Questionnaire 1.3 — Quanta, Oscillators & Waves

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).

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.

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.

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.

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.

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.

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 ✗"]

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.

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.

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.