Corrected exercises: sound and standing waves (203-SN3-RE)

$\lambda=\frac{v}{f}$: at 440 Hz, $\lambda=\frac{343}{440}\approx 0.78$ m; at 20 kHz, $\lambda=\frac{343}{20\,000}\approx 1.7$ cm. The audible range thus runs from about a centimetre to about ten metres (17 m at 20 Hz): three orders of magnitude, which is why low sounds bend around obstacles (diffraction) far better than high ones.

The sound travels there and back: $d=\frac{v\,t}{2}=\frac{(343)(1.2)}{2}\approx 206$ m.

$d=v\,t=(343)(3.0)=1029$ m, about a kilometre (hence the rule of three seconds per kilometre). Light covers this kilometre in $\frac{1029}{3.0\times 10^{8}}\approx 3\times 10^{-6}$ s: a million times shorter than the 3 s of the sound, so the flash really is an instantaneous “time zero”.

$v=331+0.6\times 35\approx 352$ m/s. In a pipe, the natural frequencies are $f=n\frac{v}{2L}$ (or $\frac{v}{4L}$): if $v$ increases by 3 % with the heat, ALL the notes rise by 3 %: the wind instrument goes sharp when the room warms up. The guitar, on the other hand, depends on $v=\sqrt{F/\mu}$ in the string: heat expands the string, the tension $F$ drops, and the guitar goes FLAT: the two families of instruments drift in opposite directions, the nightmare of summer concerts.

$\Delta P_{max}=\sqrt{2\rho v I}=\sqrt{2(1.20)(343)(1.0\times 10^{-12})}\approx 2.87\times 10^{-5}$ Pa. Then $s_{max}=\frac{\Delta P_{max}}{\rho v\omega}=\frac{2.87\times 10^{-5}}{(1.20)(343)(2\pi\times 1000)}\approx 1.11\times 10^{-11}$ m. That is about a TENTH of the diameter of an atom: at the threshold of hearing, the ear detects an air motion smaller than the molecules that carry it. There is no contradiction here, because $s_{max}$ describes the AVERAGE displacement of a very large number of molecules, not that of a single molecule; the ear is simply one of the most sensitive detectors in the whole body.

$\Delta P_{max}=\sqrt{2(1.20)(343)(1.0)}\approx 28.7$ Pa, and $s_{max}\approx 1.11\times 10^{-5}$ m, that is 11 μm. Both have been multiplied by a million, as they should be, since the intensity was multiplied by $10^{12}$ and the amplitudes go as its square root. Note the order of magnitude: even at the threshold of pain, the pressure variation is only $\frac{28.7}{1.0\times 10^{5}}\approx 3\times 10^{-4}$ of atmospheric pressure. Sound is only a tiny ripple superimposed on an enormous, constant pressure.

Look at a layer of air located at a DISPLACEMENT antinode: it moves a lot, but its two neighbours move almost as much and in the same direction, so it is neither compressed nor expanded; the excess pressure there is zero. Now look at a layer that does NOT move: its neighbours move toward it on one side and away from it on the other, or converge on it from both sides at once; this is therefore where the air is most compressed and most rarefied. A displacement node is a pressure antinode, and vice versa. This is because the excess pressure is proportional to the SPATIAL RATE OF CHANGE of the displacement, and a sinusoidal function has its maximum slope where it crosses zero.

The ear covers a range of pressures from $3\times 10^{-5}$ Pa to about 30 Pa, a factor of one MILLION in pressure and $10^{12}$ in intensity. No linear scale can represent such a range conveniently: on a graph where the threshold of pain was one metre long, the threshold of hearing would sit one micrometre from zero. Moreover, the perceived sensation grows roughly as the logarithm of the stimulus, so that the same difference in decibels corresponds to the same difference in sensation, whatever the starting level. The scale $\beta=10\log\frac{I}{I_{0}}$ compresses the $10^{12}$ into 120 convenient units and follows perception: it is both a mathematical convenience and a physiological fact.

Water: $\Delta P_{max}=\sqrt{2(1000)(1480)(1.0\times 10^{-12})}\approx 1.72\times 10^{-3}$ Pa, sixty times more than in air, while $s_{max}=\frac{1.72\times 10^{-3}}{(1000)(1480)(6283)}\approx 1.85\times 10^{-13}$ m, sixty times LESS. A dense, fast medium carries the same energy with a lot of pressure and very little motion. Impedances: $Z_{air}=(1.20)(343)\approx 412$ rayls and $Z_{water}=(1000)(1480)=1.48\times 10^{6}$ rayls, a ratio of 3600. Transmission coefficient: $\frac{4(412)(1.48\times 10^{6})}{(1.48\times 10^{6}+412)^{2}}\approx 1.1\times 10^{-3}$, barely 0.11 % of the energy transmitted, and therefore 99.89 % reflected. This is why a swimmer underwater hears almost nothing of what is shouted at the water's edge, and why ultrasound imaging requires a gel: without it, the thin layer of air between the probe and the skin would send almost the whole beam back. Impedance matching is also the job of the ossicles of the middle ear, which pass sound from the air to the fluid of the cochlea.

$I=\frac{P}{4\pi r^{2}}=\frac{0.50}{4\pi(10)^{2}}\approx 4.0\times 10^{-4}$ W/m². Level: $\beta=10\log\frac{4.0\times 10^{-4}}{10^{-12}}=10\log(4.0\times 10^{8})\approx 86$ dB: the level of a very noisy street.

At twice the distance, $I$ is divided by 4 (the $\frac{1}{r^{2}}$ law): $\beta'=\beta-10\log 4\approx 86-6.0=80$ dB. General rule: each doubling of the distance subtracts 6 dB, whatever the point source.

The intensity doubles: $\beta$ rises by $10\log 2\approx 3.0$ dB, to 89 dB. Doubling the power produces only a barely noticeable gain (3 dB is close to the smallest difference the ear can clearly detect): the logarithmic scale compresses enormously; for a sound that is subjectively “twice as loud”, about ten times the power is needed.

$\Delta\beta=50$ dB corresponds to $\frac{I_{concert}}{I_{conversation}}=10^{50/10}=10^{5}$: a hundred thousand times more intense. The ear thus covers twelve orders of magnitude between the threshold of hearing and the threshold of pain: only a logarithmic perception can embrace such a range, and the decibel is logarithmic precisely to match the ear.

Approaching source: $f'=f\,\frac{v}{v-v_{s}}=800\times\frac{343}{343-25}=800\times\frac{343}{318}\approx 863$ Hz. The source catches up with its own wavefronts: they pile up ahead of it, the wavelength is compressed there, and the ear, which receives wavefronts that are closer together, hears a higher pitch.

Receding source: $f'=800\times\frac{343}{343+25}=800\times\frac{343}{368}\approx 746$ Hz. As it goes by, the pitch drops by about $863-746\approx 117$ Hz all at once, a little more than a whole tone (the ratio $863/746$ is about $1.16$, roughly two and a half semitones): this is the characteristic “nee-yow”.

Observer approaching a stationary source: $f'=f\,\frac{v+v_{o}}{v}=800\times\frac{368}{343}\approx 858$ Hz: close to 863 Hz, but NOT equal. The asymmetry comes from the medium: sound propagates in air, which is a physical reference. Moving source: the wavelength itself changes in the air. Moving observer: the waves are untouched, it is their RELATIVE speed of reception that changes. Two different mechanisms, two different formulas: confusing them is costly on an exam.

The light from the beacon of a stationary ambulance undergoes no shift: there is no relative motion (and for light, only relative motion matters, there is no medium). The passenger in the ambulance, for their part, travels WITH the source: no relative motion either, so they hear exactly 800 Hz, while the pedestrian hears the siren rise then fall. The Doppler effect lies in the source-receiver relationship, not in the siren.

Two nodes at the ends: $\lambda_{1}=2L=1.30$ m. Speed in the string: $v=f_{1}\lambda_{1}=(220)(1.30)=286$ m/s. (Careful: this is the speed IN THE STRING; the sound emitted into the air keeps the frequency of 220 Hz but has the wavelength of the air, $\frac{343}{220}\approx 1.56$ m.)

$f_{n}=nf_{1}$: $f_{2}=440$ Hz and $f_{3}=660$ Hz. Harmonic $n$ has $n$ antinodes and $n+1$ nodes (ends included): $f_{2}$ has an extra node in the MIDDLE of the string, $f_{3}$ has two, at the thirds.

$f_{1}=\frac{v}{2L}$: halving $L$ doubles the frequency: 440 Hz. This is the OCTAVE, and the geometric reason why the twelfth fret is always at the mid-point of the string, on every guitar.

$f=\frac{1}{2L}\sqrt{\frac{F}{\mu}}$: the frequency goes as the square root of the tension, so $\frac{F'}{F}=\left(\frac{233}{220}\right)^{2}\approx 1.12$: about 12 % more tension is needed for a semitone of 6 %. The square doubles the bill: this is why tuning too high breaks strings far faster than one would expect.

FIXED end: the wall cannot move, so the displacement there is necessarily zero at every instant. For the sum of the incident and reflected waves to cancel permanently at that point, the reflected wave must at every instant be the OPPOSITE of the incident one: it therefore comes back inverted, with a phase shift of $\pi$. A dynamic reading says the same thing: the wave pulls the wall upward, the wall reacts by pulling the string downward (Newton's third law), and this reaction creates the return wave, directed downward. FREE end: nothing opposes the motion, the end can on the contrary overshoot, it behaves like an antinode and the reflection takes place WITHOUT inversion.

$y_{1}+y_{2}=A[\sin(kx-\omega t)+\sin(kx+\omega t)]=2A\sin\left(\frac{2kx}{2}\right)\cos\left(\frac{-2\omega t}{2}\right)=2A\sin(kx)\cos(\omega t)$, using the evenness of the cosine. The difference is crucial: in a travelling wave, $x$ and $t$ appear TOGETHER in a single argument $(kx-\omega t)$, and it is this coupling that makes the pattern travel. Here, they are SEPARATED into a product: $\sin(kx)$ fixes the shape in space once and for all, and $\cos(\omega t)$ only makes it breathe in amplitude. Nothing moves forward: each point oscillates in place, with its own amplitude $2A\sin(kx)$ that depends on the position and on nothing else.

Nodes where $\sin(kx)=0$, so $kx=n\pi$ and $x=n\frac{\lambda}{2}$; antinodes midway, $x=(2n+1)\frac{\lambda}{4}$. Two consecutive nodes are therefore separated by HALF a wavelength, not a whole wavelength: this is the most common mistake in the whole chapter. For the string: $f_{n}=n\frac{v}{2L}=n\frac{240}{2.40}=100n$ Hz, that is 100 Hz, 200 Hz and 300 Hz. In the third mode, $\lambda_{3}=\frac{2L}{3}=0.800$ m, the nodes are 0.400 m apart and are located at $x=0$, 0.400, 0.800 and 1.20 m: four nodes enclosing three antinodes.

Each of the two travelling waves carries a power $\frac{1}{2}\mu v\omega^{2}A^{2}$, but they travel in OPPOSITE directions: the two energy flows are equal and opposite, and their sum is zero. Another way to see it: at the nodes, the amplitude is permanently zero, so the string there is motionless and cannot transmit any power through that point, at any instant. The energy is therefore TRAPPED between two consecutive nodes, and it oscillates there: entirely kinetic when the string passes through its straight position (all points at maximum speed), entirely elastic potential when the string reaches its extreme deformation (all points momentarily at rest). Twice per period, the energy switches from one form to the other without ever crossing a node.

Since the nodes are separated by $\frac{\lambda}{2}$, we have $\lambda=2(0.100)=0.200$ m. Speed: $v=\lambda f=(0.200)(1700)=340$ m/s. Temperature: $340=331+0.6T$ gives $T=\frac{9}{0.6}=15$ °C. That is the whole trick of Kundt's tube: measuring a propagation speed, which is hard in itself, by reducing it to a simple LENGTH measurement with a ruler, on piles of motionless dust.

At open ends, the air communicates with the outside: the pressure there stays at atmospheric pressure and the displacement of the air is maximum: a displacement ANTINODE at both ends. We therefore need $L=n\frac{\lambda}{2}$: $f_{n}=n\frac{v}{2L}$. Fundamental: $f_{1}=\frac{343}{2(0.50)}=343$ Hz, then $f_{2}=686$ Hz and $f_{3}=1029$ Hz: all the harmonics.

At the closed end, the air cannot oscillate along the axis: displacement NODE; at the open end: antinode. The longest wavelength that fits a quarter of a pattern is $\lambda_{1}=4L$: $f_{1}=\frac{v}{4L}=\frac{343}{2.0}=171.5$ Hz. The following modes must keep a node on one side and an antinode on the other: only the ODD multiples fit: $f_{3}=3f_{1}=514.5$ Hz and $f_{5}=5f_{1}=857.5$ Hz.

The closed pipe sounds one OCTAVE LOWER (171.5 against 343 Hz: half) and produces only the odd harmonics. Both differences come from the same geometric constraint: replacing an antinode with a node at one end forces the pattern to fit in quarter wavelengths rather than half wavelengths, which doubles the fundamental wavelength and rules out the even ranks.

Closed: $L=\frac{v}{4f}=\frac{343}{4(27.5)}\approx 3.1$ m. Open: $L=\frac{v}{2f}\approx 6.2$ m. The builder chooses the CLOSED pipe (the bourdon): half the height, metal and cost for the same low note; the great organ bourdons are closed precisely for this saving.

The common note fixes the fundamental, but the TIMBRE comes from the mixture of superimposed harmonics. The flute, an open pipe, supplies every rank (1, 2, 3, 4...): a round, full sound. The clarinet, closed at the mouthpiece, skips the even ranks in its low register (1, 3, 5...): it is this “hollow” spectrum that gives it its woody, slightly nasal timbre. Same physics, same formulas: what the ear calls timbre, the equation calls boundary conditions.

$\lambda=\frac{v}{f}=\frac{343}{686}=0.500$ m. Since the two sources emit in phase, the path travelled is the only thing that shifts the waves: CONSTRUCTIVE interference if $\Delta=n\lambda$ (the two waves arrive in phase, the amplitude doubles and the intensity quadruples), DESTRUCTIVE if $\Delta=\left(n+\frac{1}{2}\right)\lambda$ (they arrive in opposition and cancel, for equal amplitudes).

This point is equidistant from the two loudspeakers: $\Delta=0$, so zeroth-order constructive interference. The listener hears a MAXIMUM there, whatever the frequency, which makes this central position the only one that is right for all notes at once. This is why stereo listening is judged from the apex of the isosceles triangle formed with the speakers.

First minimum: $\Delta=\frac{\lambda}{2}=0.250$ m, so $\sin\theta=\frac{0.250}{3.00}=0.0833$ and $\theta\approx 4.78°$. Position along the listening line: $y=D\tan\theta=(8.00)(0.0836)\approx 0.669$ m. So you only need to shift by 67 cm, less than the width of an armchair, to go from a maximum to an almost total silence: this kind of dead zone is perfectly audible in a real room, and many listeners wrongly blame it on faulty equipment.

The path difference can never exceed the distance between the sources, $\Delta\leq d=3.00$ m, that is $6\lambda$. The condition $\left(n+\frac{1}{2}\right)(0.500)\leq 3.00$ gives $n+\frac{1}{2}\leq 6$, so $n\leq 5.5$: the orders $n=0$ to $5$ are possible, that is SIX minima on each side of the central point. In practice, a listener walking along the listening line therefore crosses a dozen dead zones, alternating with as many reinforcements.

At 100 Hz, $\lambda=\frac{343}{100}=3.43$ m and $\frac{\lambda}{2}=1.72$ m, which is still less than $d=3.00$ m: there is only ONE minimum on each side, at $\sin\theta=\frac{1.72}{3.00}=0.572$, that is $\theta\approx 34.9°$, very far from the axis. At 50 Hz, $\frac{\lambda}{2}=3.43$ m already exceeds $d$: the cancellation condition cannot be met ANYWHERE, there is no minimum. The general principle is that the number of interference zones is set by the ratio $\frac{d}{\lambda}$: the longer the wavelength compared with the spacing of the sources, the less structured the sound field. At low frequencies, the whole room is essentially in phase, no dead zone forms and the ear even loses all sense of direction, for lack of a perceptible difference between the two ears. The subwoofer can therefore be placed wherever you like, whereas a high-frequency speaker, whose wavelength is counted in centimetres, creates a finely divided field and demands precise placement, just like the listening position.

The guitarist hears ONE pitch, the average, 442 Hz (the ear does not separate two frequencies so close together), whose intensity pulses: these are the beats, at $f_{beat}=|444-440|=4.0$ Hz, that is one intensity maximum every $\frac{1}{4.0}=0.25$ s.

At a given instant, the two waves are in phase: their amplitudes add and the sound is loud (constructive interference). But their frequencies differ: one gradually pulls ahead and, half a beat period later, they are in opposition: the amplitudes subtract and the sound almost dies out (destructive). The cycle of agreement and disagreement repeats $|f_{1}-f_{2}|$ times per second: a beat is interference in TIME.

$|f-440|=3.0$: the string is at 437 Hz or at 443 Hz. The beat gives only the DIFFERENCE in absolute value: too low or too high produce exactly the same count, the sign information is lost.

Tightening the string raises its frequency. The beats ACCELERATED: the difference from 440 Hz has grown, so the string was moving AWAY from the tuning fork as it went up: it was already above, at 443 Hz. If they had slowed down, the string would have been approaching 440 Hz from below: 437 Hz, and it would have sufficed to keep tightening until the beats fell silent. Tuning is literally a search for $f_{beat}=0$: the tuner does not listen to the note, the tuner listens for the beat to disappear.

$v=M\,v_{sound}=(1.60)(340)=544$ m/s, that is $(544)(3.6)\approx 1.96\times 10^{3}$ km/h.

During a time $t$, the plane travels $v_{plane}t$ while the wavefront emitted at the starting point has grown into a sphere of radius $v_{sound}t$. All these spheres, emitted at different instants, have exactly the same envelope: a cone with the plane at its apex. In the right triangle formed by the radius of the sphere (opposite side) and the path of the plane (hypotenuse), $\sin\theta=\frac{v_{sound}t}{v_{plane}t}=\frac{v_{sound}}{v_{plane}}=\frac{1}{M}$. Here $\sin\theta=\frac{1}{1.60}=0.625$, so $\theta\approx 38.7°$. On this cone, infinitely many wavefronts pile up at the same place: the pressure rise there is abrupt, and this is the shock wave.

The observer hears the boom at the instant the SURFACE of the cone reaches them. Since the apex of the cone is on the plane, the horizontal distance between the plane and the observer is then $x=\frac{h}{\tan\theta}=\frac{12000}{0.801}\approx 1.50\times 10^{4}$ m, about 15.0 km. The plane is therefore already fifteen kilometres FARTHER ON when the noise arrives. In time: $t=\frac{x}{v_{plane}}=\frac{14988}{544}\approx 27.6$ s after passing overhead. Nearly half a minute of silence separates the flyover from the boom.

The cone accompanies the plane permanently, for as long as it flies faster than sound: it is not an event but a regime. Its trace on the ground is a continuous line that sweeps the ground all along the flight path, the famous boom carpet, and everyone located under this strip hears the detonation in THEIR turn, when the cone sweeps over them, that is always after the plane has passed. A plane that stayed supersonic for an hour would produce a boom over a thousand kilometres of terrain, not a single crack as it crosses Mach 1. It is this spread-out nuisance, and not the noise of the engines, that led to supersonic flight being banned over populated land.

As $v_{s}\to v$, the denominator $v-v_{s}$ tends to zero and $f'\to\infty$: the formula diverges. It says exactly what the geometry of part b) shows: the plane catches up with its own wavefronts, which pile up on a single surface instead of following one another. An infinite frequency obviously has no physical meaning, and it is the sign that the linear model of the sound wave stops being valid: the pressure rise becomes too large, and we enter the domain of shock waves. Beyond that, for $v_{s}>v$, the formula simply no longer applies. An observer located AHEAD of the plane, outside the cone, hears absolutely NOTHING: no sound emitted by the aircraft can get ahead of it, since the aircraft is faster than all of them. The observer therefore sees the plane go by in silence, then hears the boom, then only afterwards the sound of the engines. The silence that precedes a supersonic plane is not an impression, it is a direct consequence of the kinematics.