Corrected exercises: waves and sound (PHYS 101)

a) The amplitude is the distance from the centre line to a crest, NOT from a trough to a crest. The curve rises to $2.0$ and falls to $-2.0$ on an axis graduated in centimetres, so $A = 2.0$ cm. The wavelength is the distance over which the shape repeats: crest at $x = 1$ m, next crest at $x = 5$ m, so $\lambda = 4.0$ m. Counting from a crest to the next trough gives $2.0$ m, which is half a wavelength, and this is the usual way of losing a factor of two.

b) On the right graph the same shape repeats after $0.50$ s, so $T = 0.50$ s. The frequency is the number of repetitions per second, $f = 1/T = 1/0.50 = 2.0$ Hz. Read the number of seconds per cycle, then invert: writing $f = 0.50$ Hz is writing the period in the frequency slot.

c) One full wavelength passes any fixed point in one period, so the wave advances $\lambda$ in a time $T$. Hence $v = \lambda / T = \lambda f = 4.0 \times 2.0 = 8.0$ m/s. This is the only formula of the chapter that connects the two graphs, and it is why you need both of them: one graph alone can never give a speed.

d) The horizontal axis of the left graph carries metres, not seconds. The $4$ read on it is therefore a length, the wavelength, and a period cannot be measured on a snapshot at all: the snapshot froze time. The student has read the right number on the wrong axis, and every later answer, frequency and speed included, inherits the error.

e) The snapshot is $y = A \sin(2\pi x / \lambda)$ with $A = 2.0$ cm and $\lambda = 4.0$ m. At $x = 1.0$ m the argument is $\pi/2$, so $y = +2.0$ cm: that point sits at a crest. Each point of the rope does nothing but oscillate up and down, exactly like the mass on a spring of the previous chapter, so a point at a crest is at the end of its travel and is momentarily AT REST, while a point crossing $y = 0$ is passing through the middle of its travel and is moving fastest. The wave moves along the rope at $8.0$ m/s; the material of the rope never does.

a) Convert first: $\mu = 5.0$ g/m $= 5.0 \times 10^{-3}$ kg/m. Then $v = \sqrt{F/\mu} = \sqrt{72 / 5.0\times 10^{-3}} = \sqrt{14400} = 120$ m/s. The wavelength follows from the source and the medium together: $\lambda = v/f = 120/240 = 0.50$ m. Leaving $\mu$ in grams per metre gives $\sqrt{14.4} = 3.8$ m/s, which is a walking pace for a wave on a taut string and should be caught at once.

b) The vibrator was not touched, so $f = 240$ Hz, unchanged. The medium changed, so the speed changed: $v = \sqrt{144 / 5.0 \times 10^{-3}} = \sqrt{28800} = 169.7$ m/s, that is $120\sqrt{2}$. The wavelength adjusts: $\lambda = 169.7/240 = 0.707$ m. Doubling the tension multiplies the speed by $\sqrt{2} \approx 1.41$ and NOT by $2$, because the tension sits under a square root.

c) Since $v = \sqrt{F/\mu}$, doubling $v$ means multiplying $F$ by $2^{2} = 4$, so $F = 4 \times 72 = 288$ N. The general habit worth keeping: a quantity under a square root needs a factor of $k^{2}$ to produce a factor of $k$.

d) At $20$ degrees, $v = 331 + 0.60 \times 20 = 343$ m/s, so $\lambda = 343/512 = 0.670$ m. At $-10$ degrees, $v = 331 + 0.60 \times (-10) = 325$ m/s, so $\lambda = 325/512 = 0.635$ m. The fork is a mechanical object vibrating at its own rate, so the frequency does not care about the weather: what the cold changes is the medium, hence the speed, hence the wavelength. The note heard is the same.

e) Only the medium changed, so only the speed changes. The frequency cannot change: the knot is a single point, and if the last point of the light string went up and down 240 times a second while the first point of the heavy string did it 200 times a second, the string would have to break at the knot. A heavier string means a larger $\mu$ at the same tension, so $v = \sqrt{F/\mu}$ drops, and $\lambda = v/f$ drops with it. The rule to carry into the optics chapters: across a boundary, frequency is what does not change.

a) The medium fixes the speed and the amplifier fixes the frequency, so $\lambda = v/f = 343/686 = 0.500$ m. Keep this number in metres: the path difference is in metres too, and the whole exercise is the ratio of the two.

b) The path difference is $\Delta = 4.75 - 4.00 = 0.75$ m, that is $\Delta/\lambda = 0.75/0.500 = 1.5$ wavelengths. One and a half is a whole number plus a half, so the crest of one wave arrives together with the trough of the other and the listener sits at a MINIMUM. He does not hear silence in a real room, because reflections fill in part of it, but the sound drops sharply and comes back if he moves a few centimetres.

c) A maximum requires $\Delta = n\lambda$ with $n$ a whole number, so $\lambda = \Delta / n = 0.75/n$. The lowest frequency comes from the largest wavelength, that is $n = 1$: $\lambda = 0.75$ m and $f = v/\lambda = 343/0.75 = 457$ Hz. Note what is being changed here: the geometry is fixed, the source is retuned. Frequencies $914$ Hz, $1372$ Hz and so on also give maxima, which is why a single listening point never tells you the frequency by itself.

d) With equal distances the path difference is zero, which is a whole number of wavelengths for EVERY wavelength at once. The equidistant surface is therefore a maximum for every frequency simultaneously, and that is why the central line between two speakers sounds full while the off-axis minima eat one frequency and leave its neighbours.

e) Reversing the wires sends out a signal in opposite phase, which acts exactly like adding half a wavelength to the path. The path difference at $P$ becomes the equivalent of $1.5 + 0.5 = 2$ wavelengths, so $P$ turns into a maximum. The equidistant point, where the difference was $0$, becomes the equivalent of half a wavelength and turns into a minimum: a stereo pair wired out of phase loses its bass on the axis, which is how the fault is diagnosed by ear.

a) With both ends clamped, $\lambda_1 = 2L = 2 \times 0.60 = 1.20$ m: the fundamental holds exactly one half-wavelength, one bump. Then $f_1 = v/\lambda_1 = 120/1.20 = 100$ Hz. The speed came from the medium, $v = \sqrt{F/\mu}$, and the string then picked the frequency out of it. This is the reversal that makes a musical instrument work: here the geometry chooses the WAVELENGTH and the medium supplies the speed, so the frequency is the output, not the input.

b) $\lambda_3 = 2L/3 = 1.20/3 = 0.40$ m and $f_3 = 120/0.40 = 300$ Hz, which is $3f_1$ as expected. Counting on the figure: $n = 3$ shows three bumps, so three antinodes, and four nodes because the two clamped ends count. The classic slip is to count three nodes, forgetting the ends, and the classic consequence is a node spacing that comes out wrong by a factor of two.

c) A light touch forces a node at that point. The midpoint is a node only for the modes that already have one there, that is the even ones: $n = 2$ at $200$ Hz and $n = 4$ at $400$ Hz survive, while $n = 1$ and $n = 3$ have an antinode at the middle and are killed. This is exactly the string harmonic of a guitarist, and it is heard as a jump of one octave, from $100$ Hz to $200$ Hz.

d) Frequencies here all go with $v$, and $v = \sqrt{F/\mu}$, so raising $F$ by $21$ percent multiplies $v$ by $\sqrt{1.21} = 1.1$. The geometry did not move, so every $\lambda_n$ is unchanged and $f_1 = 1.1 \times 100 = 110$ Hz. Tightening a string raises its pitch by the SQUARE ROOT of the tension ratio: a $21$ percent pull buys $10$ percent of pitch.

e) In mode $n$ the length $L$ holds $n$ half-wavelengths, and the nodes sit at the ends of each of them, so two neighbouring nodes are one half-wavelength apart: $\lambda_n/2 = (2L/n)/2 = L/n$. For $n = 3$ that is $0.60/3 = 0.20$ m. Measuring the node spacing is the standard laboratory way of getting the wavelength, because nodes hold still and are easy to locate, while an antinode is a blur.

a) Open at both ends: $L = \lambda_1/2$, so $\lambda_1 = 2L = 1.00$ m and $f_1 = v/\lambda_1 = 343/1.00 = 343$ Hz. Same formula as the clamped string, for the opposite reason: there both ends were nodes, here both ends are antinodes, and a half-wavelength separates two of either.

b) Closed at one end: $L = \lambda_1/4$, so $\lambda_1 = 4L = 2.00$ m and $f_1 = 343/2.00 = 171.5$ Hz. Exactly half the open pipe, which is one octave BELOW, for the same length of tube. This is why a stopped organ pipe is half as long as the open pipe that plays the same note, and it is the one number everybody remembers correctly.

c) The open pipe takes all whole multiples: $2f_1 = 686$ Hz and $3f_1 = 1029$ Hz. The closed pipe takes only the odd ones: after $171.5$ Hz come $3f_1 = 514.5$ Hz and $5f_1 = 857.5$ Hz. The even multiples are not merely quiet, they are impossible: $2f_1$ would need an antinode at the closed end, where the air cannot move.

d) Two frequencies in the ratio $600/200 = 3$, with nothing at $400$ Hz, is the signature of the odd series: the pipe is closed at one end, $200$ Hz is its fundamental and $600$ Hz its third harmonic. Then $L = v/(4f_1) = 343/800 = 0.429$ m. Had it been open, $400$ Hz would have sounded, being its second harmonic. Checking the MISSING frequency is faster than checking the present ones.

e) The pipe is a piece of geometry: its length still holds one half-wavelength, so $\lambda_1 = 1.00$ m, unchanged. What changed is the medium, hence the speed, hence $f_1 = v/\lambda_1 = 331/1.00 = 331$ Hz, twelve hertz flat. A pipe organ goes flat in the cold for this reason and cannot be corrected by anything but heat, while a guitar string, whose frequency is set by its own tension, goes the other way. Note the hierarchy of the chapter used in reverse here: a pipe does not choose a frequency, it chooses a wavelength.

a) Invert the definition: $I = I_0 \times 10^{\beta/10} = 10^{-12} \times 10^{9} = 1.0 \times 10^{-3}$ W/m$^2$. Divide the decibels by ten, never by twenty: the factor $20$ belongs to formulas written with pressure amplitudes, not with intensities, and mixing the two is the most common error of this section.

b) The source is isotropic, so at a distance $r$ its power is spread over a sphere of area $4\pi r^{2}$: $P = I \times 4\pi r^{2} = 1.0\times 10^{-3} \times 4\pi \times (2.0)^{2} = 5.0 \times 10^{-2}$ W. Fifty milliwatts of sound is a lot, which is the point: loudspeakers are inefficient, and a few watts of electrical input make a very loud room.

c) Through the intensity: the distance is multiplied by $4$, so $I$ is divided by $4^{2} = 16$, giving $I = 6.25 \times 10^{-5}$ W/m$^2$ and $\beta = 10\log_{10}(6.25 \times 10^{7}) = 78.0$ dB. Directly: dividing an intensity by $16$ subtracts $10\log_{10} 16 = 12.0$ dB, so $\beta = 90 - 12.0 = 78.0$ dB. The second route is the one to use under exam time pressure, and the number to memorise is that each doubling of distance costs $6$ dB.

d) Intensities from independent sources ADD, levels do not. Two speakers double the intensity to $2.0 \times 10^{-3}$ W/m$^2$, and $10\log_{10} 2 = 3.0$ dB, so $\beta = 93.0$ dB. Ten speakers multiply it by ten, that is $+10$ dB, so $\beta = 100$ dB. Adding $90 + 90 = 180$ dB would be louder than a rocket launch and is the signature error here.

e) $90$ dB against $45$ dB is a difference of $45$ dB, that is a ratio of $10^{45/10} = 10^{4.5} \approx 3.2 \times 10^{4}$ in intensity. The student treated a logarithmic scale as if it were proportional. Subjective loudness is yet another matter: the working rule of thumb is that about $10$ dB is needed for a sound to seem twice as loud, so $90$ dB sounds roughly eight times as loud as $45$ dB while carrying thirty thousand times the power.

a) $f_{beat} = |443 - 440| = 3$ beats per second. The ear hears one note of average pitch $441.5$ Hz throbbing three times a second, not two separate notes: that is the whole practical value of beats, they turn a difference too small to hear into a rhythm anyone can count.

b) Four beats per second means the string is at $436$ Hz or at $444$ Hz, and the beat count alone cannot decide which. The test is the tightening: raising the tension raises the string frequency, and the beat rate went UP, so the string was already ABOVE the fork, at $444$ Hz. Had it been at $436$ Hz, tightening would have brought it closer and the beats would have slowed to $2$ per second. Every tuning problem of this chapter is settled by this deliberate detuning, never by the beat count alone.

c) Source moving, observer still: $f' = f\,\dfrac{v}{v \mp v_s}$, with the minus sign while it approaches. Approaching, $f' = 900 \times \dfrac{343}{343 - 30} = 900 \times \dfrac{343}{313} = 986$ Hz. Receding, $f' = 900 \times \dfrac{343}{373} = 828$ Hz. Check the sign by physics and not by memory: approaching must raise the pitch, so the denominator must shrink.

d) Observer moving, source still: $f' = f\,\dfrac{v + v_o}{v} = 900 \times \dfrac{373}{343} = 979$ Hz. The two situations look symmetric and are not: $986$ Hz against $979$ Hz for the same $30.0$ m/s of relative motion. A moving source bunches the wavefronts themselves, changing the WAVELENGTH in the air; a moving observer meets unchanged wavefronts more often. The medium breaks the symmetry, and a question that specifies who moves is testing exactly this.

e) While the ambulance comes toward you the shift is the full approaching value, and the instant it passes it becomes the full receding value, so what you hear is a step from $986$ Hz to $828$ Hz rather than a slide, with only the brief moment of passage in between. With a headwind the air itself is moving, so the speed of sound relative to the ground changes and both shifted frequencies move with it, which is why Doppler formulas are written for a still medium and corrected afterwards.

a) FALSE. The speed of sound in air depends on the air and on almost nothing else: $v = 331 + 0.60\,T_C$, so $343$ m/s at $20$ degrees whether the siren is parked or doing $30$ m/s. What the motion changes is the spacing of the wavefronts, hence the frequency received. Corrected: a source moving toward you raises the frequency you receive and shortens the wavelength in the air, while the speed of the wave stays $343$ m/s.

b) FALSE, and it is the reverse of the truth. The frequency is imposed by the source and cannot change at a boundary, because the last particle of air and the first particle of water are in contact and must move together. The speed does change, from $343$ m/s to about $1480$ m/s, so the wavelength changes in the same proportion. Corrected: crossing into water, $f$ is unchanged, $v$ is multiplied by about $4.3$ and so is $\lambda$.

c) FALSE. The tension sits under a square root: $v = \sqrt{F/\mu}$ and $f_1 = v/(2L)$, so doubling $F$ multiplies $f_1$ by $\sqrt{2} = 1.41$, about six semitones and not the twelve of an octave. Corrected: to double the frequency of a string you must QUADRUPLE its tension, which is also why the low strings of a guitar are made heavy rather than slack.

d) FALSE, though the octave is right. The closed pipe fundamental is $v/(4L)$ against $v/(2L)$, so it is indeed an octave lower, but its series is $f_1, 3f_1, 5f_1, \ldots$, only the odd multiples, while the open pipe has all of them. For $L = 0.500$ m the closed pipe gives $171.5$, $514.5$, $857.5$ Hz and nothing at $343$ Hz. Corrected: same length, half the fundamental, and half the harmonics missing, which is why the two pipes do not sound alike even on their common notes.

e) FALSE. Decibels are logarithms, and logarithms do not add when the things they measure do. Each machine gives $I = 10^{-12} \times 10^{6} = 10^{-6}$ W/m$^2$; together, $2 \times 10^{-6}$ W/m$^2$, so $\beta = 10\log_{10}(2 \times 10^{6}) = 63.0$ dB. Corrected: doubling the acoustic power adds $3$ dB, and $120$ dB would be a million times the intensity of one machine, not twice.

a) The hanging mass is in equilibrium, so the tension equals its weight: $F = mg = 2.00 \times 9.80 = 19.6$ N. Convert the linear density, $\mu = 1.20$ g/m $= 1.20 \times 10^{-3}$ kg/m, then $v = \sqrt{F/\mu} = \sqrt{19.60 / 1.20\times 10^{-3}} = \sqrt{16333} = 128$ m/s. Writing $F = 2.00$ N, forgetting $g$, is worth the whole question and is caught by the units: newtons, not kilograms.

b) The bridges clamp the wire, so they are the nodes and the vibrating length is $L = 0.650$ m, not the whole wire. Hence $\lambda_1 = 2L = 1.300$ m and $f_1 = v/\lambda_1 = 127.80/1.300 = 98.3$ Hz. Measuring $L$ from the pulley or from the end post instead of from bridge to bridge is the classic laboratory error, and it always makes the frequency come out too low.

c) The two sounds differ by $|100 - 98.3| = 1.7$ Hz, so about $1.7$ swells of loudness per second, slow enough to count on a watch. To silence them the wire must reach $100$ Hz, that is a speed $v = 2Lf = 1.300 \times 100 = 130$ m/s, which needs $F = \mu v^{2} = 1.20 \times 10^{-3} \times 130^{2} = 20.3$ N, so $m = F/g = 2.07$ kg. A $3.5$ percent increase in mass for a $1.7$ percent rise in frequency: the square root again, and this is what makes a sonometer a sensitive instrument, since the beats vanish long before the eye can see the wire change.

d) Neighbouring nodes are half a wavelength apart, so the spacing is $L/n = 0.650/3 = 0.217$ m. The manual asks for it because a node is a point of the wire that does not move, so it is sharp and can be located with a millimetre rule, whereas an antinode is a blur several centimetres wide. Measuring across several nodes and dividing is better than measuring a single gap, since the uncertainty on the two end positions is then shared out.

e) A tension $2.0$ percent lower multiplies $v$, and therefore $f_1$, by $\sqrt{0.980} = 0.990$: the measured fundamental falls to $97.3$ Hz instead of $98.3$ Hz. If the student still uses $F = mg$ in $\mu = F/(2Lf_1)^{2}$, the numerator is too big and the frequency too small, so $\mu$ comes out about $4$ percent too LARGE. The point worth keeping: a systematic error does not scatter the points, it tilts the whole line, so it never shows up as a poor repeatability.

a) $\lambda = v/f = 1540 / (2.5 \times 10^{6}) = 6.2 \times 10^{-4}$ m, that is $0.62$ mm. That is the resolution limit, and it explains the whole design of the machine: a higher frequency gives a sharper image, which is why a $7$ MHz probe is used for a neck and a $2.5$ MHz probe for an abdomen, where the sound must go deep and high frequencies are absorbed too fast.

b) The probe is unchanged, so $f = 2.5$ MHz still. In air $\lambda = 343/(2.5\times 10^{6}) = 1.4 \times 10^{-4}$ m. The hierarchy of the chapter once more: the source kept the frequency, the medium changed the speed, and the wavelength followed. In practice almost nothing enters the body at all, which is why the gel is there: air between probe and skin reflects the burst back before it starts.

c) The burst travels to the surface and back, so it covers $2d$ in $130$ microseconds: $d = vt/2 = 1540 \times 130 \times 10^{-6} / 2 = 0.100$ m, that is $10.0$ cm. Forgetting to halve puts the reflector at $20$ cm and is the error the machine itself cannot detect. The same halving runs through sonar, radar and the bat of part e).

d) $\Delta f = 2 \times 0.40 \times 2.5\times 10^{6} / 1540 = 1.3 \times 10^{3}$ Hz, about $1.3$ kHz. The factor $2$ is two Doppler shifts in a row: the moving blood first RECEIVES a frequency raised because it advances into the wave, then re-emits that frequency while still advancing, so the probe receives it raised a second time. The shift is about one part in two thousand of the carrier, far too small to see, but it is audible as a beat against the emitted frequency, and that beat is the whistling sound of a Doppler blood monitor.

e) Bat: $\lambda = 343/40000 = 8.6$ mm, small enough to reflect off an insect a centimetre across, which a longer wave would simply flow around. Elephant: $\lambda = 343/15 = 23$ m, a wave that bends around trees and hills and is absorbed very little, so it carries for kilometres. Same physics, opposite choices: the bat buys resolution, the elephant buys range, and both are reading $\lambda = v/f$ in the direction that suits them.