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.

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

• Oscillations and simple harmonic motion

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Transverse and longitudinal waves
Wavelength, period and frequency
Wave speed on a string
Speed of sound in air
Superposition and path difference
Standing waves
Nodes and antinodes
Open and closed pipes
Sound intensity level in decibels
Inverse square law
Beats
Doppler effect

The essentials

The three quantities, and who fixes each one

• $v = f\lambda$ is the whole chapter, but it is read in one direction only: $f$ comes from the SOURCE, $v$ from the MEDIUM, and $\lambda$ is what is left over.
• On a string, $v = \sqrt{F/\mu}$ with $F$ in newtons and $\mu$ in kilograms per metre. In air, $v = 331 + 0.60\,T_C$ 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 $\lambda = v/f$ 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

\lambda

, 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.

Standing waves, pipes, levels and shifts

• String clamped at both ends, or pipe open at both ends: $\lambda_n = 2L/n$ and $f_n = n\,v/(2L)$ , every whole $n$ . Neighbouring nodes are $\lambda/2 = L/n$ apart.
• Pipe closed at one end: $\lambda_n = 4L/n$ with $n$ ODD only, so $f_1 = v/(4L)$ and then $3f_1$ , $5f_1$ , and nothing in between.
• Two sources, one listening point: count the path difference in wavelengths. $\Delta = n\lambda$ is loud, $\Delta = \left(n+\frac{1}{2}\right)\lambda$ is quiet.
• Level: $\beta = 10\log_{10}(I/I_0)$ with $I_0 = 10^{-12}$ W/m $^2$ , and $I = P/(4\pi r^{2})$ for a point source. Beats: $f_{beat} = |f_1 - f_2|$ . Doppler: $f' = f\,\dfrac{v \pm v_o}{v \mp v_s}$ .

Three numbers to hold instead of three formulas: doubling an intensity adds $3$ dB, multiplying it by ten adds $10$ dB, and doubling a distance costs $6$ dB.

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 $\lambda = 4.0$ 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 $A^{2}$

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.”

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

“ $v = \sqrt{F/\mu}$ , so doubling $F$ multiplies the speed and the frequency by $\sqrt{2} = 1.41$ : $141$ Hz. To reach $200$ Hz the tension must be QUADRUPLED, $288$ N.”

Why: The tension sits under a square root. A quantity under a root needs a factor $k^{2}$ 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 $\lambda = v/f$ .”

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 $2 \times 171.5 = 343$ Hz.”

What to write

“A closed pipe takes only the ODD multiples: after $171.5$ Hz come $3f_1 = 514.5$ Hz and $5f_1 = 857.5$ Hz.”

Two pipes of the same length: the open one has a rung at every multiple of

343

Hz, the closed one starts an octave lower and skips every second rung.

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: $2 \times 10^{-6}$ W/m $^2$ gives $\beta = 10\log_{10}(2\times 10^{6}) = 63$ dB.”

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. 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 $f' = 900 \times \frac{343}{313} = 986$ Hz.”

What to write

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

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.

8. 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 $d = 1540 \times 130\times 10^{-6} = 0.20$ m.”

What to write

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

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 → $v = \sqrt{F/\mu}$ first, then $\lambda = v/f$ if a source frequency is given

Example: $\sqrt{72/0.0050} = 120$ m/s
If sound in air, with a temperature → $v = 331 + 0.60\,T_C$ , then everything else

Example: $331 + 0.60\times 20 = 343$ m/s
If a string clamped at both ends, or a pipe open at both ends → $\lambda_n = 2L/n$ for every whole $n$ , then $f_n = v/\lambda_n$

Example: $L = 0.60$ m, $v = 120$ m/s: $f_1 = 100$ Hz
If a pipe closed at one end, or one frequency missing from a measured series → $\lambda_n = 4L/n$ 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 $\Delta/\lambda$ : 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 of intensities into decibels, $10\log_{10}$ of it

Example: distance $\times 4$ : $-10\log_{10}16 = -12.0$ dB
If something is moving and a received frequency is asked → Doppler, with the sign chosen so that approaching RAISES the pitch

Example: $900\times\frac{343}{313} = 986$ Hz
If nothing moves and two close frequencies sound together → beats, $f_{beat} = |f_1 - f_2|$ , 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 $\lambda$ 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 Get the speed from the MEDIUM first: $v = \sqrt{F/\mu}$ for a string, $v = 331 + 0.60\,T_C$ for air. Nothing else can be computed before it, and the units are checked here, once.
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 Turn the drawing into a WAVELENGTH, $\lambda_n = 2L/n$ or $\lambda_n = 4L/n$ with $n$ odd. Never write a frequency directly from the length.
4 Divide only now: $f_n = v/\lambda_n$ , 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 $L = \lambda_1/2$ , hence $\lambda_1 = 1.300$ m. With $v = \sqrt{F/\mu} = \sqrt{19.60 / 1.20\times 10^{-3}} = 128$ m/s, the fundamental is $f_1 = v/\lambda_1 = 98.3$ Hz.”

The trap: Writing $f_1 = v/(2L)$ 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 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 $\mu$ was left in grams per metre.

$\sqrt{72/5.0} = 3.8$ 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, ten times 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 typical problem, taken apart

Sonometer: from a hanging mass to a beat count

A wire of linear density $\mu = 1.20$ 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/s $^2$ .

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 third harmonic between the bridges: four nodes counting the two ends, three antinodes, and a node spacing of

L/3 = 0.217

Step 1

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

Why

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

Step 2

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

Why

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

Step 3

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

Why

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

Step 4

Third harmonic: neighbouring nodes are $\lambda_3/2 = L/3 = 0.217$ 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: $|100 - 98.3| = 1.7$ per second. To silence them, $v = 2Lf = 130$ m/s, so $F = \mu v^{2} = 20.3$ 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 $f_1 = 98.3$ 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

• $v = f\lambda$ , read as: $f$ from the source, $v$ from the medium, $\lambda$ from the quotient.
• $v = \sqrt{F/\mu}$ on a string; $v = 331 + 0.60\,T_C$ in air, so $343$ m/s at $20$ degrees Celsius.
• Clamped string and open pipe: $\lambda_n = 2L/n$ , every $n$ . Pipe closed at one end: $\lambda_n = 4L/n$ , odd $n$ only.
• Two neighbouring nodes are $\lambda/2$ apart, whatever the harmonic.
• $\Delta = n\lambda$ is loud, $\Delta = \left(n + \frac{1}{2}\right)\lambda$ is quiet.
• $\beta = 10\log_{10}(I/I_0)$ with $I_0 = 10^{-12}$ W/m $^2$ : $\times 2$ gives $+3$ dB, $\times 10$ gives $+10$ dB, distance $\times 2$ gives $-6$ dB.
• Beats $|f_1 - f_2|$ ; Doppler $f' = f\,\dfrac{v \pm v_o}{v \mp v_s}$ , sign chosen so that approaching raises the pitch.
• Echo: halve the round trip time before multiplying by $v$ .

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 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.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Oscillations and simple harmonic motion Next sheet Geometric optics: mirrors and lenses

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Revision sheet: waves and sound (PHYS 101)

The essentials

The three quantities, and who fixes each one

Standing waves, pipes, levels and shifts

The mistakes that cost marks

1. Reading the period off a snapshot graph

2. Taking the amplitude from a trough to a crest

3. Doubling the tension to double the frequency

4. Letting the frequency change at a boundary

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

6. Adding two sound levels in decibels

7. Using the moving source formula when the listener moves

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

Which method to choose

Which formula, from the shape of the question

How the answer is expected to be written

Writing a standing wave question, string or pipe

Check before you hand in

The typical problem, taken apart

Sonometer: from a hanging mass to a beat count

Learn by heart

Frequently asked questions

What is the difference between a wavelength and a period?

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

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

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

Corrected exercises: Waves and sound, PHYS 101 at McGill

See also

Looking for a PHYS 101 tutor in Montreal?