Wave equation — derivation for string: Worked examples

This page is the workout room for Wave equation — derivation for string. The parent note built the equation $\frac{\partial ^{2} y}{\partial t ^{2}} = v^{2} \frac{\partial ^{2} y}{\partial x ^{2}}, v = \frac{T}{μ} .$ Here we throw every kind of question at it — normal numbers, doubling and halving, a value that goes to zero, a degenerate string, a real-world word problem, and an exam-style twist. Nothing on the exam should feel new after this.

Before we start, one plain-language reminder of the symbols, so line one is readable from zero:

The scenario matrix

Every question this topic can throw is one of these cells. The examples below are tagged with the cell they hit.

Cell	What it tests	Covered by
A. Direct value	Plug $T, μ$ into $v = T / μ$	Ex 1
B. Scaling up	Multiply $T$ (or $μ$ ) by a factor — how does $v$ move?	Ex 2
C. Scaling down	Divide/halve a quantity — the inverse direction	Ex 3
D. Zero / degenerate input	$T \to 0$ , or $μ \to 0$ , or a slack string	Ex 4
E. Limiting behaviour	What happens as $T \to \infty$	Ex 4
F. Given a wave, find $v$	Read $v$ off $y = A sin (k x - ω t)$ via $v = ω / k$	Ex 5
G. Left- vs right-mover / sign	Does $f (x - v t)$ vs $g (x + v t)$ still solve it? Sign of $v$	Ex 6
H. Real-world word problem	Guitar / rope with a hanging mass	Ex 7
I. Exam twist (combined)	Two effects at once (mass and tension change) + time-of-travel	Ex 8

Worked examples

Ex 1 — Cell A: Direct value

Ex 2 — Cell B: Scaling tension up

Ex 3 — Cell C: Scaling density down (inverse direction)

Ex 4 — Cells D & E: Zero, degenerate, and limiting inputs

Figure — Wave equation — derivation for string

Worked example Statement

Explore the extremes of $v = T / μ$ : (a) A completely slack string, $T \to 0$ . (b) An infinitely rigid pull, $T \to \infty$ . (c) A massless idealised string, $μ \to 0$ .

Forecast: In which case does the wave stop moving, and in which does it become infinitely fast?

(a) $T \to 0$ : $v = 0/ μ = 0$ . Why this step? With no tension there is no restoring tug (parent Step 2: the vertical force is $T \times$ curvature — kill $T$ and the force is zero). A bent slack rope just sags; nothing propagates. Look at the red curve on the figure flattening to the axis as $T \to 0$ .
(b) $T \to \infty$ : $v = T / μ \to \infty$ . Why this step? Bigger tug ⇒ bigger restoring force ⇒ faster response. The curve rises without bound. (Physically capped by real materials, but the equation says $\infty$ .)
(c) $μ \to 0$ : $v = T / μ \to \infty$ as well. Why this step? Zero mass means $F = ma$ gives infinite acceleration for any force — the "shape" reacts instantly. This is why an idealised massless string is treated as transmitting tension instantly.

Verify: Sanity by a concrete tiny case: $T = 64, μ = 0.0001 \Rightarrow v = 640000 = 800 m/s$ — indeed huge as $μ$ shrinks ✓. And $T = 0.01, μ = 0.04 \Rightarrow v = 0.25 = 0.5 m/s$ — indeed crawling as $T$ shrinks ✓.

Ex 5 — Cell F: Given a wave, read off $v$

Worked example Statement

A wave on a string is $y (x, t) = 0.02 sin (5 x - 20 t)$ (SI units). Find its speed. If $μ = 0.05 kg/m$ , what tension holds this string?

Forecast: Which two numbers in the formula set the speed — the $0.02$ , the $5$ , or the $20$ ?

Match to the standard travelling wave $y = A sin (k x - ω t)$ : amplitude $A = 0.02$ , wavenumber $k = 5 rad/m$ , angular frequency $ω = 20 rad/s$ . Why this step? $k$ is the coefficient of $x$ (how many radians of wiggle per metre) and $ω$ the coefficient of $t$ (radians of wiggle per second). The amplitude $A$ never affects speed — recall the parent's mistake box. See Travelling wave function y = A sin(kx - omega t).
Use $v = ω / k$ (the Dispersion relation omega = vk): $v = 20/5 = 4 m/s$ . Why this step? The parent's Ex 3 showed sines solve the wave equation only if $ω^{2} = v^{2} k^{2}$ , i.e. $v = ω / k$ . This is the tool that turns a written wave into a speed.
Now invert $v = T / μ \Rightarrow T = μ v^{2} = 0.05 \times 4^{2}$ . Why this step? Squaring both sides removes the root; multiplying by $μ$ isolates $T$ .
$T = 0.05 \times 16 = 0.8 N$ .

Verify: $T / μ = 0.8/0.05 = 16 = 4 m/s = ω / k$ ✓. Amplitude $0.02$ never entered — as promised.

Ex 6 — Cell G: Left-mover vs right-mover (sign of $v$ )

Worked example Statement

The parent showed $y = f (x - v t)$ (a shape sliding right) solves the equation. Does $y = g (x + v t)$ (sliding left) also solve it? Take the concrete pulse $y (x, t) = \frac{1}{1 + ( x + 3 t ) ^{2}}$ and check; here the number $3$ plays the role of speed.

Forecast: Will the $+$ sign break the equation, or does the equation not care about direction?

Set the inside as one variable $u = x + v t$ with $v = 3$ , so $y = \frac{1}{1 + u ^{2}}$ . Why this step? Bundling the argument lets us differentiate once with respect to $u$ and reuse it — the chain rule trick from the parent's forecast section.
Time derivatives: $\frac{\partial u}{\partial t} = + v$ , so $\frac{\partial ^{2} y}{\partial t ^{2}} = y^{''} (u) \cdot v^{2}$ . Why this step? Two applications of the chain rule; crucially the $+ v$ gets squared, so its sign disappears. The $(+ v)^{2}$ and $(- v)^{2}$ are identical — that's the whole point.
Space derivatives: $\frac{\partial u}{\partial x} = 1$ , so $\frac{\partial ^{2} y}{\partial x ^{2}} = y^{''} (u)$ . Why this step? No hidden $v$ in $x$ , so the second $x$ -derivative is just $y^{''} (u)$ .
Compare: $\frac{\partial ^{2} y}{\partial t ^{2}} = v^{2} y^{''} (u) = v^{2} \frac{\partial ^{2} y}{\partial x ^{2}}$ ✓. Why this step? It matches $y_{tt} = v^{2} y_{xx}$ identically, no matter the sign of $v$ . So both directions solve it — this is why the general solution is $f (x - v t) + g (x + v t)$ : a right-mover plus a left-mover, exactly what reflects to make Standing waves on a string.

Verify: At the sample point $x = 0, t = 0$ : symbolic differentiation (see VERIFY) gives $y_{tt} = v^{2} y_{xx}$ numerically for $v = 3$ ✓. The red arrow on the figure shows the pulse moving left as $t$ grows, since $x + 3 t = const$ needs $x$ to decrease.

Ex 7 — Cell H: Real-world word problem (hanging mass)

Worked example Statement

A rope of mass $0.2 kg$ and length $2 m$ runs horizontally over a pulley; a $5 kg$ block hangs from its end, stretching it (take $g = 10 m/s^{2}$ ). How fast does a pluck travel along the horizontal part?

Forecast: The tension comes from the hanging block — do you use the block's mass or its weight?

Tension = weight of the hanging block: $T = m g = 5 \times 10 = 50 N$ . Why this step? The block hangs in equilibrium, so the rope's tension exactly balances gravity on it (Newton's Second Law with $a = 0$ ). Use the force $m g$ , not the mass — a classic trap.
Linear density of the rope: $μ = \frac{mass}{length} = \frac{0.2}{2} = 0.1 kg/m$ . Why this step? $μ$ is a property of the rope, computed from its own mass and length — nothing to do with the block.
Speed: $v = T / μ = 50/0.1 = 500 \approx 22.4 m/s$ . Why this step? Same master formula; the physics was just to find $T$ and $μ$ correctly.

Verify: $500 = 22.36 \dots$ ; squaring back $22.3 6^{2} \approx 500 = 50/0.1$ ✓. Units: $N / (kg/m) = m/s$ ✓.

Ex 8 — Cell I: Exam twist (two changes + travel time)

Worked example Statement

A wire has $v_{1} = 100 m/s$ over a length $L = 5 m$ . For a new experiment you use a wire of double the mass density and pull it with triple the tension. (a) Find the new speed $v_{2}$ . (b) How long does a pulse now take to cross the $5 m$ ?

Forecast: Two changes tug in opposite directions — will $v$ go up or down overall?

Combine both effects in one ratio. Since $v = T / μ$ , $\frac{v _{2}}{v _{1}} = \frac{T _{2} / μ _{2}}{T _{1} / μ _{1}} = \frac{3 T _{1}}{2 μ _{1}} \cdot \frac{μ _{1}}{T _{1}} = \frac{3}{2} .$ Why this step? Taking the ratio cancels the original $T_{1}, μ_{1}$ , so we only need the factors $3$ and $2$ . This avoids ever knowing the actual numbers.
So $v_{2} = 100 3/2 = 100 1.5 \approx 122.5 m/s$ . Why this step? $1.5 > 1$ , so tripling tension wins over doubling density — the wire ends up faster. The root, again, tames the change.
Travel time (b): $t = \frac{L}{v _{2}} = \frac{5}{122.5} \approx 0.0408 s$ . Why this step? Distance $=$ speed $\times$ time, rearranged. Time and speed are inversely linked, so the faster wire crosses sooner.

Verify: $v_{2}^{2} / v_{1}^{2} = (122.47)^{2} /10 0^{2} \approx 1.5 = 3/2$ ✓. And $v_{2} \cdot t = 122.47 \times 0.04083 \approx 5 m = L$ ✓. See Energy carried by a wave for what changes besides speed when $T$ and $μ$ change.

Recall Rapid re-run (hide answers)

Ratio when $T \times 4$ , $μ$ fixed? ::: $v \times 2$ Ratio when $μ \times \frac{1}{4}$ , $T$ fixed? ::: $v \times 2$ $T \to 0$ gives $v =$ ? ::: $0$ (no restoring force) Read $v$ from $y = A sin (5 x - 20 t)$ ? ::: $v = ω / k = 20/5 = 4 m/s$ Hanging $5 kg$ sets tension to? ::: $T = m g = 50 N$ , not $5$ Does $g (x + v t)$ solve the equation? ::: Yes — $v$ gets squared, sign irrelevant

Connections

Wave equation — derivation for string
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
Newton's Second Law
Partial derivatives and curvature
Energy carried by a wave
Transverse vs Longitudinal Waves