplitude \(a\), at angles \(0\), \(60^{\circ}\), \(120^{\circ}\) and \(180^{\circ}\).

Step 2: Horizontal sum
\(a\left(1+\dfrac{1}{2}-\dfrac{1}{2}-1\right)=0\)

Step 3: Vertical sum
\(a\left(0+\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}+0\right)=\sqrt{3}\,a\)

Step 4: Intensity
Resultant amplitude is \(\sqrt{3}\,a\), and \(I\propto a^{2}\), so \(I=3I_0\). Hence \(n=3\), option (C).

Final Answer:
The resultant intensity is \(3I_0\), so n is 3. \[ \boxed{\text{(C) }3} \]