# 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:**
1. 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.
2. **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.)
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 $v = \omega/k$. ✓
4. **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.
5. **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. ✓
6. **g)** is a function of $u = z - 2t$ alone, $h(u) = \cos(u)e^{u}$ — a pure right-mover at
$v = 2$ (take $g \equiv 0$). ✓
7. **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 $v^2 = 6.25$. ✓
8. 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.
```qm-demo traveling-wave-superposition
```
Solution
**Answer: a) and f).**
**Why, step by step:**
1. 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.
2. **a)** $k = 1\ \text{m}^{-1}$: $f = \dfrac{100 \times 1}{2\pi} \approx 15.92\ \text{Hz}$. ✓
3. **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
$f = \dfrac{100 \times 3950}{2\pi} \approx 62\,870\ \text{Hz} \approx 62.87\ \text{kHz}$. ✓
4. **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.
```mermaid
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 ✗"]
```
Solution
**Answer: g) $k = \dfrac{n\pi}{L}$, where $n = 1, 2, 3\dots$**
**Why, step by step:**
1. 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 $\sin(kz)$.
2. 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.)
3. 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:
$\lambda_1 = 2L$.
4. 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.
```qm-demo standing-wave-modes
```