Visual walkthrough — Superposition principle

We are chasing this one boxed fact: $A_{\text{res}}=2A\cos \left(\frac{\varphi }{2}\right)$

Everything below earns every symbol before using it.

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Step 1 — What is a wave, drawn as a moving shape?

WHAT. A wave on a string is a shape that slides sideways without changing form. We write one as $y_1=A\sin (kx-\omega t)$

Let's read every symbol on the picture:

• $y_1$ — the up/down height of the string at a place and a time. Positive = string lifted up, negative = pushed down.
• $x$ — position along the string (how far along we look).
• $t$ — time.
• $A$ — the amplitude: the biggest height the wave ever reaches. It is the distance from the flat line up to a crest.
• $k$ — the wavenumber: how many radians of the sine wave fit into each metre of string. Big $k$ = tightly packed wiggles.
• $\omega$ — the angular frequency: how fast the wiggle bobs up and down in time.

WHY a sine? Because a sine is the simplest repeating shape — one smooth hump up, one down, forever. The combination $(kx-\omega t)$ is the wave's "clock reading": as $t$ grows, you'd have to move $x$ forward to keep the clock reading the same, so the whole shape travels to the right.

PICTURE.

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Step 2 — A second wave, one step behind

WHAT. Now send a second identical wave down the same string — same $A$, same $k$, same $\omega$ — but launched a little later. That delay is written as a phase difference $\varphi$: $y_2=A\sin (kx-\omega t+\varphi )$

Reading the new symbol:

• $\varphi$ — the phase difference, measured in radians. It is a head start: the $+\varphi$ shifts the wave so its crests sit a bit ahead (or behind) wave 1's crests.

WHY phase and not distance? Because a wave repeats every $2\pi$ radians of its clock. Talking in phase means "how far around the repeating cycle is wave 2 compared to wave 1?" — that single number captures the offset no matter the wavelength.

PICTURE. The red wave is wave 2, shifted by $\varphi$.

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Step 3 — Superposition: just add the heights

WHAT. The Superposition principle says the medium does both instructions at once. At every point the true height is $y=y_1+y_2=A\sin (kx-\omega t)+A\sin (kx-\omega t+\varphi )$

WHY are we allowed to just add? Because the Wave equation is linear — the parent note proved that if $y_1$ and $y_2$ each solve it, so does $y_1+y_2$. No cross-terms, no fighting. Add the pushes, get the truth.

PICTURE. Black + red, added point-by-point, gives the blue resultant.

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Step 4 — The tool we need: turning a sum of sines into a product

WHAT. Adding two sines looks messy. We reach for one exact identity: $\sin P+\sin Q=2\sin \left(\frac{P+Q}{2}\right)\cos \left(\frac{P-Q}{2}\right)$

WHY this tool and not another? We could expand each sine with angle-addition formulas, but this "sum-to-product" identity is purpose-built: it hands us a single sine (a wave) multiplied by a constant cosine (an amplitude). That split is exactly the thing we want to read off — the "wave part" and the "size part" fall out separately.

Let's name the two clock readings so we can plug in cleanly:

• $P=kx-\omega t$ (wave 1's phase)
• $Q=kx-\omega t+\varphi$ (wave 2's phase)

Then the two ingredients of the identity are: $\frac{P+Q}{2}=\frac{(kx-\omega t)+(kx-\omega t+\varphi )}{2}=kx-\omega t+\frac{\varphi }{2}$ $\frac{P-Q}{2}=\frac{(kx-\omega t)-(kx-\omega t+\varphi )}{2}=-\frac{\varphi }{2}$

PICTURE. Where those halves come from geometrically — the average clock, and half the gap.

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Step 5 — Assemble the resultant wave

WHAT. Feed those two pieces into the identity: $y=2A\cos (-\frac{\varphi }{2})\sin (kx-\omega t+\frac{\varphi }{2})$

Cosine is even — $\cos (-\theta )=\cos (\theta )$ — so the minus sign harmlessly disappears: $\boxed{y=\underset{\text{amplitude}}{\underbrace{2A\cos \left(\frac{\varphi }{2}\right)}}\underset{\text{a travelling wave}}{\underbrace{\sin (kx-\omega t+\frac{\varphi }{2})}}}$

Reading the two boxed groups:

• $\sin (kx-\omega t+\frac{\varphi }{2})$ — a wave with the same $k$ and $\omega$ as before. So the result is still a wave of the same frequency, just nudged to sit halfway between the two originals.
• $2A\cos (\frac{\varphi }{2})$ — a constant (it has no $x$ or $t$ in it). It multiplies the wave, so it is the new amplitude: $A_{\text{res}}=2A\cos \left(\frac{\varphi }{2}\right)$

WHY split like this? Because now the phase $\varphi$ lives in exactly one place — the amplitude. Turn the phase knob and only the height changes; the wave keeps rolling along untouched.

PICTURE. The blue resultant with its envelope amplitude $2A\cos (\varphi /2)$ marked.

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Step 6 — Case $\varphi =0$: constructive (they double up)

WHAT. Set $\varphi =0$: $A_{\text{res}}=2A\cos (0)=2A\times 1=2A$

WHY. With zero phase gap the crests sit exactly on top of each other. Every push-up adds to another push-up — the string swings twice as high. This is constructive interference.

PICTURE. Two crests aligned, resultant reaches $2A$.

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Step 7 — Case $\varphi =\pi$: destructive (they vanish)

WHAT. Set $\varphi =\pi$: $A_{\text{res}}=2A\cos \left(\frac{\pi }{2}\right)=2A\times 0=0$

WHY. Half a cycle apart means wave 2's trough sits exactly over wave 1's crest. Every up-push meets an equal down-push; they cancel everywhere and always. The string stays flat. This is destructive interference.

PICTURE. Crest over trough, resultant is the flat line.

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Step 8 — The in-between and the sign of the cosine

WHAT. For a general $\varphi$ between $0$ and $\pi$, the amplitude smoothly slides from $2A$ down to $0$. But keep going past $\pi$ and something subtle happens: $\cos (\varphi /2)$ goes negative (for $\pi <\varphi <3\pi$, i.e. $\varphi /2$ between $\frac{\pi }{2}$ and $\frac{3\pi }{2}$).

WHY does a negative amplitude make sense? A negative sign in front of a sine just flips it upside-down — which is the same as a sine shifted by half a cycle. So a "negative amplitude" is a positive amplitude with an extra flip. The physically-measured amplitude (the biggest height you'd actually see) is the magnitude: $A_{\text{res}}=2A|\cos (\frac{\varphi }{2})|$

This covers ALL cases:

• $\varphi =0$ → $|\cos 0|=1$ → $2A$ (max).
• $\varphi =\pi$ → $|\cos \frac{\pi }{2}|=0$ → $0$ (min).
• $\varphi =2\pi$ → $|\cos \pi |=1$ → $2A$ again — because $2\pi$ is a full cycle, indistinguishable from $\varphi =0$.
• Any $\varphi$ → smoothly between $0$ and $2A$, repeating every $2\pi$.

PICTURE. The amplitude curve $2A|\cos (\varphi /2)|$ across a full range of phase, its peaks and zeros marked.

Recall Why

$|\cdot |$ and not just the raw cosine? Amplitude means "biggest size", which is never negative ::: The raw $2A\cos (\varphi /2)$ can be negative; that sign is a half-cycle flip absorbed into the wave, so the measured amplitude is its absolute value.

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The one-picture summary

Read left to right: two identical waves offset by $\varphi$ → superpose (add heights) → sum-to-product identity → a single travelling wave whose amplitude is $2A\cos (\varphi /2)$. Turn the phase knob and the amplitude slides between $2A$ (constructive) and $0$ (destructive).

Recall Feynman retelling — the whole walkthrough in plain words

Two identical ripples chase each other down a string, one a little behind — that "little behind" is the phase $\varphi$. Superposition says: at every spot, just add the two heights. When I add two sines of the same shape, there's a neat trick that splits the answer into one sine (still a rolling wave, sitting halfway between the two) times a fixed number out front. That fixed number is the new amplitude, and it turns out to be $2A\cos (\varphi /2)$. If the ripples march in step ($\varphi =0$) the cosine is $1$ and the wave doubles to $2A$. If one is half a cycle behind ($\varphi =\pi$) the cosine is $0$ and they wipe each other out flat. In between, the cosine slides the amplitude smoothly between those extremes — and because a wave repeats every full cycle, the pattern comes right back to $2A$ at $\varphi =2\pi$. That's the entire story: add the heights, use the trick, read off the knob.

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Connections

• Superposition principle — the parent law this page derives a picture-proof of.
• Interference of waves — the constructive/destructive cases (Steps 6–7) are interference.
• Beats — what happens when the two waves have slightly different $\omega$ instead of just a phase gap.
• Standing waves — same addition, but the second wave travels the opposite way.
• Phasor method — the geometric shortcut for this same amplitude formula.
• Simple Harmonic Motion — each point on the string executes SHM; superposition adds two SHMs.
• Wave equation — its linearity is why Step 3 (just add) is legal.