Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, pi/3, 2pi/3 a

Q: Why can we treat wave amplitudes as vectors? Waves are scalar functions, aren't they?

While wave displacement at a point is a scalar quantity, harmonic waves of the same frequency can be represented as rotating vectors (phasors) because their superposition mathematically mirrors vector addition. The projection of these phasors onto an axis gives the actual physical displacement.

Q: What if the frequencies of the waves were different? Could we still use this phasor method?

No. The phasor method relies on the vectors rotating at the exact same angular frequency $\omega$, meaning their relative phase angles remain constant over time. If the frequencies are different, the vectors rotate at different speeds, and the phase differences change continuously, leading to beats rather than a steady interference pattern.

Q: Why did the waves with phase $0$ and $\pi$ cancel out completely?

A phase difference of $\pi$ radians is exactly $180^\circ$. This means when one wave is at its positive peak (crest), the other is at its negative peak (trough). Since their amplitudes are equal, they undergo perfect destructive interference and cancel out completely.

Q: Is the intensity always proportional to the square of the amplitude?

Yes, for all mechanical and electromagnetic waves, the energy transported by the wave (and hence its intensity) is proportional to the square of the amplitude of oscillation ($I \propto A^2$).

Q: Can we solve this using standard trigonometric identities instead of vectors?

Absolutely! You can write the sum as $y = A_0 \sin(\omega t) + A_0 \sin(\omega t + \pi/3) + A_0 \sin(\omega t + 2\pi/3) + A_0 \sin(\omega t + \pi)$. Using trig identities like $\sin(\theta) + \sin(\theta + \pi) = 0$, you will get the same result, but it is much more tedious and prone to algebraic errors compared to the elegant phasor method.

Question

Four harmonic waves of equal frequencies and equal intensities $I_0$ have phase angles $0$, $\frac{\pi}{3}$, $\frac{2\pi}{3}$ and $\pi$. When they are superposed, the intensity of the resulting wave is $nI_0$. The value of $n$ is

Sigma Percentile · Accepted Answer

### Introduction to Wave Superposition

When multiple waves travel through the same medium, they pass through each other without being disrupted.

At any point of intersection, the net displacement of the medium is simply the vector sum of the individual wave displacements.

This is the celebrated **Principle of Superposition**.

In this problem, we are dealing with four harmonic waves of equal frequency and equal intensity $I_0$.

Since the intensity of a wave is directly proportional to the square of its amplitude ($I \propto A^2$), equal intensity implies that all four waves have the same amplitude, which we will denote as $A_0$.

$$
I_0 = k A_0^2
$$

Our goal is to find the resultant intensity when these four waves, each having a specific phase angle, are superposed.

### The Phasor Representation: Mapping Waves to Vectors

Instead of dealing with complex trigonometric functions like sines and cosines, we can represent each wave as a rotating vector called a **phasor** on a two-dimensional plane.

The length of the vector represents the amplitude of the wave, and the angle it makes with the positive $x$-axis represents its phase angle.

Let's write down the four phasor vectors:

*   $\vec{A}_1$ has magnitude $A_0$ and phase angle $0$ (along the $+x$-axis).
*   $\vec{A}_2$ has magnitude $A_0$ and phase angle $\frac{\pi}{3}$ ($60^\circ$).
*   $\vec{A}_3$ has magnitude $A_0$ and phase angle $\frac{2\pi}{3}$ ($120^\circ$).
*   $\vec{A}_4$ has magnitude $A_0$ and phase angle $\pi$ (along the $-x$-axis).

### The Power of Symmetry: The Cancellation of $A_1$ and $A_4$

Before jumping into heavy calculations, let's look for symmetry.

Notice the pair $\vec{A}_1$ and $\vec{A}_4$:

$$
\vec{A}_1 = A_0 \hat{i}
$$
$$
\vec{A}_4 = -A_0 \hat{i}
$$

These two vectors are collinear, equal in magnitude, but point in exactly opposite directions.

Their phase difference is $\pi$ radians ($180^\circ$), which corresponds to perfect **destructive interference**.

When we add them together, they completely cancel each other out:

$$
\vec{A}_{14} = \vec{A}_1 + \vec{A}_4 = 0
$$

This beautiful cancellation simplifies our problem immensely! We are now left with only two active waves.

### Resolving the Survivors: Adding $A_2$ and $A_3$

We are left with $\vec{A}_2$ at $60^\circ$ and $\vec{A}_3$ at $120^\circ$.

The angle $	heta$ between these two vectors is:

$$
	heta = 120^\circ - 60^\circ = 60^\circ
$$

To find the resultant amplitude $A_{	ext{res}}$ of these two vectors, we use the standard vector addition formula:

$$
A_{	ext{res}} = \sqrt{A_0^2 + A_0^2 + 2 A_0 A_0 \cos(60^\circ)}
$$

Since $\cos(60^\circ) = \frac{1}{2}$, we substitute this value into the equation:

$$
A_{	ext{res}} = \sqrt{A_0^2 + A_0^2 + 2 A_0^2 \left(\frac{1}{2}ight)}
$$
$$
A_{	ext{res}} = \sqrt{3 A_0^2} = \sqrt{3} A_0
$$

Thus, the resultant amplitude of the superposed waves is $\sqrt{3} A_0$.

### From Amplitude to Intensity: The Final Leap

Now, let's relate this resultant amplitude back to the intensity of the wave.

Using the relation $I \propto A^2$, the resultant intensity $I_{	ext{res}}$ is:

$$
I_{	ext{res}} = k A_{	ext{res}}^2
$$

Substitute $A_{	ext{res}} = \sqrt{3} A_0$:

$$
I_{	ext{res}} = k (\sqrt{3} A_0)^2 = 3 (k A_0^2)
$$

Since $I_0 = k A_0^2$, we get:

$$
I_{	ext{res}} = 3 I_0
$$

Comparing this with the given expression $I_{	ext{res}} = n I_0$, we find:

$$
**n = 3**
$$

### Conclusion

By mapping the waves to a phasor diagram, we avoided tedious trigonometric expansions and solved the problem using simple vector geometry.

The symmetry of the phase angles allowed us to cancel out two of the waves immediately, leaving a straightforward calculation for the remaining two.

This elegant method is a staple for solving complex wave interference problems in JEE Advanced!

Animated Solution for Physics - Waves: Four harmonic waves of equal frequencies and equal intensities $I_{0}$ have phase angles $0$ , $\frac{π}{3}$ , $\frac{2 π}{3}$ and $π$ . When they are superposed, the intensity of the resulting wave is $n I_{0}$ . The value of $n$ is

Enter Numerical Value:

Visualized Solution

Visualizing the Wave Amplitudes as Phasors

The Principle of Superposition

Analyzing the Opposite Pairs

Resultant of $A_{1}$ and $A_{4}$

Analyzing the Remaining Pair $A_{2}$ and $A_{3}$

Resultant Amplitude Formula

Calculating $A_{res}$

Relating Amplitude to Intensity

Substituting $A_{res}$ into Intensity Relation

Finding the Value of $n$

Generalizing for $N$ Phasors

The Sigma Insight: Interference of Waves

Introduction to Wave Superposition

The Phasor Representation

The Power of Symmetry

Resolving the Survivors

From Amplitude to Intensity

Conclusion

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