Visual walkthrough — Wave equation — derivation for string

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Step 0 — The words before the symbols

Before any formula, let us fix the picture of a string and name the two things that describe it.

Nothing here is a derivative yet — just a picture with four labels: $x$, $y$, $T$, $\mu$.

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Step 1 — Zoom in on one tiny piece

WHAT. We cut out a very short stretch of string, from position $x$ to position $x+dx$. The symbol $dx$ just means "a tiny extra bit of $x$" — so small that the piece is almost straight.

WHY. Newton's second law, $F=ma$, is a law about one object. A whole wiggly string is too complicated. But a piece so short it is nearly straight is a simple object we can push around. If the law holds for every tiny piece, it holds for the whole string.

PICTURE.

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Step 2 — The two pulls on the ends

WHAT. The tension tugs the piece at both cut ends, pulling outward along the string.

WHY. A piece of string feels nothing from the middle of itself — only from the neighbours yanking on its two ends. Those two yanks are what could accelerate it. So we must look at exactly them.

PICTURE. Notice in the figure that the string is tilted at each end, and the two tilts are not the same — that difference is the whole game.

Each pull has size $T$ but points along the string, which is slanted. We split each pull into a horizontal part and a vertical (sideways) part. Call the tilt angle $\theta$ — the angle the string makes with the horizontal. Then:

At the right end the string tilts up-right, so its sideways pull is $+T\sin \theta (x+dx)$ (upward). At the left end the string tilts down-left, so its sideways pull is $-T\sin \theta (x)$ (downward). The net sideways pull is their sum: $F_y=T\sin \theta (x+dx)-T\sin \theta (x).$

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Step 3 — Why the horizontal pulls cancel (a needed side-trip)

WHAT. We check the horizontal direction and show it contributes nothing.

WHY. If the piece were being yanked sideways and dragged along $x$, the problem would be two-dimensional and horrible. We want to prove the horizontal parts kill each other so we can forget them.

PICTURE.

For small tilts, $\cos \theta \approx 1$, so the horizontal part at each end is $T\cos \theta \approx T$ — the same size at both ends but pointing in opposite directions. Opposite and equal ⇒ they cancel.

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Step 4 — Turn angle into slope

WHAT. We swap the angle $\theta$ for the slope of the string.

WHY the slope? We can measure an angle, but we cannot easily do calculus on it. The slope, on the other hand, is directly a derivative of $y$ — which lets us bring in the machinery of Partial derivatives and curvature. So we translate geometry into a derivative we can compute.

Why this particular tool — $\tan$? On a right triangle, $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{\text{rise}}{\text{run}}$. That ratio is the steepness of the string, and steepness of $y$ versus $x$ is exactly the derivative $\partial y/\partial x$. So $\tan$ is the one trig function that equals a slope.

PICTURE.

For small tilts, the sine and the tangent of the angle are almost equal (look at how the two curves hug near $\theta =0$ in the figure), so:

Now the vertical force becomes, with $y^{\prime }\equiv \partial y/\partial x$: $F_y=T[\underset{\text{slope at right end}}{\underbrace{y^{\prime }(x+dx)}}-\underset{\text{slope at left end}}{\underbrace{y^{\prime }(x)}}].$

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Step 5 — A change of slope is curvature

WHAT. We rewrite "slope at the right minus slope at the left" as one thing: how fast the slope changes.

WHY. The bracket $[y^{\prime }(x+dx)-y^{\prime }(x)]$ is a difference of the slope at two nearby spots. Anything of the form "(value here $+dx$) minus (value here)" divided by $dx$ is a derivative — that is the very definition of a derivative. Doing it to the slope $y^{\prime }$ gives the derivative of the slope, i.e. the second derivative.

PICTURE. The figure shows the string over the piece: where the slope turns (the string bends), the two end-slopes differ. A straight piece (no bend) has equal end-slopes and thus zero net force.

So the net sideways pull is now written purely with curvature: $\boxed{F_y=T\frac{{\partial }^{2}y}{\partial x^2}dx}$ Read it aloud: net sideways force = tension × curvature × length. More bend ⇒ more force. A straight piece feels no net pull — exactly the point of the figure.

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Step 6 — Newton's second law on the piece

WHAT. We set force equal to mass times acceleration.

WHY. This is Newton's Second Law — the only law that connects a force to motion. We built the left side (force) and the mass in Step 1; now we need the acceleration.

The acceleration is how fast the sideways position $y$ speeds up in time, which is the second time-derivative ${\partial }^{2}y/\partial t^2$. (First time-derivative = velocity; second = acceleration — same idea as position→speed→acceleration.)

PICTURE.

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Step 7 — Cancel $dx$ and read off the speed

WHAT. Divide both sides by $dx$; it vanishes.

WHY it matters that $dx$ cancels. The size of the piece we chose was arbitrary. If the answer still depended on $dx$, our derivation would be nonsense. It cancels — so the law is the same for every piece, hence for the whole string.

$\frac{{\partial }^{2}y}{\partial t^2}=\frac{T}{\mu }\frac{{\partial }^{2}y}{\partial x^2}.$

Compare this with the standard shape $y_{tt}=v^2{y}_{xx}$. The number multiplying the curvature must be a speed squared:

This is the same $v$ used in Wave speed on a string — v = sqrt(T over mu) and behind the dispersion relation.

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Step 8 — The degenerate cases (never leave the reader stranded)

WHAT. Test the extreme inputs to make sure the picture is complete.

Every sign of curvature bends the piece back toward straight — that restoring tug is precisely what makes an oscillation, and a travelling chain of oscillations is a wave (see Travelling wave function y = A sin(kx - omega t)).

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The one-picture summary

The whole derivation is one arrow of logic:

$\underset{{\partial }^{2}y/\partial x^2}{\underbrace{\text{curvature}}}\stackrel{\times T}{\to }\underset{F_y}{\underbrace{\text{net force}}}\stackrel{÷dm}{\to }\underset{{\partial }^{2}y/\partial t^2}{\underbrace{\text{acceleration}}}⟹v=\sqrt{\frac{T}{\mu }}$

Recall Feynman: tell the whole walkthrough to a friend

Take a long tight rope. Snip out a teeny piece so short it looks straight. The rope on either side tugs it along its own two ends. If that little piece is bent — even slightly — the two tugs point in just-slightly-different directions, so they don't fully cancel and leave a small leftover pull sideways. How much leftover pull? It grows with how tightly the rope is pulled ($T$) and with how sharply the piece is bent (its curvature). That leftover pull, divided by how heavy the piece is ($\mu$ times its length), gives how hard it accelerates up or down. Write that out and the length of the piece cancels — proof it's true for the whole rope. Line the tugs up over time and you get a shape that runs along the rope at a fixed speed $v=\sqrt{T/\mu }$: tighter and lighter means faster. Bend, tug, shake — that's the entire wave equation.

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Connections

• Wave equation — derivation for string (parent)
• Newton's Second Law
• Partial derivatives and curvature
• Wave speed on a string — v = sqrt(T over mu)
• Travelling wave function y = A sin(kx - omega t)
• Dispersion relation omega = vk
• Standing waves on a string
• Transverse vs Longitudinal Waves
• Energy carried by a wave