# Questionnaire 1.3 — Quanta, Oscillators & Waves > Nine questions sweeping the week-one arc: black bodies, the photoelectric effect, the > de Broglie wavelength, energy and forces on a slope, the harmonic oscillator (right-side-up > and upside-down), and which frequencies a string will actually stand still for. ## Key Ideas A **black body** absorbs every photon that lands on it, at every wavelength — it is by definition the best possible absorber of light. By Kirchhoff's law, a perfect absorber in thermal equilibrium is also the best possible thermal _emitter_: "black" means it reflects nothing, not that it emits nothing. The **Stefan–Boltzmann law** fixes how much a black body radiates in total: $$ P = \sigma A T^4 $$ Power grows as the _fourth_ power of temperature, so a hotter black body always emits more light than a colder one — double the temperature and the output multiplies by sixteen. **Wien's displacement law** fixes _where_ the emission peaks: $$ \lambda_{\text{max}} = \frac{b}{T}, \qquad b \approx 2.898\times 10^{-3}\ \text{m·K} $$ Hotter bodies peak at _shorter_ wavelengths. Since blue light is shorter-wavelength than red, the color ladder runs red-hot → white-hot → blue-hot as temperature climbs. In the **photoelectric effect**, light hands energy to electrons one photon at a time, each photon carrying $E = hf = hc/\lambda$. The fastest ejected electron has $KE_{\max} = hc/\lambda - W$, where $W$ is the metal's work function. The **stopping voltage** — the reverse voltage that just barely chokes off the photocurrent — satisfies $eV_{\text{stop}} = KE_{\max}$. It therefore tracks the light's _wavelength_ (per-photon energy) and is completely indifferent to the light's _intensity_, which only sets how many photons arrive per second, i.e. the size of the current, not the energy per electron. The **de Broglie wavelength** of a particle with momentum $p$ is $$ \lambda = \frac{h}{p} = \frac{h}{mv} $$ Small mass and small speed mean a _large_ quantum wavelength: an electron creeping along at $1\ \text{m/s}$ has $\lambda \approx 0.73\ \text{mm}$ — practically macroscopic. **Gravitational potential energy** near the Earth's surface changes only with _vertical rise_: $\Delta U = mgh$. Walking a distance $d$ up a slope inclined at $\theta$ gains height $h = d\sin\theta$ — the path length itself never enters, only how much altitude it buys. On an incline, the **weight component along the slope** is $mg\sin\theta$; the perpendicular component $mg\cos\theta$ is carried by the normal force. Holding an object stationary on a frictionless slope therefore takes an uphill push of exactly $mg\sin\theta$ — not the full weight $mg$. A **parabolic potential** is a spring in disguise: matching $V(z) = \tfrac{1}{2}kz^2$ to a given coefficient reads off $k$ (so $V = 10\,z^2$ means $k = 20\ \text{N/m}$ — don't forget the $\tfrac{1}{2}$). A mass $m$ in that potential oscillates at $$ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} $$ in hertz; $\sqrt{k/m}$ alone is the angular frequency $\omega$ in rad/s, a different unit. The simple harmonic oscillator is **isochronous**: it has exactly _one_ frequency, independent of amplitude and initial conditions. A mass on a spring has no harmonics, no overtones, no frequency ladder — that richness belongs to extended systems like strings. An **inverted parabolic potential** $V(z) = -\tfrac{1}{2}kz^2$ is a hilltop instead of a valley. The equation of motion flips sign, $\ddot z = +\omega^2 z$, and the solutions become growing and decaying **exponentials** $e^{\pm\omega t}$: the equilibrium at the top is unstable, and the least nudge runs away instead of swinging back. A **standing wave** on a string of length $L$ fixed at both ends must fit a whole number of half-wavelengths between the walls: $\lambda_n = 2L/n$. With wave speed $v$, the allowed frequencies are $$ f_n = n\,\frac{v}{2L}, \qquad n = 1, 2, 3, \dots $$ Any frequency of this form rings; any other frequency — half-integer multiples like $v/4L$, or arbitrary real/complex multiples — cannot keep both ends pinned. ## Exercises **E1 (easy).** For the following statements about a black body, choose those that are true. (More than one may be true. No partial credit — you must choose all those that are true and only those that are true.) a) A black body is the best possible absorber of light. b) A colder black body emits more light than a hotter one. c) A black body that is glowing blue is colder than a black body that is glowing red.
Solution **Answer: a) only.** **Why, step by step:** 1. **a) is true by definition.** A black body absorbs _all_ radiation incident on it, at every wavelength — nothing is reflected or transmitted. No absorber can do better than "all of it." 2. **b) is false by the Stefan–Boltzmann law.** Total emitted power is $P = \sigma A T^4$ — a steeply _increasing_ function of temperature. Hotter emits more, at every wavelength in fact, not just in total. 3. **c) is false by Wien's law.** The spectral peak sits at $\lambda_{\text{max}} = b/T$: higher temperature pushes the peak to _shorter_ wavelengths. Blue is shorter-wavelength than red, so a blue-glowing body is the **hotter** one (think: a blue-white star outshines and out-cooks a red dwarf). **Why the tempting options fail:** - **b)** may tempt via "black = dark = doesn't glow" — but blackness is about _absorbing_, and a perfect absorber is also the best emitter (Kirchhoff); its output then grows as $T^4$. - **c)** rides on everyday color language — "red-hot" sounds extreme, and cool colors vs. warm colors in art run exactly backwards from physics. **See it:** slide the temperature up — the whole curve rises (hotter emits more everywhere, sinking b) while the peak marches from red toward blue (blue-hot is hotter, sinking c). ```qm-demo blackbody-spectrum ```
**E2 (medium).** For the following statements about the photoelectric effect, choose those that are true. (More than one may be true. No partial credit — you must choose all those that are true and only those that are true.) a) In the photoelectric effect experiment, if the light is changed to have a shorter wavelength, the magnitude of the voltage on the collecting plate will have to increase to cut off current in the circuit. b) In the photoelectric effect experiment, if the intensity (i.e., power per unit area) of the light is increased, the magnitude of the voltage on the collecting plate will have to increase approximately in proportion to the intensity to cut off current in the circuit.
Solution **Answer: a) only.** **Why, step by step:** 1. The current is cut off when the reverse voltage stops even the _fastest_ photoelectrons: $eV_{\text{stop}} = KE_{\max}$. 2. Einstein's photoelectric relation gives $KE_{\max} = \dfrac{hc}{\lambda} - W$. Shorter wavelength → more energy per photon → faster electrons → a larger stopping voltage is needed. **a) is true.** 3. Intensity is photons _per second_, not energy _per photon_. Doubling the intensity doubles the number of ejected electrons (more current) but leaves each electron's maximum kinetic energy — and hence $V_{\text{stop}}$ — untouched. **b) is false.** 4. Historically, b) is exactly what the classical wave picture predicted (more intense wave → more energy pumped into each electron), and its experimental failure is what the photon model was invented to explain. **Why the tempting option fails:** - **b)** feels right because "stronger light = more energetic light" — but intensity buys _more photons_, never _better photons_. Only wavelength sets the energy each electron can receive. **See it:** work the intensity slider and keep your eye on the two bars — the current bar follows it, the $V_{\text{stop}}$ bar never twitches. Only the wavelength slider moves that one. ```qm-demo photoelectric-stopping-voltage ```
**E3 (easy).** What is the corresponding wavelength, in meters, of an electron moving at $1\ \text{m/s}$? Enter your answer to three significant figures in the form `1.23e45`, where the number after the "e" represents the power of ten. (Just enter a number — do not attempt to enter any units.)
Solution **Answer: `7.27e-4`** (i.e. $\lambda \approx 7.27\times 10^{-4}\ \text{m}$). **Why, step by step:** 1. The de Broglie relation: $\lambda = \dfrac{h}{p} = \dfrac{h}{m_e v}$. 2. Momentum of the electron: $p = m_e v = (9.109\times 10^{-31}\ \text{kg})(1\ \text{m/s}) = 9.109\times 10^{-31}\ \text{kg·m/s}$. 3. Wavelength: $\lambda = \dfrac{6.626\times 10^{-34}}{9.109\times 10^{-31}} = 7.274\times 10^{-4}\ \text{m}$. 4. To three significant figures: $7.27\times 10^{-4}\ \text{m}$ — about $0.7$ millimeters. A slow electron has a nearly macroscopic wavelength; this is why matter-wave effects demand slow, light particles. **Why the tempting slips fail:** - Using the _proton_ mass ($1.67\times 10^{-27}$ kg) gives $3.97\times 10^{-7}$ m — the formula punishes grabbing the wrong particle's mass. - Computing $h/(m_e v^2)$ or mixing in kinetic energy ($\lambda = h/\sqrt{2mE}$ is the same thing only if you use $E = \tfrac{1}{2}mv^2$ consistently) are the classic algebra detours; at $v = 1$ m/s they happen to coincide numerically, which hides the error for the next problem where $v \neq 1$. **See it:** drag the speed slider down toward $1\ \text{m/s}$ and watch the wavelength readout balloon — $\lambda \propto 1/v$, so slower means longer, into the sub-millimeter range. ```qm-demo de-broglie-speed ```
**E4 (medium).** For E4 and E5, consider a hill with a slope of 30 degrees relative to the horizontal direction. I push a ball of mass $2\ \text{kg}$ directly up the slope, and as I do so, I take ten steps, walking up the slope, with each step being $50\ \text{cm}$ long. Presuming that the downward (i.e., vertical) force, in newtons, exerted by gravity on the ball is $9.8$ times the mass of the ball in kg, by how many joules has the potential energy of the ball increased? a) $0\ \text{J}$ b) $49\ \text{J}$ c) $56.6\ \text{J}$ d) $58.8\ \text{J}$ e) $84.9\ \text{J}$ f) $98\ \text{J}$ g) Cannot be determined; need more information. h) None of the above.
Solution **Answer: b) $49\ \text{J}$.** **Why, step by step:** 1. Distance walked _along the slope_: $d = 10 \times 0.5\ \text{m} = 5\ \text{m}$. 2. Gravitational potential energy cares only about _vertical rise_: $h = d\sin 30^\circ = 5 \times 0.5 = 2.5\ \text{m}$. 3. Weight of the ball: $mg = 2 \times 9.8 = 19.6\ \text{N}$. 4. $\Delta U = mgh = 19.6\ \text{N} \times 2.5\ \text{m} = 49\ \text{J}$. **Why the tempting options fail:** - **f) $98\ \text{J}$** $= 19.6 \times 5$: treats the whole 5 m slope path as if it were vertical height — the central trap of the question. - **e) $84.9\ \text{J}$** $= 19.6 \times 5\cos 30^\circ$: resolves along the _horizontal_ instead of the vertical — gravity doesn't care how far sideways you went. - **c) $56.6\ \text{J}$** $= 19.6 \times 5\tan 30^\circ$: a tan-for-sin slip (rise-over-run applied to the wrong side of the triangle). - **d) $58.8\ \text{J}$** $= 19.6 \times 5 \times 0.6$: the 3–4–5-triangle reflex, $\sin 30^\circ$ misremembered as $0.6$ instead of exactly $0.5$. - **a) $0\ \text{J}$**: the ball ended up 2.5 m higher — its potential energy cannot be unchanged. - **g)**: no missing information — friction, speed, and how hard I pushed are all irrelevant to a potential energy _difference_, which depends only on the endpoints' heights. **See it:** hold $d$ at $5\ \text{m}$ and flatten the slope toward $0^\circ$. The walk stays five metres long the whole way while the dashed rise — and $\Delta U$ with it — drains to zero. ```qm-demo incline-decomposition ```
**E5 (easy).** Presuming the ball is free to move, what force, pushing uphill along the direction of the slope, do I have to exert on the ball just to hold it stationary on the hill? a) $-19.6\ \text{N}$ b) $-9.8\ \text{N}$ c) $0\ \text{N}$ d) $9.8\ \text{N}$ e) $19.6\ \text{N}$ f) Cannot be determined; need more information. g) None of the above.
Solution **Answer: d) $9.8\ \text{N}$.** **Why, step by step:** 1. The ball's weight is $mg = 19.6\ \text{N}$, pointing straight down. 2. Resolve it relative to the slope: the component _along_ the slope (pulling the ball downhill) is $mg\sin 30^\circ = 19.6 \times 0.5 = 9.8\ \text{N}$; the component _into_ the slope is $mg\cos 30^\circ \approx 17.0\ \text{N}$. 3. The slope's normal force automatically balances the into-the-slope component. The only thing left for me to balance is the downhill pull. 4. So I must push uphill with $9.8\ \text{N}$. The question defines the positive direction as "pushing uphill along the slope," so the answer is $+9.8\ \text{N}$. **Why the tempting options fail:** - **e) $19.6\ \text{N}$**: pushes with the ball's entire weight, forgetting that the normal force already carries the $mg\cos\theta$ part — you'd launch the ball uphill. - **b) $-9.8\ \text{N}$**: right magnitude, wrong sign — a negative answer here would mean pushing _downhill_, which would only help gravity. - **c) $0\ \text{N}$**: true on flat ground; on a frictionless slope an unheld ball rolls down. - **f)**: nothing is missing — statics on a frictionless incline needs only $m$, $g$, and $\theta$. **See it:** the same slope again, now reading the force arrows. Sweep $\theta$ and compare the long arrow the surface absorbs for free against the short dashed one that is yours to cancel. ```qm-demo incline-decomposition ```
**E6 (medium).** Consider a one-dimensional parabolic potential of the form $V(z) = 10\,z^2$, acting on a mass of $0.5\ \text{kg}$. What is the oscillation frequency of this mass? Choose the most correct answer. (Note: when we call something a "potential" here, we are using it as shorthand for "potential energy" — a quantity in joules, not an electrostatic potential in volts. The "$10$" therefore carries appropriate units and is not a pure number.) a) $25.2\ \text{mHz}$ b) $158\ \text{mHz}$ c) $993\ \text{mHz}$ d) $1.01\ \text{Hz}$ e) $6.32\ \text{Hz}$ f) $6.37\ \text{Hz}$ g) $39.7\ \text{Hz}$ h) $40\ \text{Hz}$ i) $251\ \text{Hz}$ j) Cannot be determined; need more information. k) None of the above.
Solution **Answer: d) $1.01\ \text{Hz}$.** **Why, step by step:** 1. Match the potential to the standard spring form $V(z) = \tfrac{1}{2}k z^2$: $\tfrac{1}{2}k = 10\ \text{J/m}^2 \Rightarrow k = 20\ \text{N/m}$. (The factor of 2 is the first trap.) 2. Angular frequency: $\omega = \sqrt{k/m} = \sqrt{20/0.5} = \sqrt{40} \approx 6.32\ \text{rad/s}$. 3. Convert to hertz — cycles per second, one cycle being $2\pi$ radians: $f = \dfrac{\omega}{2\pi} = \dfrac{6.32}{6.28} \approx 1.01\ \text{Hz}$. 4. Sanity check: about one bounce per second for a half-kilogram mass on a 20 N/m spring — entirely believable. **Why the tempting options fail:** - **e) $6.32\ \text{Hz}$** is $\omega$ itself — right number, wrong unit (rad/s, not Hz). - **h) $40\ \text{Hz}$** is $k/m$ with the square root forgotten. - **g) $39.7\ \text{Hz}$** is $2\pi\omega$ — multiplying by $2\pi$ instead of dividing. - **i) $251\ \text{Hz}$** is $2\pi\,k/m$ — both mistakes at once. - **c) $993\ \text{mHz}$** is $1/f = 0.993\ \text{s}$ — the _period_ dressed up as a frequency; suspiciously close to the right answer, which is what makes it evil. - **b) $158\ \text{mHz}$** is $1/\omega$; **a) $25.2\ \text{mHz}$** is $1/(2\pi\omega)$ — reciprocals of the wrong quantities. - **j)**: the note fixes the units of the "10," so everything needed is on the table. **See it:** one road from $V(z)$ to hertz, with every distractor a marked wrong exit. ```mermaid flowchart TD v["V(z) = 10·z² (joules, z in meters)"] -->|"match ½k·z² → k = 2 × 10"| k["k = 20 N/m"] v -. "read k = 10 (forgot the ½)" .-> badk["k = 10 ✗ — wrong from step one"] k -->|"ω = √(k/m) = √40"| om["ω ≈ 6.32 rad/s"] k -. "forget the √" .-> h40["40 Hz ✗ — that's k/m"] om -->|"÷ 2π rad per cycle"| f["f ≈ 1.01 Hz ✓"] om -. "report as-is" .-> e632["6.32 Hz ✗ — rad/s, not Hz"] om -. "× 2π instead" .-> g397["39.7 Hz ✗"] f -. "flip it" .-> c993["0.993 — the period in s, not a frequency ✗"] ```
**E7 (medium).** In the previous problem, the solutions to the governing differential equation are oscillatory. What is the nature of the solutions if the potential is inverted (i.e., turned "upside down" or, equivalently, multiplied by $-1$)? a) still oscillatory b) a constant c) linear d) quadratic e) exponential f) logarithmic g) Cannot be determined; need more information. h) None of the above.
Solution **Answer: e) exponential.** **Why, step by step:** 1. Newton's law with $V(z) = -10\,z^2$: the force is $F = -\dfrac{dV}{dz} = +20\,z$ — it points _away_ from $z = 0$ and grows with distance. 2. The equation of motion becomes $\ddot z = +\dfrac{k}{m} z = +40\,z$. Compare the upright potential's $\ddot z = -40\,z$: one sign has flipped, and that sign is the entire story. 3. With a **minus** sign, trial solution $e^{i\omega t}$ works and you get sines and cosines — oscillation. With a **plus** sign, the trial solution $e^{\lambda t}$ gives $\lambda^2 = 40$, so $\lambda = \pm\sqrt{40}$, and the general solution is $z(t) = A\,e^{+\sqrt{40}\,t} + B\,e^{-\sqrt{40}\,t}$ — real exponentials. 4. Physically: the mass now sits on a hilltop. Any nudge is amplified — the displacement runs away exponentially instead of being pulled back. This is an _unstable_ equilibrium. **Why the tempting options fail:** - **a) still oscillatory**: flipping $V$ looks cosmetic ("same parabola, upside down"), but it reverses the force from restoring to repelling — oscillation requires the pull-back. - **b) a constant**: $z = 0$ forever _is_ one solution (balanced exactly on the summit), but it is a single unstable special case, not the nature of the general solution. - **d) quadratic**: confuses the shape of $V(z)$ in _space_ with the behavior of $z(t)$ in _time_. - **g)**: the equation is fully specified; its solution family is a known closed form. **See it:** drag $c$ from $+10$ down through zero and watch the force arrow on the mass flip outward — the trace stops turning back and leaves the frame before a cycle can finish. ```qm-demo restoring-vs-runaway ```
**E8 (medium).** Consider the simple harmonic oscillator of a mass $m$ on a spring with spring constant $k$. What oscillation frequency or frequencies is/are possible in this system? ("Frequency" here is in cycles per second or Hz, not angular frequency in radians per second.) Choose the correct answer or answers. (There might be more than one correct answer, though it is also possible there is only one. You must choose all correct answers and no incorrect ones.) a) $\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ b) $\sqrt{\dfrac{k}{m}}$ c) $\dfrac{n}{2\pi}\sqrt{\dfrac{k}{m}}$ where $n$ is any positive integer d) $2\pi\sqrt{\dfrac{k}{m}}$ e) $n\sqrt{\dfrac{k}{m}}$ where $n$ is any positive integer f) $2\pi n\sqrt{\dfrac{k}{m}}$ where $n$ is any positive integer g) $\dfrac{1}{2\pi}\sqrt{\dfrac{m}{k}}$ h) $\sqrt{\dfrac{m}{k}}$ i) $\dfrac{n}{2\pi}\sqrt{\dfrac{m}{k}}$ where $n$ is any positive integer j) $2\pi\sqrt{\dfrac{m}{k}}$ k) $n\sqrt{\dfrac{m}{k}}$ where $n$ is any positive integer l) $2\pi n\sqrt{\dfrac{m}{k}}$ where $n$ is any positive integer m) None of the above.
Solution **Answer: a) only — $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$.** **Why, step by step:** 1. The equation of motion $m\ddot x = -kx$ has the general solution $x(t) = A\cos(\omega t + \phi)$ with $\omega = \sqrt{k/m}$ — and _no other frequency appears for any choice of $A$ or $\phi$_. 2. In hertz: $f = \omega/2\pi = \dfrac{1}{2\pi}\sqrt{k/m}$. That is option a). 3. The crucial physics: a mass on a spring is a **single-mode** system. Unlike a string or an organ pipe, it has no shape to subdivide — there is nothing that could vibrate at a second, higher frequency. Amplitude doesn't change the frequency either (isochronism). So exactly one frequency is possible, and only a) is correct. **Why the tempting options fail:** - **c)** is the seductive one: it _contains_ a) as its $n = 1$ case, so it feels safe to tick. But c) claims _every_ integer multiple is possible — that the oscillator could also run at $2f$, $3f$, … It can't. Ticking c) asserts frequencies the system does not have. - **b)** is $\omega$, the angular frequency in rad/s — the question explicitly asks for Hz. - **d)** is off by a factor of $(2\pi)^2$ from a) — multiplied instead of divided. - **g)–l)** all invert the ratio to $\sqrt{m/k}$, which has units of time per radian; **j)** $2\pi\sqrt{m/k}$ is exactly the _period_ $T$, the reciprocal of the answer. - **m)** fails because a) is sitting right there. **See it:** kick the mass with any amplitude you like — the trace changes size but never pace; there is exactly one frequency and no knob that adds harmonics. ```qm-demo mass-spring-oscillation ```
**E9 (hard).** Now consider a string tied to two walls a distance $L$ apart. Given a particular wave velocity $v$ on this string, which of the following oscillatory frequencies will produce a standing wave? Choose the correct answer or answers. Notes: - There might be more than one correct answer, though it is also possible there is only one. You must choose all correct answers and no incorrect ones. - This question does not ask if any of these is a _universal formula_ for these frequencies; it just asks if these frequencies will produce a standing wave. Possibly none of these is the universal formula for the possible frequencies, but still one or more of them may correspond to frequencies that give standing waves. - Even if you think one case is included in a more general one — e.g., $\frac{v}{L}$ is a special case of $n\frac{v}{L}$ where $n = 1, 2, 3, \dots$ — but you think both are correct, you should check both. a) $\dfrac{v}{4L}$ b) $\dfrac{v}{2L}$ c) $\dfrac{v}{L}$ d) $n\dfrac{v}{L}$ where $n = 1, 2, 3, \dots$ e) $c\dfrac{v}{L}$ where $c$ is any complex number f) Cannot be determined; need more information. g) None of the above.
Solution **Answer: b), c), and d).** **Why, step by step:** 1. Both ends are tied down, so the string must fit a whole number of half-wavelengths between the walls: $L = n\,\dfrac{\lambda_n}{2}$, i.e. $\lambda_n = \dfrac{2L}{n}$. 2. With $f = v/\lambda$, the allowed frequencies are the harmonic ladder $f_n = n\,\dfrac{v}{2L}$ for $n = 1, 2, 3, \dots$ — the universal formula. Now test each option for membership in that ladder. 3. **b) $\dfrac{v}{2L}$**: the $n = 1$ rung — the fundamental. ✓ 4. **c) $\dfrac{v}{L}$**: equals $2\cdot\dfrac{v}{2L}$, the $n = 2$ rung — the second harmonic. ✓ 5. **d) $n\dfrac{v}{L}$**: each member equals $2n\cdot\dfrac{v}{2L}$ — the even rungs ($n = 2, 4, 6, \dots$ of the true ladder). It is _not_ the universal formula (it misses the odd harmonics), but the question only asks whether these frequencies produce standing waves — and every one of them does. ✓ 6. **a) $\dfrac{v}{4L}$**: would need $n = \tfrac{1}{2}$ — half a rung. At that frequency the string would have a quarter wavelength between the walls: one end pinned, the other at maximum swing. A wall can't swing. ✗ 7. **e)** sweeps in _every_ complex multiple — $c = 0.3$, $c = \pi$, $c = i$ — almost none of which land on a rung (and a complex frequency isn't a physical driving frequency at all). ✗ **Why the tempting options fail:** - **Skipping d)** is the main trap: "d) isn't the formula I memorized ($n\frac{v}{2L}$), so it must be wrong." The notes warn against exactly this — a set of frequencies can be valid without being complete. - **a)** borrows the right answer to a _different_ problem: $\frac{v}{4L}$ is the fundamental of a string (or air column) fixed at one end and **free** at the other. Both ends fixed here. - **f)** fails because $v$ and $L$ are given as parameters — the ladder is fully determined in terms of them. **See it:** step through the mode number and watch which shapes the two pinned ends allow — whole half-wavelengths only; there is no rung between $0$ and the fundamental for $\frac{v}{4L}$ to stand on. ```qm-demo standing-wave-modes ```