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.
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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.
On your formula sheet: this chapter's lines, and when each one holds
•v=fλ: The medium sets v, the source sets f, and λ follows. Into another medium, f stays and λ changes.
•v=FT/μ: μ is the mass per unit LENGTH in kg/m, not the mass of the string.
•λm=2L/m: A string fixed at both ends, m=1, 2, 3 and so on, which gives fm=m2Lv.
•v1sinθ2=v2sinθ1: Angles measured from the NORMAL, and the speeds CROSSED: the wave bends toward the normal on the side where it is slower. Divided by c, it is n1sinθ1=n2sinθ2.
•vs=330: In m/s, the sheet's typical value for air. When a statement gives another one (343 m/s at 20 °C), the statement wins.
•β=10log10(I0I)=20log10(A0A): Ten with intensities, twenty with amplitudes, because I grows like A2: doubling the intensity adds 3 dB, doubling the amplitude adds 6 dB.
•I0=1.0×10−12: In W/m², the threshold of hearing, the reference for 0 dB.
•fn=nf1=n2Lv: A tube OPEN at both ends: every harmonic, n=1, 2, 3, 4 and so on.
•fn=nf1=n4Lv: A tube CLOSED at one end: odd harmonics only, n=1, 3, 5, 7 and so on.
•fb=∣f1−f2∣: Two close frequencies: the ear hears their average pitch, swelling fb times per second.
•f′=fvwave±vsourcevwave: Moving source: minus when the source approaches (higher pitch), plus when it moves away.
•f′=fvwavevwave±vobs: 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=fλ is the whole chapter, but it is read in one direction only: f comes from the SOURCE, v from the MEDIUM, and λ is what is left over.
•On a string, v=F/μ with F in newtons and μ in kilograms per metre. In air, v=331+0.60TC in metres per second, so 343 m/s at 20 degrees Celsius.
•Snapshot, y against x: you read the amplitude and the WAVELENGTH on it. History of one point, y against t: you read the amplitude and the PERIOD. A period cannot be measured on a snapshot, which froze time.
•Crossing into a new medium: f is unchanged, v is the one of the new medium, and λ=v/f follows. Each point of the medium only oscillates in place, exactly like the mass on a spring of the previous chapter.
The same drawing twice: on the left the horizontal axis carries metres and the bracket measures λ, on the right it carries seconds and the same bracket measures T.
A point sitting at a crest is at the end of its travel and is momentarily at rest; a point crossing y=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: v1sinθ2=v2sinθ1, 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/sinθ1=λ2/sinθ2, and the common frequency cancels out of λ=v/f.
•Slower on the far side: the wave turns TOWARD the normal. Swell at 12 m/s reaching shallower water at 7.0 m/s, 40∘ from the normal, leaves at 22.0∘, its wavelength going from 120 m to 70 m with the same 10 s period.
•Checks: v2=v1 gives θ2=θ1, no bending; θ1=0 gives θ2=0, slowed but not turned. The same line, with v=c/n, is the refraction of light.
Each crest bends where it crosses the depth line: the direction of travel turns from θ1=40∘ to θ2=22.0∘, the crests close up from 120 m to 70 m, and the period stays 10 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 v is the disturbance and its energy; the medium only oscillates about its place.
Rope: the points move across the direction of travel. Air: the layers move along it, bunching into compressions. On both, λ 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 and fn=nv/(2L), every whole n. Neighbouring nodes are λ/2=L/n apart.
•Pipe closed at one end: λn=4L/n with n ODD only, so f1=v/(4L) and then 3f1, 5f1, and nothing in between.
•Two sources, one listening point: count the path difference in wavelengths. Δ=nλ is loud, Δ=(n+21)λ is quiet.
•Level: β=10log10(I/I0)=20log10(A/A0) with I0=1.0×10−12 W/m2, the second form for an amplitude ratio, and I=P/(4πr2) for a point source. Beats: fbeat=∣f1−f2∣. Doppler: f′=fv∓vsv±vo.
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, 2L and 4L.
Four numbers to hold instead of four formulas: doubling an intensity adds 3 dB, doubling an amplitude adds 6 dB, multiplying an intensity by ten adds 10 dB, and doubling a distance costs 6 dB.
Two sources, one listener: the path difference
•Two sources in step reach a point P by paths r1 and r2. Only the path difference Δ=r2−r1 matters, and only once it is counted in wavelengths.
•Δ/λ 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: loud for every frequency at once. Inverting one source swaps every loud point with a quiet one.
The two waves arriving at P: 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 f1 and f2 sounded together are heard as ONE note at the average (f1+f2)/2, whose loudness swells and fades ∣f1−f2∣ times a second.
•Count loudness MAXIMA: the two notes fall back into step ∣f1−f2∣ times a second. With 256 Hz and 260 Hz that is 4 swells per second, heard as one note near 258 Hz, never as a hum at 4 Hz and never as 2 swells per second.
•A beat count alone gives two candidates, fref±fbeat. 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.
Two close notes added: the sum keeps the fast oscillation, but its loudness swells and fades, one maximum every 1/∣f1−f2∣ 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 4, so T=4 s and f=0.25 Hz.”
What to write
“The horizontal axis carries metres, so the 4 is λ=4.0 m. The period is read on the y against t graph: T=0.50 s, f=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 A2
What not to write
“The curve runs from −2 to +2, so the amplitude is 4.0 cm.”
What to write
“The amplitude is measured from the centre line, so A=2.0 cm. The 4.0 cm is the peak to peak value.”
The amplitude is measured from the centre line to a crest (green), A=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 72 N to 144 N, so the fundamental goes from 100 Hz to 200 Hz.”
What to write
“v=F/μ, so doubling F multiplies the speed and the frequency by 2=1.41: 141 Hz. To reach 200 Hz the tension must be QUADRUPLED, 288 N.”
The fundamental grows as F (green): doubling F from 72 N gives 141 Hz, not the 200 Hz of a straight line (red); 200 Hz needs four times the tension.
Why: The tension sits under a square root. A quantity under a root needs a factor k2 to produce a factor k, 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 4.3 times faster in water, so its frequency is multiplied by 4.3.”
What to write
“The frequency is imposed by the source and does not change: v is multiplied by 4.3, and so is λ=v/f.”
The same 512 Hz on both sides of the boundary: the wave runs 4.3 times faster in water, so its wavelength is 4.3 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 171.5 Hz, so the next note is 2×171.5=343 Hz.”
What to write
“A closed pipe takes only the ODD multiples: after 171.5 Hz come 3f1=514.5 Hz and 5f1=857.5 Hz.”
Same length: the open pipe has a rung at every multiple of 343 Hz; the closed one starts an octave lower, at 171.5 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 60 dB each, so together 60+60=120 dB.”
What to write
“Intensities add, levels do not: 2×10−6 W/m2 gives β=10log10(2×106)=63 dB.”
Doubling the intensity moves the level by 3 dB only, from 60 to 63 dB; 120 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 3 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 10log102=3 dB.”
What to write
“I goes as A2, so the level rises by 20log102=6.0 dB. The 3 dB belongs to a doubled INTENSITY.”
One gain read twice: +3 dB is an intensity times 2 but an amplitude times only 1.41; doubling the amplitude (boxed) quadruples the intensity and gives +6 dB.
Why: The sheet writes the level twice, 10log10(I/I0)=20log10(A/A0), and the two agree only because log10(A2)=2log10A. 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 12 dB the amplitude is divided by 1012/20=3.98, about 4, and the intensity by about 16.
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 30.0 m/s toward a parked 900 Hz siren, so f′=900×313343=986 Hz.”
What to write
“The source is still and I am moving, so f′=900343343+30=979 Hz.”
A moving source bunches the fronts and shortens λ, f′=fv/(v−vs); a moving listener meets unchanged fronts more often, f′=f(v+vo)/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 130 microseconds, so d=1540×130×10−6=0.20 m.”
What to write
“The burst goes AND returns, so d=vt/2=0.100 m.”
The burst goes down AND comes back, so the 130 microseconds cover 2d: the surface is at 0.100 m, and d=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=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/μ first, then λ=v/f if a source frequency is given
Example: 72/0.0050=120 m/s
If sound in air, with a temperature → v=331+0.60TC, then everything else
Example: 331+0.60×20=343 m/s
If a wave crossing, at an angle, a line where its speed changes → v1sinθ2=v2sinθ1, angles from the normal; f unchanged, so λ2=λ1v2/v1
Example: sinθ2=127.0sin40∘: θ2=22.0∘
If a string clamped at both ends, or a pipe open at both ends → λn=2L/n for every whole n, then fn=v/λn
Example: L=0.60 m, v=120 m/s: f1=100 Hz
If a pipe closed at one end, or one frequency missing from a measured series → λn=4L/n with n odd only
Example: 200 Hz and 600 Hz but no 400 Hz: L=343/800=0.429 m
If two sources and one listening point → count Δ/λ: a whole number is loud, a whole number plus a half is quiet
Example: 0.75/0.500=1.5, so a minimum
If a distance, a number of sources, or an answer wanted in decibels → turn the RATIO into decibels: 10log10 of an intensity ratio, 20log10 of an amplitude ratio
Example: distance ×4: −10log1016=−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: 900×313343=986 Hz
If nothing moves and two close frequencies sound together → beats, fbeat=∣f1−f2∣, and the ambiguity is lifted by detuning on purpose
Example: 443−440=3 beats per second
The first question is never which formula, it is which of v, f and λ 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.
1Get the speed from the MEDIUM first: v=F/μ for a string, v=331+0.60TC for air. Nothing else can be computed before it, and the units are checked here, once.
2Draw 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.
3Turn the drawing into a WAVELENGTH, λn=2L/n or λn=4L/n with n odd. Never write a frequency directly from the length.
4Divide only now: fn=v/λn, 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=λ1/2, hence λ1=1.300 m. With v=F/μ=19.60/1.20×10−3=128 m/s, the fundamental is f1=v/λ1=98.3 Hz.”
The trap: Writing f1=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 order of magnitude of the speed
A wave on a taut string runs at tens to hundreds of metres per second, and sound in air at about 343 m/s. A speed of a few metres per second means μ was left in grams per metre.
72/5.0=3.8 m/s instead of 120 m/s.
The frequency that must not have moved
Circle the frequency given in the statement. Unless the source itself was changed, it must appear untouched in your answer, whatever the medium does.
512 Hz outdoors at 20 and at −10 degrees: same 512 Hz, two different wavelengths.
The ratio of the harmonics
Divide every frequency you found by the smallest. A string or an open pipe gives whole numbers, a closed pipe gives odd numbers only. Anything else is a slip.
171.5, 514.5, 857.5 Hz: ratios 1, 3, 5, so a closed pipe.
The decibel arithmetic done in your head
Doubling the intensity adds 3 dB, doubling the amplitude adds 6 dB, ten times the intensity adds 10 dB, doubling the distance costs 6 dB. If your level moved by more than that, you added levels instead of intensities.
90 dB at 2.0 m gives 78 dB at 8.0 m, two doublings of distance.
The angle after refraction moved the right way
Into a slower medium the angle from the normal must DECREASE, into a faster one it increases. An angle that moved the wrong way means the two speeds were swapped in the formula of the sheet.
12 m/s to 7.0 m/s: 40∘ becomes 22.0∘; swapping the speeds gives sinθ2=1.10, which no angle has.
The typical problem, taken apart
Sonometer: from a hanging mass to a beat count
A wire of linear density μ=1.20 g/m passes over two bridges L=0.650 m apart, then over a pulley, and is tensioned by a hanging mass m=2.00 kg. Take g=9.80 m/s2.
Find the fundamental frequency, the distance between two neighbouring nodes in the third harmonic, the number of beats heard against a 100 Hz reference, and the mass that would silence those beats.
The vibrating length is bridge to bridge, L=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=19.6 N.
Why
The hanging mass is in equilibrium, so the tension is its WEIGHT. Writing F=2.00 N is worth the whole question and is caught by the unit.
Step 2
μ=1.20×10−3 kg/m, so v=19.60/1.20×10−3=128 m/s.
Why
The speed belongs to the medium and is computed before anything else. The conversion of μ is done here, once, where it can still be checked.
Step 3
λ1=2L=1.300 m, so f1=v/λ1=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 L to the pulley always makes the frequency come out too low.
Step 4
Third harmonic: neighbouring nodes are λ3/2=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 per second. To silence them, v=2Lf=130 m/s, so F=μv2=20.3 N and m=F/g=2.07 kg.
Why
The square root again: 3.5 percent more mass for 1.7 percent more frequency. The beats vanish long before the eye sees anything change in the wire.
The conclusion, written out
“The tension is 19.6 N and the waves run at 128 m/s, so f1=98.3 Hz. In the third harmonic the nodes are 0.217 m apart. Against the 100 Hz reference the wire gives 1.7 beats per second, and hanging 2.07 kg instead of 2.00 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=fλ, read as: f from the source, v from the medium, λ from the quotient.
•v=F/μ on a string; v=331+0.60TC in air, so 343 m/s at 20 degrees Celsius.
•Refraction: v1sinθ2=v2sinθ1, angles from the normal; the slower side is closer to the normal, and f does not change.
•Clamped string and open pipe: λn=2L/n, every n. Pipe closed at one end: λn=4L/n, odd n only.
•Two neighbouring nodes are λ/2 apart, whatever the harmonic.
•Beats ∣f1−f2∣; Doppler f′=fv∓vsv±vo, sign chosen so that approaching raises the pitch.
•Echo: halve the round trip time before multiplying by v.
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.
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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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.