Principle of Superposition of Waves

The Superposition Principle

A medium can carry many disturbances at once. A single point on a stretched string, a single layer of air in a pipe, a single patch of water surface — each can be pulled simultaneously by more than one passing wave. The question superposition answers is precise: given that two or more disturbances arrive together, how does the medium respond at that instant.

The answer, stated by NCERT, is that each wave moves as if the others were not present, and the net displacement of any particle is the algebraic sum of the displacements that each wave would have produced alone. The word algebraic is doing real work here. Displacements carry sign — a crest is positive, a trough is negative — and the sum respects those signs, so two disturbances can reinforce one another or wipe one another out entirely.

When two or more waves overlap in a medium, the resultant displacement at any point and any instant is the algebraic sum of the displacements due to each individual wave.

Two Pulses Crossing

The cleanest demonstration uses two pulses travelling in opposite directions along a string. As they meet, the wave pattern in the overlap region is unlike either pulse on its own. Yet the moment they separate, each pulse re-emerges unchanged in shape and speed. The waves pass through one another; they do not collide or scatter.

NCERT highlights the dramatic special case: two pulses of equal size but opposite sign. At the instant of complete overlap, every positive displacement of one pulse is matched by an equal negative displacement of the other, and the string is momentarily flat — zero displacement everywhere. An instant later both pulses reappear, intact. Cancellation of displacement does not mean the energy has vanished; it is stored as kinetic energy in the moving string and reappears as the pulses separate.

Mathematical Statement

Let $y_1(x,t)$ and $y_2(x,t)$ be the displacements that two disturbances would individually produce in the medium. If they arrive in a region together, the net displacement is simply their sum:

$$y(x,t) = y_1(x,t) + y_2(x,t)$$

For any number of overlapping waves with wave functions $y_1 = f_1(x-vt)$, $y_2 = f_2(x-vt)$, and so on, the disturbance in the medium is the sum of all of them:

$$y = \sum_{i=1}^{n} f_i(x-vt)$$

This linear additivity is what defines superposition. It is the property that makes interference possible, and NCERT names it as basic to the phenomenon of interference.

Superposing Two Harmonic Waves

The most examined case is two harmonic waves on a stretched string that share the same angular frequency $\omega$, the same wave number $k$ (hence the same wavelength and speed), and the same amplitude $a$, both travelling in the $+x$ direction. They differ only in their initial phase $\phi$:

$$y_1(x,t) = a\sin(kx - \omega t), \qquad y_2(x,t) = a\sin(kx - \omega t + \phi)$$

By superposition the net displacement is $y(x,t) = a\sin(kx - \omega t) + a\sin(kx - \omega t + \phi)$. Applying the identity $\sin A + \sin B = 2\sin\!\frac{A+B}{2}\cos\!\frac{A-B}{2}$ gives a single harmonic wave of the same frequency and wavelength:

$$y(x,t) = \underbrace{2a\cos\tfrac{\phi}{2}}_{\text{resultant amplitude}}\;\sin\!\Big(kx - \omega t + \tfrac{\phi}{2}\Big)$$

Two results follow at once. The resultant is itself a travelling harmonic wave with an extra initial phase of $\phi/2$, and its amplitude depends entirely on the phase difference:

$$A(\phi) = 2a\cos\frac{\phi}{2}$$

Constructive vs Destructive Interference

The amplitude $A(\phi) = 2a\cos(\phi/2)$ is largest when $\cos(\phi/2) = \pm 1$ and zero when $\cos(\phi/2) = 0$. These two extremes are the two named regimes of interference. The table sets them side by side.

Here $n = 0, 1, 2, \dots$ Constructive interference is the case where the amplitudes add up in the resultant; destructive interference is where they subtract out. For two equal waves exactly out of phase the cancellation is complete and permanent — the resultant displacement is zero at every point and at every instant.

When the Amplitudes Are Unequal

The $2a\cos(\phi/2)$ formula assumes equal amplitudes. When the two waves have different amplitudes $a_1$ and $a_2$ but the same frequency and a constant phase difference $\phi$, superposition still yields a single harmonic wave, now with amplitude given by the general expression:

$$A = \sqrt{a_1^{2} + a_2^{2} + 2a_1 a_2 \cos\phi}$$

This reduces to the familiar special cases. In phase ($\phi = 0$, $\cos\phi = +1$) the amplitudes add to $A = a_1 + a_2$, the maximum. Antiphase ($\phi = \pi$, $\cos\phi = -1$) gives $A = |a_1 - a_2|$, the minimum, which is non-zero unless the amplitudes are equal. Setting $a_1 = a_2 = a$ recovers $A = 2a\cos(\phi/2)$ exactly. Because intensity scales as amplitude squared, this is also the route to $I_{\max} \propto (a_1 + a_2)^2$ and $I_{\min} \propto (a_1 - a_2)^2$.

Why Superposition Matters

Superposition is not a stand-alone topic; it is the engine behind the rest of the Waves chapter. Two specific applications recur in NEET, and both are nothing more than this same algebraic-sum rule applied under different conditions.

The 2020 NEET beats problem listed below is solved entirely on the superposition idea that two near-equal frequencies produce a slow throb at the difference frequency. Recognising that beats and standing waves are special cases of superposition — rather than separate formulas to memorise — is what turns this section into a multiplier across the chapter.