Questionnaire 1.2.7 — The Classical Wave Equation

Three questions on the classical wave equation: recognizing which functions solve it (d'Alembert's form, and the traps that merely look wavy), pairing wavevectors with frequencies through the dispersion relation without dropping a $2\pi$ , and how fixed ends quantize $k$ .

Key Ideas

The wave equation

\frac{\partial^2 \phi}{\partial z^2} - \frac{1}{v^2}\,\frac{\partial^2 \phi}{\partial t^2} = 0

The classical wave equation is linear with real coefficients and owns a single speed $v$ , fixed by the medium. Linearity means solutions superpose — but only solutions of the same equation, i.e. riding at the same $v$ . Complex-valued solutions are perfectly legal (take real parts at the end); on an infinitely long string with no boundary conditions, growing exponentials and unbounded ramps are legal too. "Wave" does not mean "sinusoid".

d'Alembert's general solution

\phi(z,t) = f(z - vt) + g(z + vt)

The d'Alembert solution says a function solves the wave equation exactly when it is a sum of a right-mover $f(z-vt)$ and a left-mover $g(z+vt)$ — any twice-differentiable shapes whatsoever (ramps, parabolas, exponentials, sinusoids), but both riding at the same single $v$ , with the time dependence entering only through the linear combinations $z \mp vt$ (rigid translation at constant speed). Every failure mode is a violation of one clause: two different speeds in one sum, an argument nonlinear in $t$ , or a product of a left- and right-mover instead of a sum.

Standing waves

A standing wave is a product of a pure-space factor and a pure-time factor, such as $\sin(kz)\cos(\omega t)$ , and it is a solution whenever $\omega = vk$ : the product-to-sum identity unfolds it into $\tfrac{1}{2}\sin(kz-\omega t) + \tfrac{1}{2}\sin(kz+\omega t)$ — equal counter-propagating waves. The look-alike trap is a product of a left-mover and a right-mover, e.g. $\sin(z-vt)\cos(z+vt)$ : there each factor already mixes $z$ and $t$ , and the product is not a sum of movers — it fails the equation.

Monochromatic waves and the Helmholtz equation

For a wave oscillating at one angular frequency $\omega$ — time dependence $e^{\pm i\omega t}$ , $\cos(\omega t)$ , $\sin(\omega t)$ , or combinations — separating variables as $\phi(z,t) = Z(z)\,T(t)$ gives $\partial^2\phi/\partial t^2 = -\omega^2\phi$ , and the wave equation collapses to the Helmholtz equation for the spatial part alone: $d^2 Z/dz^2 + k^2 Z = 0$ with $k^2 = \omega^2/v^2$ . Its solutions $\sin(kz)$ , $\cos(kz)$ are the profiles standing waves are built from — the formal reason a (pure space) × (pure time) product is a legitimate solution whenever $\omega = vk$ .

The dispersion relation on a string

\omega = vk \qquad\Longleftrightarrow\qquad f = \frac{\omega}{2\pi} = \frac{vk}{2\pi} = \frac{v}{\lambda}

The dispersion relation ties the wavevector magnitude $k = 2\pi/\lambda$ to the angular frequency $\omega = vk$ in rad/s; the frequency in hertz is $f = \omega/2\pi$ . Two classic slips: quoting $\omega$ when asked for $f$ (off by $2\pi$ ), and treating $k$ as $1/\lambda$ (also off by $2\pi$ , from the other side). Units slip too: a centimetre is smaller than a metre, so a metre holds a hundred times more waves — $1\ \text{cm}^{-1} = 100\ \text{m}^{-1}$ , the conversion runs "up", not "down".

Fixed-end quantization

k_n = \frac{n\pi}{L}, \qquad \lambda_n = \frac{2L}{n}, \qquad n = 1, 2, 3, \dots

Quantization by boundary conditions: pinning a rope at $z=0$ and $z=L$ forces the standing profile $\sin(kz)$ to vanish at both walls, so $\sin(kL) = 0$ and $kL = n\pi$ . Each mode adds half a wavelength between the walls ( $L = n\lambda/2$ ); the fundamental fits half a wave, $\lambda_1 = 2L$ . Contrast a closed ring (periodic boundary conditions), which must fit whole wavelengths and so allows only $k = 2n\pi/L$ — applying the ring rule to a pinned rope throws away every odd mode, fundamental included. The medium's speed $v$ never enters the allowed $k$ 's; it only sets the mode frequencies $f_n = vk_n/2\pi$ .

Exercises

E1 (hard). Which of the following waves are possible solutions to the classical wave equation for a wave on an infinitely long string, assuming there are no boundary conditions. The wave equation is given by

\frac{\partial^2 \phi}{\partial z^2} - \frac{1}{v^2}\,\frac{\partial^2 \phi}{\partial t^2} = 0

where $v$ is the magnitude of the velocity of the wave. SIX of the following choices are possible solutions. Choose ALL of the options that are correct for at least some value of the wave velocity (possibly different in each case).

a) $\phi(z,t) = 6\left(\sin(z - 4t) + \sin(z + 4t)\right)$ b) $\phi(z,t) = \cos(0.1t)\,\sin(11.1z)$ c) $\phi(z,t) = i\cos(z - vt^2)$ d) $\phi(z,t) = \exp(i(3z - 5t)) + \exp(i(2z + 4t))$ e) $\phi(z,t) = \exp(2z - t) + \exp(2z + t)$ f) $\phi(z,t) = 4z - 4vt - 5\exp(z + vt)$ g) $\phi(z,t) = \cos(z - 2t)\exp(z - 2t)$ h) $\phi(z,t) = \sin(z - vt)\cos(z + vt)$ i) $\phi(z,t) = (2z + 5t)^2$ j) $\phi(z,t) = (az)^2 - (bt)^2$ where $a$ and $b$ are real constants with appropriate physical dimensions

Solution

Answer: a), b), e), f), g), i).

Why, step by step:

The master test is d'Alembert's theorem: $\phi$ solves the equation for some $v$ iff it can be written $f(z - vt) + g(z + vt)$ — any twice-differentiable shapes, but one single $v$ in both terms. Nothing else matters: not whether it oscillates, stays bounded, or is real.
a) is already in that form with $v = 4$ : $f = 6\sin(z-4t)$ , $g = 6\sin(z+4t)$ . ✓ (It is also the standing wave $12\sin z\cos 4t$ — see step 3.)
b) is a standing wave, space factor × time factor. Product-to-sum: $\cos(0.1t)\sin(11.1z) = \tfrac{1}{2}\sin(11.1z - 0.1t) + \tfrac{1}{2}\sin(11.1z + 0.1t)$ , two movers sharing $v = \omega/k = 0.1/11.1 \approx 0.009$ . Or directly: $\phi_{zz} = -k^2\phi$ and $\phi_{tt} = -\omega^2\phi$ , so the equation reads $(-k^2 + \omega^2/v^2)\phi = 0$ , satisfied at $v = \omega/k$ . ✓
e) regroups as $e^{2(z - t/2)} + e^{2(z + t/2)}$ — both movers at $v = \tfrac{1}{2}$ . Check: $\phi_{zz} = 4\phi$ , $\phi_{tt} = \phi$ , and $4 - 1/v^2 = 0$ at $v = \tfrac{1}{2}$ . ✓ Blowing up as $z \to \infty$ is fine: no boundary conditions were imposed.
f) is $4(z - vt) - 5e^{(z+vt)}$ : a right-moving linear ramp plus a left-moving exponential, same $v$ in both. Shape is irrelevant; only the arguments $z \mp vt$ matter. ✓
g) is a function of $u = z - 2t$ alone, $h(u) = \cos(u)e^{u}$ — a pure right-mover at $v = 2$ (take $g \equiv 0$ ). ✓
i) is $\bigl(2(z + \tfrac{5}{2}t)\bigr)^2$ , a left-moving parabola with $v = 2.5$ . Check: $\phi_{zz} = 8$ , $\phi_{tt} = 50$ , and $8 - 50/v^2 = 0$ at $v^2 = 6.25$ . ✓
That is six — matching the question's count — so c), d), h), j) must all fail; the autopsy of each is below.

Why the tempting options fail:

c) The argument $z - vt^2$ is not linear in $t$ : this profile accelerates, and d'Alembert demands rigid translation at constant speed. Plugging in leaves $i\left[-\cos u - \tfrac{2}{v}\sin u + 4t^2\cos u\right]$ with $u = z - vt^2$ , and the explicit $t^2$ can never cancel the rest for all $t$ . The $i$ out front is a red herring — the equation is linear with real coefficients, so complex solutions are fine (d's individual terms prove it); the $t^2$ is the killer.
d) Each exponential is a solution — but of different equations: $e^{i(3z-5t)}$ needs $v = 5/3$ , $e^{i(2z+4t)}$ needs $v = 2$ . Superposition only holds between solutions of the same equation: the sum requires $(-9 + 25/v^2) = 0$ and $(-4 + 16/v^2) = 0$ simultaneously, i.e. $v^2 = 25/9$ and $v^2 = 4$ at once. One string, one $v$ . ✗
h) A product of a right-mover and a left-mover, not a sum. Product-to-sum: $\sin(z-vt)\cos(z+vt) = \tfrac{1}{2}\sin(2z) - \tfrac{1}{2}\sin(2vt)$ , so $\phi_{zz} = -2\sin 2z$ (no time dependence to balance it) and $\phi_{tt} = 2v^2\sin(2vt)$ (no $z$ dependence), leaving $-2\sin 2z - 2\sin 2vt \neq 0$ . Contrast b): a standing wave is (pure space) × (pure time); here each factor already mixes $z$ and $t$ , which is fatal.
j) $\phi_{zz} = 2a^2$ and $\phi_{tt} = -2b^2$ , so the equation's left side is $2a^2 + 2b^2/v^2 > 0$ — no $v$ can help; the minus sign between the squares is the killer. Both cousins work: $(az)^2 + (bt)^2$ (with $v = b/a$ ) and $(az - bt)^2$ (a mover of speed $b/a$ ). The bait is the factorization $(az)^2 - (bt)^2 = (az - bt)(az + bt)$ — a product of movers, which h) just showed is no solution.

See it: option d)'s autopsy, live — the demo starts as two movers at different speeds ( $v_1 = 1$ right, $v_2 = 2$ left, d's situation). The top trace always looks like a perfectly good wave, yet no setting of the equation speed $v$ flattens the residual $\phi_{zz} - \phi_{tt}/v^2$ underneath. Make the speeds equal and match $v$ to them: the residual dies, and the solution that remains is a standing wave — which is how a) and b) earn their checkmarks.

E2 (medium). Consider the string from the previous question. If the wave velocity on the string is given to be $100\ \text{m/s}$ , which of the following combinations of wavevector magnitude, $k$ , and frequency, $f$ , are possible. Choose the correct answers (there may be more than one).

[Note for (f) that if you have one mark on a ruler for every centimeter, you have 100 for every meter, so $1\ \text{cm}^{-1} = 100\ \text{m}^{-1}$ .]

a) $f \approx 15.92\ \text{Hz}$ and $k = 1\ \text{m}^{-1}$ b) $f = 100\ \text{Hz}$ and $k = 1\ \text{m}^{-1}$ c) $f = 100\pi\ \text{Hz}$ and $k = \pi\ \text{m}^{-1}$ d) $f = 2\pi\ \text{kHz}$ and $k = 100\ \text{m}^{-1}$ e) $f = 10\ \text{kHz}$ and $k = 100\ \text{m}^{-1}$ f) $f \approx 62.87\ \text{kHz}$ and $k \approx 39.50\ \text{cm}^{-1}$ g) Cannot be determined; need more information. h) None of the above.

Solution

Answer: a) and f).

Why, step by step:

A harmonic wave $\sin(kz - \omega t)$ has the d'Alembert form only if $\omega = vk$ , so the frequency in hertz must satisfy $f = \dfrac{\omega}{2\pi} = \dfrac{vk}{2\pi}$ with $v = 100\ \text{m/s}$ . Every option is a one-line check against this.
a) $k = 1\ \text{m}^{-1}$ : $f = \dfrac{100 \times 1}{2\pi} \approx 15.92\ \text{Hz}$ . ✓
f) first the units: $k \approx 39.50\ \text{cm}^{-1} = 3950\ \text{m}^{-1}$ (the note — 100 marks per metre for every mark per centimetre, so cm⁻¹ → m⁻¹ multiplies by 100). Then $f = \dfrac{100 \times 3950}{2\pi} \approx 62\,870\ \text{Hz} \approx 62.87\ \text{kHz}$ . ✓
g) fails because nothing is undetermined: given $v$ , the pairs $(k, f)$ form the one-parameter family $f = vk/2\pi$ , and each option hands you both members to test. h) fails because a) and f) work.

Why the tempting options fail:

b), c), e) all satisfy $f = vk$ without the $2\pi$ : $100 = 100 \times 1$ , $100\pi = 100 \times \pi$ , $10^4 = 100 \times 100$ . That is, each quotes the angular frequency $\omega = vk$ (rad/s) and calls it $f$ (Hz) — or equivalently treats $k$ as $1/\lambda$ when it is $2\pi/\lambda$ . Correct values: 15.92 Hz, 50 Hz, and 1.59 kHz respectively.
d) is $2\pi$ -flavoured bait parked next to e): $f = 2\pi\ \text{kHz} \approx 6.28\ \text{kHz}$ would need $k = 2\pi f/v \approx 395\ \text{m}^{-1}$ , not $100\ \text{m}^{-1}$ — and read the other way, $k = 100\ \text{m}^{-1}$ demands $f \approx 1.59\ \text{kHz}$ . It matches no single misreading; it just looks like someone's $\omega \leftrightarrow f$ conversion.
The stealth trap inside f): converting cm⁻¹ downward ( $\div 100$ , "centi means small") gives $0.395\ \text{m}^{-1}$ and $f \approx 6.3\ \text{Hz}$ — nowhere near the listed value, so the wrong conversion makes you reject the right answer. Marks per centimetre are denser on a metre: $1\ \text{cm}^{-1} = 100\ \text{m}^{-1}$ .

See it: one road from $k$ to $f$ ; every wrong option is a labelled wrong exit.

flowchart TD
    k["given: v = 100 m/s and k"] -->|"ω = vk"| om["ω = 100·k rad/s"]
    om -->|"f = ω/2π"| f["f = vk/2π ✓"]
    f --> okA["k = 1 m⁻¹ → f = 15.92 Hz ✓ (a)"]
    f --> okF["k = 39.50 cm⁻¹ = 3950 m⁻¹ → f = 62.87 kHz ✓ (f)"]
    om -.->|"report ω as if it were f"| trap1["(b) 100 Hz · (c) 100π Hz · (e) 10 kHz ✗ — all pass f = vk, none pass f = vk/2π"]
    k -.->|"convert cm⁻¹ by ÷100"| trap2["39.50 cm⁻¹ → 0.395 m⁻¹ ✗ — a metre holds 100× more waves: 1 cm⁻¹ = 100 m⁻¹"]
    f -.->|"no consistent misreading"| trap3["(d) 2π kHz would need k ≈ 395 m⁻¹, not 100 m⁻¹ ✗"]

E3 (medium). Consider a rope strung between two walls that are a distance $L$ apart. For a standing wave to occur (i.e. there are boundary conditions such that $f(z,t)$ is always equal to $0$ when $z = 0$ and $z = L$ ), what does the magnitude of the k vector need to be? Choose the correct option from the list below that is also the most general possible result (i.e., that includes all possible correct answers).

a) $k = \frac{1}{2L}$ b) $k = \frac{1}{L}$ c) $k = \frac{\pi}{L}$ d) $k = \frac{2\pi}{L}$ e) $k = \frac{n}{2L}$ , where $n = 1, 2, 3\dots$ f) $k = \frac{n}{L}$ , where $n = 1, 2, 3\dots$ g) $k = \frac{n\pi}{L}$ , where $n = 1, 2, 3\dots$ h) $k = \frac{2n\pi}{L}$ , where $n = 1, 2, 3\dots$ i) Cannot be determined; need more information. j) None of the above.

Solution

Answer: g) $k = \dfrac{n\pi}{L}$ , where $n = 1, 2, 3\dots$

Why, step by step:

A standing wave of wavevector $k$ has the profile $A\sin(kz) + B\cos(kz)$ , oscillating in time. Pinning the rope at $z = 0$ for all $t$ kills the cosine: $B = 0$ , leaving $\sin(kz)$ .
Pinning it at $z = L$ then demands $\sin(kL) = 0$ , i.e. $kL = n\pi$ : $k_n = \frac{n\pi}{L}, \qquad n = 1, 2, 3, \dots$ ( $n = 0$ is the rope lying flat — no wave.)
In wavelength language: $\lambda_n = 2\pi/k_n = 2L/n$ , so the walls must frame a whole number of half-wavelengths, $L = n\lambda/2$ . The fundamental ( $n = 1$ ) fits half a wave: $\lambda_1 = 2L$ .
This is the most general family — every allowed $k$ and nothing else — and it is quantization by boundary conditions in its original classical home: confining a wave turns the continuum of allowed $k$ 's into a discrete ladder.

Why the tempting options fail:

c), d) are genuine modes ( $n = 1$ and $n = 2$ ) but single rungs, not the ladder — the question asks for the most general result. d) is the extra-seductive one: the classic two-walls standing-wave picture is usually drawn with one full wavelength between the walls ( $k = 2\pi/L$ , $\omega = 2\pi v/L$ ), which makes it feel canonical — but that drawing is just the $n = 2$ mode, not the rule.
h) $2n\pi/L$ is whole-wavelength counting — the rule for a closed ring (periodic boundary conditions), where the wave must return to itself. A pinned rope only needs nodes at the ends, which half a wavelength already delivers; h) keeps the even modes and silently discards the fundamental and every other odd mode.
e), f) (and their single-value versions a), b)) carry correct-looking wavelength counting but equate $k$ with $1/\lambda$ : from $\lambda_n = 2L/n$ one gets $1/\lambda = n/2L$ — and then $k = 2\pi/\lambda = n\pi/L$ , not $n/2L$ . The missing $2\pi$ is the same slip as E2's.
i) fails because the geometry alone quantizes $k$ : tension, density, and $v$ set the mode frequencies $f_n = vk_n/2\pi$ , but the allowed $k$ 's come from the boundary conditions only.

See it: drag $k$ between the integers of $kL/\pi$ — the rope's right end waves in the air off its wall pin, and only at $kL = n\pi$ does it land. Watch the fundamental: half a wavelength between the walls ( $\lambda = 2L$ ) — exactly the mode the ring rule $k = 2n\pi/L$ of option h) would forbid.