Superposition and interference of waves

Two loudspeakers play the same steady note, and you walk across the room: at some spots the sound swells, at others it almost disappears — even though both speakers are pushing out sound the whole time. Nothing is being created or destroyed; the two wave patterns are simply adding up differently at different points. That single idea, called the principle of superposition, is the quiet workhorse behind interference, and it also underlies beats and stationary waves, which you meet separately as its two most important special cases. This article builds the superposition idea from a plain statement of the principle, through the qualitative picture of constructive and destructive interference, to the one piece of trigonometry that ties the special cases together: how the resultant amplitude of two equal, same-frequency waves depends on their phase difference. The treatment stays exactly where the ISC syllabus keeps it — a wave-sketch, amplitude-only picture, not the intensity or path-difference mathematics of optical interference.

The core insight

When two or more waves overlap, the resultant displacement at every instant is just the algebraic sum of the individual displacements, and for two equal-amplitude waves of the same frequency this sum ranges from full reinforcement (in phase) to full cancellation (exactly out of phase) depending only on their phase difference.

The core idea

Principle of superposition. When two or more waves overlap at a point in a medium, the resultant displacement at that point at any instant is simply the sum of the displacements each wave would have produced individually, as if the other were absent:

y_{resultant} = y_1 + y_2 (+\, y_3 + \dots)

where: $y_1, y_2, \dots$ are the individual displacements each wave alone would cause at that point and instant, and $y_{resultant}$ is what is actually observed. This holds for waves of any kind, transverse or longitudinal, as long as the medium responds linearly. Crucially, each wave passes through the region of overlap and continues afterward completely unaffected by the other, exactly as before — superposition only describes what happens during the overlap, not a permanent collision.

Interference — the qualitative picture. Consider two waves of the same frequency and (for the clean, textbook case) the same amplitude, arriving at a point from two sources.

If the waves arrive in phase (crest meets crest), the displacements add: the resultant amplitude is the sum of the two amplitudes — constructive interference, a louder or larger-amplitude disturbance.
If they arrive exactly out of phase (crest meets trough), the displacements oppose: the resultant amplitude is the difference of the two amplitudes, zero if the amplitudes are equal — destructive interference, a quieter or reduced disturbance.
At an in-between phase difference, the resultant amplitude lies between these two extremes, as the derivation below makes precise.

Interference in space, beats in time. Simple interference, as covered here, considers two waves of the same frequency meeting at different points in space, each point having its own fixed phase relationship and hence a fixed resultant amplitude that does not change with time. This is the opposite situation to beats, where two slightly different frequencies meet at one point and the loud-soft pattern shifts in time instead. Stationary waves are a third case: two identical waves travelling in opposite directions superpose to build a fixed pattern of nodes and antinodes in space. All three — interference, beats, stationary waves — are the same superposition principle applied to a different relationship between the component waves.

Figure

Two equal sine waves shown superposing crest on crest to double the amplitude and crest on trough to cancel to zero

Derivation AResultant amplitude of two equal, same-frequency waves at a phase difference

Two waves of equal amplitude $y_m$ and the same angular frequency $\omega$ meet at a point with a constant phase difference $\phi$ between them. The goal is to show, using nothing beyond the trigonometric sum-to-product identity a student already knows, that the resultant is itself a simple oscillation of amplitude $R = 2y_m\cos(\phi/2)$ — a single formula that reproduces full reinforcement at $\phi=0$ , full cancellation at $\phi=\pi$ , and every in-between case, and that underlies both worked examples in the Pass-1 article.

Step 1. Write the two waves explicitly. Take $y_1 = y_m\sin(\omega t)$ and $y_2 = y_m\sin(\omega t+\phi)$ , where $\phi$ is the fixed phase by which wave 2 leads wave 1. Both have the same amplitude $y_m$ and the same $\omega$ , which is exactly the "same source strength, same frequency" case the qualitative picture describes.

Step 2. Add them by superposition. $y_{resultant} = y_1+y_2 = y_m\big[\sin(\omega t)+\sin(\omega t+\phi)\big]$ . This sum of two sines is exactly the form the identity $\sin A+\sin B = 2\sin\!\left(\dfrac{A+B}{2}\right)\cos\!\left(\dfrac{A-B}{2}\right)$ is built for, with $A=\omega t$ and $B=\omega t+\phi$ .

Step 3. Apply the identity. $\dfrac{A+B}{2} = \omega t+\dfrac{\phi}{2}$ and $\dfrac{A-B}{2} = -\dfrac{\phi}{2}$ , so, using $\cos(-x)=\cos x$ , $y_{resultant} = 2y_m\cos\!\left(\dfrac{\phi}{2}\right)\sin\!\left(\omega t+\dfrac{\phi}{2}\right)$ .

Step 4. Read off the result. This is a single sine wave of the same angular frequency $\omega$ , phase-shifted by $\phi/2$ , whose amplitude is $R = 2y_m\cos(\phi/2)$ . At $\phi=0$ (in phase), $R = 2y_m\cos 0 = 2y_m$ , the sum of amplitudes — constructive interference. At $\phi=\pi$ (exactly out of phase), $R = 2y_m\cos(\pi/2) = 0$ — destructive interference. For any $\phi$ in between, $R$ takes an intermediate value between $0$ and $2y_m$ , confirming the qualitative picture above with one compact formula.

Worked example

Two waves of equal amplitude $y_m=0.03\ \text{m}$ and the same frequency arrive at a point with a phase difference of $60^\circ$ ( $\phi=\pi/3$ ). Find the resultant amplitude.

Step 1. Identify the values needed for the formula from the derivation: $y_m=0.03\ \text{m}$ and $\phi=60^\circ$ , so $\phi/2=30^\circ$ .

Step 2. Apply $R=2y_m\cos(\phi/2)$ : $R = 2\times0.03\times\cos(30^\circ) = 0.06\times0.866 = 0.052\ \text{m}$ .

Step 3. Check it against the two extremes already known: at $\phi=0$ this formula gives $2y_m=0.06\ \text{m}$ (matching the fully in-phase case), and at $\phi=180^\circ$ it gives $0$ (matching the fully out-of-phase case); $0.052\ \text{m}$ at $60^\circ$ sits sensibly between them, closer to full reinforcement since $60^\circ$ is a smaller phase gap than $180^\circ$ .

The resultant amplitude is $\mathbf{0.052\ m}$ , roughly $87\%$ of the fully-constructive value.

Competitive stretch

For the same $y_m=0.03\ \text{m}$ , at $\phi=90^\circ$ : $R=2(0.03)\cos(45^\circ)=0.06\times0.707=0.0424\ \text{m}$ . Notice $R$ falls steadily as $\phi$ increases from $0$ to $\pi$ — the resultant amplitude is a smooth, single-valued function of phase difference alone, which is exactly why interference patterns can be mapped out point by point.

Common traps

Thinking superposition means the waves collide and get permanently altered — each wave passes through the overlap region completely unaffected and continues on its own way afterward; only the resultant displacement during the overlap is different.
Assuming destructive interference always means total silence — perfect cancellation to zero only happens when the two amplitudes are exactly equal and the phase difference is exactly $\pi$ ; otherwise, as $R=2y_m\cos(\phi/2)$ shows, the resultant is merely reduced, not eliminated.
Confusing interference (same-frequency waves meeting at fixed points in space, giving a time-independent pattern at each point) with beats (slightly different frequencies, giving a pattern that changes in time at every point) — same underlying superposition principle, different setup.
Trying to compute intensity (power per unit area) as if it simply adds for two interfering waves — at this syllabus level only the qualitative, amplitude-based wave picture is required, not the quantitative intensity treatment that belongs to Class 12 optics.

Ask the tutor: Want to see how this same superposition idea plays out for beats or stationary waves, or try a resultant-amplitude problem with unequal source amplitudes?

Why doesn't destructive interference just make the waves disappear?

The energy carried by both waves is never destroyed — superposition only redistributes displacement, not energy. At a point of perfect destructive interference the displacement stays zero at every instant, but each source is still radiating energy, which shows up as extra amplitude (extra energy) at nearby points of constructive interference instead. Nothing about superposition violates energy conservation; it only changes how the same total is distributed in space.

Is the resultant amplitude formula R equals 2 y m cos of phi by 2 something I need to derive on an ISC exam?

The ISC syllabus keeps interference at the qualitative, wave-sketch level, so a full trigonometric derivation is unlikely to be demanded outright. But the formula is a direct, honest extension of the in-phase and out-of-phase cases the syllabus does expect you to know cold, and understanding where it comes from makes both extremes (sum of amplitudes, and zero) feel like special cases of one idea rather than two separate facts to memorise.

How is this different from what I'll see in optics next year?

Here, amplitude is the whole story: you add displacements and read off a resultant amplitude from a phase difference. In Class 12 optics, interference is treated through intensity (which depends on amplitude squared) and path difference (which converts a physical distance into a phase difference using the wavelength), giving the full quantitative treatment of fringe patterns. The underlying superposition principle is identical — only the extra tools, intensity and path difference, are added later.

Superposition and interference of waves

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads