PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: waves and sound (PHYS 101)

The course notes already tell you that a wave carries energy without carrying matter. This sheet answers a different question: on an actual PHYS 101 midterm, where do the marks go, and what single gesture keeps each one.

Everything is built on one hierarchy. The source imposes the frequency, the medium imposes the speed, and the wavelength is the quotient of the two. Hold it and the boundary questions, the string questions and the pipe questions stop being three chapters.

Two lines of the PHYS 101 formula sheet get their own place here: the refraction of a wave written with speeds, and the decibel written with an amplitude, twenty times the logarithm instead of ten.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The source chooses the frequency, the medium chooses the speed, and the wavelength is nothing but their quotient. Every trap in this chapter comes from letting the wrong one of the three change.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

Before this chapter, you need

The essentials

On your formula sheet: this chapter's lines, and when each one holds

  • • v=fλv=f\lambda: The medium sets vv, the source sets ff, and λ\lambda follows. Into another medium, ff stays and λ\lambda changes.
  • • v=FT/μv=\sqrt{F_{T}/\mu}: μ\mu is the mass per unit LENGTH in kg/m, not the mass of the string.
  • • λm=2L/m\lambda_{m}=2L/m: A string fixed at both ends, m=1m=1, 22, 33 and so on, which gives fm=mv2Lf_{m}=m\frac{v}{2L}.
  • • v1sin⁡θ2=v2sin⁡θ1v_{1}\sin\theta_{2}=v_{2}\sin\theta_{1}: Angles measured from the NORMAL, and the speeds CROSSED: the wave bends toward the normal on the side where it is slower. Divided by cc, it is n1sin⁡θ1=n2sin⁡θ2n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}.
  • • vs=330v_{s}=330: In m/s, the sheet's typical value for air. When a statement gives another one (343 m/s at 2020 °C), the statement wins.
  • • β=10log⁡10(II0)=20log⁡10(AA0)\beta=10\log_{10}(\frac{I}{I_{0}})=20\log_{10}(\frac{A}{A_{0}}): Ten with intensities, twenty with amplitudes, because II grows like A2A^{2}: doubling the intensity adds 33 dB, doubling the amplitude adds 66 dB.
  • • I0=1.0×10−12I_{0}=1.0\times10^{-12}: In W/m², the threshold of hearing, the reference for 00 dB.
  • • fn=nf1=nv2Lf_{n}=nf_{1}=n\frac{v}{2L}: A tube OPEN at both ends: every harmonic, n=1n=1, 22, 33, 44 and so on.
  • • fn=nf1=nv4Lf_{n}=nf_{1}=n\frac{v}{4L}: A tube CLOSED at one end: odd harmonics only, n=1n=1, 33, 55, 77 and so on.
  • • fb=∣f1−f2∣f_{b}=|f_{1}-f_{2}|: Two close frequencies: the ear hears their average pitch, swelling fbf_{b} times per second.
  • • f′=fvwavevwave±vsourcef'=f\frac{v_{wave}}{v_{wave}\pm v_{source}}: Moving source: minus when the source approaches (higher pitch), plus when it moves away.
  • • f′=fvwave±vobsvwavef'=f\frac{v_{wave}\pm v_{obs}}{v_{wave}}: Moving observer: plus when the observer approaches the source, minus when moving away.

The sheet prints each equation, never the condition under which it holds: copy the condition next to its line on your own sheet before the exam.

The three quantities, and who fixes each one

  • • v=fλv = f\lambda is the whole chapter, but it is read in one direction only: ff comes from the SOURCE, vv from the MEDIUM, and λ\lambda is what is left over.
  • • On a string, v=F/μv = \sqrt{F/\mu} with FF in newtons and μ\mu in kilograms per metre. In air, v=331+0.60 TCv = 331 + 0.60\,T_C in metres per second, so 343343 m/s at 2020 degrees Celsius.
  • • Snapshot, yy against xx: you read the amplitude and the WAVELENGTH on it. History of one point, yy against tt: you read the amplitude and the PERIOD. A period cannot be measured on a snapshot, which froze time.
  • • Crossing into a new medium: ff is unchanged, vv is the one of the new medium, and λ=v/f\lambda = v/f follows. Each point of the medium only oscillates in place, exactly like the mass on a spring of the previous chapter.
one wavelengthx (m)one periodt (s)y (cm)y (cm)
The same drawing twice: on the left the horizontal axis carries metres and the bracket measures λ\lambda, on the right it carries seconds and the same bracket measures TT.

A point sitting at a crest is at the end of its travel and is momentarily at rest; a point crossing y=0y = 0 is the one moving fastest. That is the oscillator chapter read again, one oscillator per position.

Refraction: a wave that crosses a boundary at an angle

  • • The formula sheet writes it with speeds, under Waves: v1sin⁡θ2=v2sin⁡θ1v_1\sin\theta_2 = v_2\sin\theta_1, both angles measured from the NORMAL to the boundary.
  • • It comes from the crests alone: they cannot break at the boundary, so they cut it at the same spacing from both sides, λ1/sin⁡θ1=λ2/sin⁡θ2\lambda_1/\sin\theta_1 = \lambda_2/\sin\theta_2, and the common frequency cancels out of λ=v/f\lambda = v/f.
  • • Slower on the far side: the wave turns TOWARD the normal. Swell at 1212 m/s reaching shallower water at 7.07.0 m/s, 40∘40^{\circ} from the normal, leaves at 22.0∘22.0^{\circ}, its wavelength going from 120120 m to 7070 m with the same 1010 s period.
  • • Checks: v2=v1v_2 = v_1 gives θ2=θ1\theta_2 = \theta_1, no bending; θ1=0\theta_1 = 0 gives θ2=0\theta_2 = 0, slowed but not turned. The same line, with v=c/nv = c/n, is the refraction of light.
θ₁θ₂beachshallow waterv₂ = 7.0 m/sλ₂ = 70 mdeep waterv₁ = 12 m/sλ₁ = 120 mT = 10 s on both sidescrests seen from above, drawn to scale
Each crest bends where it crosses the depth line: the direction of travel turns from θ1=40∘\theta_1 = 40^{\circ} to θ2=22.0∘\theta_2 = 22.0^{\circ}, the crests close up from 120120 m to 7070 m, and the period stays 1010 s.

This is why waves break almost parallel to a beach whatever their direction offshore: every depth contour turns them a little more toward the normal.

Transverse and longitudinal waves

  • • On a rope each point moves ACROSS the direction of travel: the wave is transverse. In air each layer moves back and forth ALONG it: sound is longitudinal, a train of compressions and rarefactions.
  • • The wavelength is read the same way on both: the distance between two neighbouring points in the same state, crest to crest on the rope, compression to compression in air.
  • • In both cases what travels at vv is the disturbance and its energy; the medium only oscillates about its place.
λwave travelspoints move acrosstransverse: a ropeλcompressionrarefactionlayers move alonglongitudinal: sound in air
Rope: the points move across the direction of travel. Air: the layers move along it, bunching into compressions. On both, λ\lambda is the distance between two points in the same state.

Standing waves, pipes, levels and shifts

  • • String clamped at both ends, or pipe open at both ends: λn=2L/n\lambda_n = 2L/n and fn=n v/(2L)f_n = n\,v/(2L), every whole nn. Neighbouring nodes are λ/2=L/n\lambda/2 = L/n apart.
  • • Pipe closed at one end: λn=4L/n\lambda_n = 4L/n with nn ODD only, so f1=v/(4L)f_1 = v/(4L) and then 3f13f_1, 5f15f_1, and nothing in between.
  • • Two sources, one listening point: count the path difference in wavelengths. Δ=nλ\Delta = n\lambda is loud, Δ=(n+12)λ\Delta = \left(n+\frac{1}{2}\right)\lambda is quiet.
  • • Level: β=10log⁡10(I/I0)=20log⁡10(A/A0)\beta = 10\log_{10}(I/I_0) = 20\log_{10}(A/A_0) with I0=1.0×10−12I_0 = 1.0\times 10^{-12} W/m2^2, the second form for an amplitude ratio, and I=P/(4πr2)I = P/(4\pi r^{2}) for a point source. Beats: fbeat=∣f1−f2∣f_{beat} = |f_1 - f_2|. Doppler: f′=f v±vov∓vsf' = f\,\dfrac{v \pm v_o}{v \mp v_s}.
NNAstring clamped at both endsλ₁ = 2LNAApipe open at both endsλ₁ = 2LNApipe closed at the right endλ₁ = 4LL
The ends decide: a clamped end or a closed end is a node N, an open end is an antinode A, so the fundamental has λ1=2L\lambda_1 = 2L, 2L2L and 4L4L.

Four numbers to hold instead of four formulas: doubling an intensity adds 33 dB, doubling an amplitude adds 66 dB, multiplying an intensity by ten adds 1010 dB, and doubling a distance costs 66 dB.

Two sources, one listener: the path difference

  • • Two sources in step reach a point PP by paths r1r_1 and r2r_2. Only the path difference Δ=r2−r1\Delta = r_2 - r_1 matters, and only once it is counted in wavelengths.
  • • Δ/λ\Delta/\lambda a whole number: crest meets crest, the waves add, loud. A whole number plus a half: crest meets trough, they cancel, quiet.
  • • On the line equidistant from the two sources Δ=0\Delta = 0: loud for every frequency at once. Inverting one source swaps every loud point with a quiet one.
Δ = 1.0 λ: loudΔ = 1.5 λ: quietfrom S₁from S₂sum at P
The two waves arriving at PP: one whole wavelength of path difference puts them in step and the sum doubles; one and a half puts crest on trough and the sum vanishes.

Beats: one note that throbs

  • • Two close frequencies f1f_1 and f2f_2 sounded together are heard as ONE note at the average (f1+f2)/2(f_1 + f_2)/2, whose loudness swells and fades ∣f1−f2∣|f_1 - f_2| times a second.
  • • Count loudness MAXIMA: the two notes fall back into step ∣f1−f2∣|f_1 - f_2| times a second. With 256256 Hz and 260260 Hz that is 44 swells per second, heard as one note near 258258 Hz, never as a hum at 44 Hz and never as 22 swells per second.
  • • A beat count alone gives two candidates, fref±fbeatf_{ref} \pm f_{beat}. Detune on purpose: if tightening a string makes the beats faster, it was already above the reference.
  • • Nothing moves in beats, while the Doppler effect needs motion: two steady sources and a slow throb mean beats, one moving source or listener means Doppler.
1 / |f₁ − f₂|f₁f₂togetherloudest
Two close notes added: the sum keeps the fast oscillation, but its loudness swells and fades, one maximum every 1/∣f1−f2∣1/|f_1 - f_2| seconds.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading the period off a snapshot graph

the whole question, since the frequency and the speed both follow from it

What not to write

“The curve repeats every 44, so T=4T = 4 s and f=0.25f = 0.25 Hz.”

What to write

“The horizontal axis carries metres, so the 44 is λ=4.0\lambda = 4.0 m. The period is read on the yy against tt graph: T=0.50T = 0.50 s, f=2.0f = 2.0 Hz.”

Why: Both graphs are sine curves and look identical. Only the label on the horizontal axis says which quantity repeats, and it is the first thing to read, before any number.

2. Taking the amplitude from a trough to a crest

1 point, and every energy answer after it, since the energy goes with A2A^{2}

What not to write

“The curve runs from −2-2 to +2+2, so the amplitude is 4.04.0 cm.”

What to write

“The amplitude is measured from the centre line, so A=2.0A = 2.0 cm. The 4.04.0 cm is the peak to peak value.”

12345678−3−2−1123A = 2.0 cm, from the centre linewrong: 4.0 cm, peak to peakx (m)y (cm)
The amplitude is measured from the centre line to a crest (green), A=2.0A = 2.0 cm; trough to crest (red) is twice that.

Why: The amplitude is defined as the maximum DISPLACEMENT from equilibrium, the same definition as for the mass on a spring. A factor of two here becomes a factor of four in any energy.

3. Doubling the tension to double the frequency

2 to 3 points, and the credibility of the whole answer

What not to write

“The tension goes from 7272 N to 144144 N, so the fundamental goes from 100100 Hz to 200200 Hz.”

What to write

“v=F/μv = \sqrt{F/\mu}, so doubling FF multiplies the speed and the frequency by 2=1.41\sqrt{2} = 1.41: 141141 Hz. To reach 200200 Hz the tension must be QUADRUPLED, 288288 N.”

5010015020025030050100150200250wrong: 200 Hz141 Hz200 Hz needs 288 N72 N, 100 HzF (N)f₁ (Hz)
The fundamental grows as F\sqrt{F} (green): doubling FF from 7272 N gives 141141 Hz, not the 200200 Hz of a straight line (red); 200200 Hz needs four times the tension.

Why: The tension sits under a square root. A quantity under a root needs a factor k2k^{2} to produce a factor kk, which is also why a guitar is tuned with small turns of the peg.

4. Letting the frequency change at a boundary

the whole question, and the same error returns in the optics chapters

What not to write

“Sound goes about 4.34.3 times faster in water, so its frequency is multiplied by 4.34.3.”

What to write

“The frequency is imposed by the source and does not change: vv is multiplied by 4.34.3, and so is λ=v/f\lambda = v/f.”

λ = 0.670 mλ = 2.89 mair, 343 m/swater, about 1480 m/sf = 512 Hz on both sideswrong: f multiplied by 4.3 in water
The same 512512 Hz on both sides of the boundary: the wave runs 4.34.3 times faster in water, so its wavelength is 4.34.3 times longer, and the frequency does not move.

Why: The last particle of air and the first particle of water are in contact: if they oscillated at different rates the medium would have to tear apart at the boundary.

5. Giving a closed pipe the even harmonics it does not have

2 points, and the identification of the pipe in any question that gives two of its frequencies

What not to write

“The pipe closed at one end sounds 171.5171.5 Hz, so the next note is 2×171.5=3432 \times 171.5 = 343 Hz.”

What to write

“A closed pipe takes only the ODD multiples: after 171.5171.5 Hz come 3f1=514.53f_1 = 514.5 Hz and 5f1=857.55f_1 = 857.5 Hz.”

34368610291372171.5514.5857.51200.5wrong: 2 × 171.5open pipeclosed pipesame length L = 0.500 m, frequencies in Hz
Same length: the open pipe has a rung at every multiple of 343343 Hz; the closed one starts an octave lower, at 171.5171.5 Hz, and every even rung (red) is missing.

Why: The closed end must be a node and the open end an antinode. Fitting a whole extra half wavelength would put an antinode at the closed end, where the air cannot move at all.

6. Adding two sound levels in decibels

the whole question, and the answer is absurd by a factor of a million

What not to write

“Two machines at 6060 dB each, so together 60+60=12060 + 60 = 120 dB.”

What to write

“Intensities add, levels do not: 2×10−62 \times 10^{-6} W/m2^2 gives β=10log⁡10(2×106)=63\beta = 10\log_{10}(2\times 10^{6}) = 63 dB.”

020406080100120dBone machine: 60 dBtwo together: 63 dBwrong: 60 + 60 = 120 dB10⁻⁶ and 2 × 10⁻⁶ W/m²1 W/m²10⁶ times one machine
Doubling the intensity moves the level by 33 dB only, from 6060 to 6363 dB; 120120 dB would be a million times the intensity of one machine.

Why: A decibel is a logarithm. Logarithms add where the quantities they measure MULTIPLY, so doubling the power adds 33 dB and never doubles the level.

7. Taking ten times the logarithm of an amplitude ratio

1 to 2 points, and every level computed afterwards from a voltage or a pressure

What not to write

“The microphone amplitude doubles, so the level rises by 10log⁡102=310\log_{10}2 = 3 dB.”

What to write

“II goes as A2A^{2}, so the level rises by 20log⁡102=6.020\log_{10}2 = 6.0 dB. The 33 dB belongs to a doubled INTENSITY.”

gain in dBintensityamplitude3× 2× 1.416× 4× 210× 10× 3.1612× 16× 40not × 2
One gain read twice: +3+3 dB is an intensity times 22 but an amplitude times only 1.411.41; doubling the amplitude (boxed) quadruples the intensity and gives +6+6 dB.

Why: The sheet writes the level twice, 10log⁡10(I/I0)=20log⁡10(A/A0)10\log_{10}(I/I_0) = 20\log_{10}(A/A_0), and the two agree only because log⁡10(A2)=2log⁡10A\log_{10}(A^{2}) = 2\log_{10}A. Read what the ratio compares before choosing: a microphone voltage or a pressure is an amplitude, a power per square metre is an intensity. To lower a level by 1212 dB the amplitude is divided by 1012/20=3.9810^{12/20} = 3.98, about 44, and the intensity by about 1616.

8. Using the moving source formula when the listener moves

2 points, and the whole point of the question, which is that the two cases differ

What not to write

“I drive at 30.030.0 m/s toward a parked 900900 Hz siren, so f′=900×343313=986f' = 900 \times \frac{343}{313} = 986 Hz.”

What to write

“The source is still and I am moving, so f′=900 343+30343=979f' = 900\,\dfrac{343 + 30}{343} = 979 Hz.”

the source movesthe listener movesfronts bunched ahead:λ really shortersame λ in the air,met more oftenwrong: one formula for both
A moving source bunches the fronts and shortens λ\lambda, f′=f v/(v−vs)f' = f\,v/(v - v_s); a moving listener meets unchanged fronts more often, f′=f (v+vo)/vf' = f\,(v + v_o)/v.

Why: A moving source bunches the wavefronts and really shortens the wavelength in the air; a moving listener meets unchanged wavefronts more often. The medium is there, so the two are not symmetric.

9. Forgetting to halve the round trip time of an echo

2 points, and a depth wrong by a factor of two on every echo of the exercise

What not to write

“The echo comes back after 130130 microseconds, so d=1540×130×10−6=0.20d = 1540 \times 130\times 10^{-6} = 0.20 m.”

What to write

“The burst goes AND returns, so d=vt/2=0.100d = vt/2 = 0.100 m.”

ddown and back: 2d = vtd = 0.100 mreflecting surfacewrong: d = vt puts it here, 0.20 mprobe
The burst goes down AND comes back, so the 130130 microseconds cover 2d2d: the surface is at 0.1000.100 m, and d=vtd = vt would put it twice as deep (red).

Why: Ultrasound, sonar and a bat all measure a round trip. Nothing in the numbers betrays the error, so the habit of writing 2d=vt2d = vt FIRST is the only protection.

Which method to choose

Which formula, from the shape of the question

What the statement hands you: a medium, a geometry, or a motion

  • If a rope or a string, with a tension and a mass per unit length → v=F/μv = \sqrt{F/\mu} first, then λ=v/f\lambda = v/f if a source frequency is given

    Example: 72/0.0050=120\sqrt{72/0.0050} = 120 m/s

  • If sound in air, with a temperature → v=331+0.60 TCv = 331 + 0.60\,T_C, then everything else

    Example: 331+0.60×20=343331 + 0.60\times 20 = 343 m/s

  • If a wave crossing, at an angle, a line where its speed changes → v1sin⁡θ2=v2sin⁡θ1v_1\sin\theta_2 = v_2\sin\theta_1, angles from the normal; ff unchanged, so λ2=λ1v2/v1\lambda_2 = \lambda_1 v_2/v_1

    Example: sin⁡θ2=7.012sin⁡40∘\sin\theta_2 = \frac{7.0}{12}\sin 40^{\circ}: θ2=22.0∘\theta_2 = 22.0^{\circ}

  • If a string clamped at both ends, or a pipe open at both ends → λn=2L/n\lambda_n = 2L/n for every whole nn, then fn=v/λnf_n = v/\lambda_n

    Example: L=0.60L = 0.60 m, v=120v = 120 m/s: f1=100f_1 = 100 Hz

  • If a pipe closed at one end, or one frequency missing from a measured series → λn=4L/n\lambda_n = 4L/n with nn odd only

    Example: 200200 Hz and 600600 Hz but no 400400 Hz: L=343/800=0.429L = 343/800 = 0.429 m

  • If two sources and one listening point → count Δ/λ\Delta/\lambda: a whole number is loud, a whole number plus a half is quiet

    Example: 0.75/0.500=1.50.75/0.500 = 1.5, so a minimum

  • If a distance, a number of sources, or an answer wanted in decibels → turn the RATIO into decibels: 10log⁡1010\log_{10} of an intensity ratio, 20log⁡1020\log_{10} of an amplitude ratio

    Example: distance ×4\times 4: −10log⁡1016=−12.0-10\log_{10}16 = -12.0 dB

  • If something is moving and a received frequency is asked → Doppler, with the sign chosen so that approaching RAISES the pitch

    Example: 900×343313=986900\times\frac{343}{313} = 986 Hz

  • If nothing moves and two close frequencies sound together → beats, fbeat=∣f1−f2∣f_{beat} = |f_1 - f_2|, and the ambiguity is lifted by detuning on purpose

    Example: 443−440=3443 - 440 = 3 beats per second

The first question is never which formula, it is which of vv, ff and λ\lambda the statement has already fixed. Two of the three are always given, one way or another, and the third is the answer.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing a standing wave question, string or pipe

When to use it: The statement gives a length and a medium and asks for a frequency, a harmonic, or the distance between two nodes.

  1. 1 Get the speed from the MEDIUM first: v=F/μv = \sqrt{F/\mu} for a string, v=331+0.60 TCv = 331 + 0.60\,T_C for air. Nothing else can be computed before it, and the units are checked here, once.
  2. 2 Draw the mode asked for and write the boundary condition in words: a node at a clamped end and at a closed end, an antinode at a free end and at an open end.
  3. 3 Turn the drawing into a WAVELENGTH, λn=2L/n\lambda_n = 2L/n or λn=4L/n\lambda_n = 4L/n with nn odd. Never write a frequency directly from the length.
  4. 4 Divide only now: fn=v/λnf_n = v/\lambda_n, then check that the frequencies found are in the ratio the drawing predicts.

Concluding sentence

“The bridges clamp the wire, so each of them is a node and L=λ1/2L = \lambda_1/2, hence λ1=1.300\lambda_1 = 1.300 m. With v=F/μ=19.60/1.20×10−3=128v = \sqrt{F/\mu} = \sqrt{19.60 / 1.20\times 10^{-3}} = 128 m/s, the fundamental is f1=v/λ1=98.3f_1 = v/\lambda_1 = 98.3 Hz.”

The trap: Writing f1=v/(2L)f_1 = v/(2L) from memory on a pipe closed at one end: the formula is right for a string and for an open pipe, and wrong by a factor of two here.

Marking: Usually 1 point for the speed, 1 for the boundary condition stated in words, 1 for the wavelength and 1 for each frequency asked.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The typical problem, taken apart

Sonometer: from a hanging mass to a beat count

A wire of linear density μ=1.20\mu = 1.20 g/m passes over two bridges L=0.650L = 0.650 m apart, then over a pulley, and is tensioned by a hanging mass m=2.00m = 2.00 kg. Take g=9.80g = 9.80 m/s2^2.

Find the fundamental frequency, the distance between two neighbouring nodes in the third harmonic, the number of beats heard against a 100100 Hz reference, and the mass that would silence those beats.

L = 0.650 mm = 2.00 kgbridgebridgewire, μ = 1.20 g/m
The vibrating length is bridge to bridge, L=0.650L = 0.650 m, not the whole wire; the hanging mass, through the pulley, is what sets the tension.

Step 1

F=mg=2.00×9.80=19.6F = mg = 2.00 \times 9.80 = 19.6 N.

Why

The hanging mass is in equilibrium, so the tension is its WEIGHT. Writing F=2.00F = 2.00 N is worth the whole question and is caught by the unit.

Step 2

μ=1.20×10−3\mu = 1.20\times 10^{-3} kg/m, so v=19.60/1.20×10−3=128v = \sqrt{19.60 / 1.20\times 10^{-3}} = 128 m/s.

Why

The speed belongs to the medium and is computed before anything else. The conversion of μ\mu is done here, once, where it can still be checked.

Step 3

λ1=2L=1.300\lambda_1 = 2L = 1.300 m, so f1=v/λ1=127.80/1.300=98.3f_1 = v/\lambda_1 = 127.80/1.300 = 98.3 Hz.

Why

The bridges are the clamped ends, so the vibrating length is bridge to bridge, not the whole wire. Measuring LL to the pulley always makes the frequency come out too low.

Step 4

Third harmonic: neighbouring nodes are λ3/2=L/3=0.217\lambda_3/2 = L/3 = 0.217 m apart.

Why

A node does not move, so it is sharp under a rule, while an antinode is a blur. This is the quantity the laboratory actually measures.

Step 5

Beats: ∣100−98.3∣=1.7|100 - 98.3| = 1.7 per second. To silence them, v=2Lf=130v = 2Lf = 130 m/s, so F=μv2=20.3F = \mu v^{2} = 20.3 N and m=F/g=2.07m = F/g = 2.07 kg.

Why

The square root again: 3.53.5 percent more mass for 1.71.7 percent more frequency. The beats vanish long before the eye sees anything change in the wire.

The conclusion, written out

“The tension is 19.619.6 N and the waves run at 128128 m/s, so f1=98.3f_1 = 98.3 Hz. In the third harmonic the nodes are 0.2170.217 m apart. Against the 100100 Hz reference the wire gives 1.71.7 beats per second, and hanging 2.072.07 kg instead of 2.002.00 kg silences them.”

The classic mistake on this problem: Using the whole length of the wire instead of the bridge to bridge distance: the fundamental drops below 98 Hz, the beat count doubles, and the correcting mass comes out far too large.

Learn by heart

  • • v=fλv = f\lambda, read as: ff from the source, vv from the medium, λ\lambda from the quotient.
  • • v=F/μv = \sqrt{F/\mu} on a string; v=331+0.60 TCv = 331 + 0.60\,T_C in air, so 343343 m/s at 2020 degrees Celsius.
  • • Refraction: v1sin⁡θ2=v2sin⁡θ1v_1\sin\theta_2 = v_2\sin\theta_1, angles from the normal; the slower side is closer to the normal, and ff does not change.
  • • Clamped string and open pipe: λn=2L/n\lambda_n = 2L/n, every nn. Pipe closed at one end: λn=4L/n\lambda_n = 4L/n, odd nn only.
  • • Two neighbouring nodes are λ/2\lambda/2 apart, whatever the harmonic.
  • • Δ=nλ\Delta = n\lambda is loud, Δ=(n+12)λ\Delta = \left(n + \frac{1}{2}\right)\lambda is quiet.
  • • β=10log⁡10(I/I0)=20log⁡10(A/A0)\beta = 10\log_{10}(I/I_0) = 20\log_{10}(A/A_0) with I0=1.0×10−12I_0 = 1.0\times 10^{-12} W/m2^2: I×2I \times 2 gives +3+3 dB, A×2A \times 2 gives +6+6 dB, I×10I \times 10 gives +10+10 dB, distance ×2\times 2 gives −6-6 dB.
  • • Beats ∣f1−f2∣|f_1 - f_2|; Doppler f′=f v±vov∓vsf' = f\,\dfrac{v \pm v_o}{v \mp v_s}, sign chosen so that approaching raises the pitch.
  • • Echo: halve the round trip time before multiplying by vv.

Frequently asked questions

What is the difference between a wavelength and a period?

A wavelength is a distance and a period is a time. The wavelength is how far you travel along the rope before the shape repeats, and you read it on a snapshot, where the horizontal axis carries metres. The period is how long you wait at one fixed point before that point returns to the same place, and you read it on a graph whose horizontal axis carries seconds. The two graphs look identical, so always read the axis before the number.

Why does a wave keep its frequency when it enters a new medium?

Because the frequency is imposed by the source, and the boundary is a single point where the two media are in contact. The last particle of the first medium pushes the first particle of the second one, so they must oscillate at the same rate, otherwise the material would tear apart at the boundary. What the new medium does change is the speed, and the wavelength adjusts to match.

Why do ocean waves reach the beach almost parallel to it?

Because waves slow down as the water gets shallower, and a wave that slows down while crossing a line at an angle turns toward the perpendicular to that line. The period of the swell, set far out at sea, does not change; the speed drops, so the crests close up and the part of a crest still in deep water catches up with the part already slowed. Every depth contour turns the crests a little more, so they end up nearly parallel to the shore.

Why does a pipe closed at one end play only the odd harmonics?

The closed end blocks the air, so it must be a node, while the open end is free and must be an antinode. The shortest pattern that fits is a quarter of a wavelength, and every pattern that also fits adds another half wavelength, which gives one quarter, three quarters, five quarters, and so on. The even multiples would require the air to move at the closed end, which it cannot do.

Why can two sounds of sixty decibels not make one hundred and twenty?

Because a decibel is a logarithm of an intensity, and logarithms add only when the quantities they measure are multiplied. Two machines deliver twice the intensity of one, and doubling an intensity adds about three decibels, so two sixty decibel machines give sixty three decibels together. One hundred and twenty decibels would be a million times the intensity of a single machine.

Practise it

Corrected exercises: Waves and sound, PHYS 101 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

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  • 100 points
  • 150 minutes
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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/phys101-waves-sound. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

See also

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Get in touch for a first session. Waves and sound are worked as a method here: name the medium, get the speed, let the geometry choose the wavelength, and divide only at the end.

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