Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: sound and standing waves (203-SN3-RE)

Sound brings two tools that are not handled like the rest of physics, in a chapter of Waves and Modern Physics, 203-SN3-RE (formerly 203-NYC): a logarithmic scale, where quantities no longer add, and standing waves, where the only question that matters is which ends impose a node.

This sheet gathers what really costs marks on tests: decibels that get added, the inverse-square fall-off with distance, the odd harmonics of the closed pipe, the two Doppler formulas that are not symmetric, and beats mistaken for a sum. It is written for CEGEP students in the Natural Science program (Sciences de la nature).

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The thread of the chapter

Nothing adds up in this chapter: not the decibels, which are logarithms, not the beat frequencies, which subtract, not the superposed waves, of which only nodes remain. Everything is decided by ratios and by the ends.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

Intensity and decibels: a scale that does not add

• Speed of sound in air: about $343$ m/s at $20\,^{\circ}$ C, and $v\approx 331+0.6\,T$ with $T$ in degrees Celsius. Always $\lambda=\dfrac{v}{f}$ .
• Point source: $I=\dfrac{P}{4\pi r^{2}}$ in W/m $^{2}$ , so a fall-off as $\dfrac{1}{r^{2}}$ .
• Sound level: $\beta=10\log\dfrac{I}{I_{0}}$ with $I_{0}=10^{-12}$ W/m $^{2}$ . Decibels NEVER add: it is the intensities that add.
• Two rules that save you a calculation: doubling the intensity adds $3$ dB; doubling the distance subtracts $6$ dB.
• $\Delta P_{max}=\rho v\omega s_{max}$ and $I=\dfrac{\Delta P_{max}^{2}}{2\rho v}$ . Displacement and pressure are in QUADRATURE: a displacement node is a pressure antinode.

The four marked points are at

1

2

4

and

8

m: at each doubling, the level loses exactly

6

dB, never half of its value.

Two loudspeakers of $60$ dB each give $63$ dB, not $120$ dB. To gain $10$ dB, you need TEN times the power, which is why an amplifier twice as powerful is barely noticeable.

Standing waves: everything is decided at the ends

• Superposition: $A\sin(kx-\omega t)+A\sin(kx+\omega t)=2A\sin(kx)\cos(\omega t)$ . Space and time are SEPARATED: nothing travels, and no energy is carried.
• Nodes are spaced $\dfrac{\lambda}{2}$ apart, never $\lambda$ ; antinodes are midway between two nodes.
• String fixed at both ends, or pipe OPEN at both ends: $L=n\dfrac{\lambda_{n}}{2}$ , so $f_{n}=n\dfrac{v}{2L}$ , with ALL integers $n$ .
• Pipe CLOSED at one end: node at the closed end, antinode at the open end, $L=n\dfrac{\lambda_{n}}{4}$ and $f_{n}=n\dfrac{v}{4L}$ with $n$ ODD only. Its fundamental is an octave lower than that of the open pipe of the same length.
• INVERTED reflection at a fixed end, non-inverted at a free end.

The open pipe fits HALF a wavelength, the closed pipe fits only a QUARTER: for equal length, the closed pipe sounds an octave lower.

A pipe takes two seconds: look at its two ends, put a node at each closed end and an antinode at each open end, and the wavelength of the fundamental can be read off the drawing.

Vibrating string: the natural modes

• A string fixed at both ends has a node at each end: mode $n$ has $n$ antinodes and $n+1$ nodes, ends included.
• $\lambda_{n}=\dfrac{2L}{n}$ and $f_{n}=n\dfrac{v}{2L}$ , with $v=\sqrt{\dfrac{F}{\mu}}$ : tightening the string raises all its natural frequencies.
• Halving the vibrating length doubles the fundamental: this is the octave of the twelfth fret.

Mode

n

fits

n

antinodes between

n+1

nodes, ends included:

\lambda_{n}=\dfrac{2L}{n}

and

f_{n}=nf_{1}

Doppler, beats and shock waves

• Moving source, stationary observer: $f'=f\dfrac{v}{v\mp v_{s}}$ , the minus sign on APPROACH, which raises the frequency.
• Moving observer, stationary source: $f'=f\dfrac{v\pm v_{o}}{v}$ , the plus sign on approach. The two cases are NOT symmetric: the air is the reference.
• Beats: two nearby sounds give an intensity that pulses at $f_{beat}=|f_{1}-f_{2}|$ , around the AVERAGE pitch.
• Interference of two in-phase sources: constructive if the path difference is $n\lambda$ , destructive if it is $\left(n+\dfrac{1}{2}\right)\lambda$ ; in the far field $\Delta=d\sin\theta$ .
• Supersonic flight: $M=\dfrac{v_{plane}}{v_{sound}}$ and the shock cone satisfies $\sin\theta=\dfrac{1}{M}$ .

On the left, the source advances and squeezes its wavefronts together ahead of it: the wavelength is shorter there. On the right, the wavefronts are intact and it is the observer who crosses them more often.

The supersonic boom is not a single event at the crossing of Mach 1: the cone accompanies the plane permanently and sweeps the ground all along its flight.

Two sources in phase: the path difference decides

• At a point $P$ , the path difference is $\Delta=|S_{2}P-S_{1}P|$ : for two sources in phase, it alone decides.
• Constructive if $\Delta=n\lambda$ , destructive if $\Delta=\left(n+\dfrac{1}{2}\right)\lambda$ ; far from the sources, $\Delta\approx d\sin\theta$ .
• Since $\Delta\le d$ , the number of minima is finite, and there are none if $\dfrac{\lambda}{2}>d$ : a subwoofer creates no dead zone.

Sources in phase,

d=2\lambda

: reinforcement on the blue lines (

\Delta=0

and

\lambda

), silence on the orange ones (

\Delta=\frac{\lambda}{2}

and

\frac{3\lambda}{2}

), and no other silence since

\Delta\le d

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. Adding decibels

the whole question, 5 points

What not to write

“Two loudspeakers of $60$ dB each give $120$ dB.”

What to write

“It is the INTENSITIES that add: $10^{-6}+10^{-6}=2\times 10^{-6}$ W/m $^{2}$ , so $\beta=10\log\dfrac{2\times 10^{-6}}{10^{-12}}=63$ dB. Doubling the intensity adds $3$ dB, never double.”

Two speakers of

60

dB: the intensities add, giving

63

dB (green).

120

dB (red), the addition of the logarithms, is the level of a jet taking off.

Why: The decibel is a logarithm: adding two logarithms amounts to MULTIPLYING the intensities, which makes no physical sense here. $120$ dB is the level of a jet taking off, not of two living-room speakers.

2. Dividing the sound level when the distance doubles

3 points, and a result unrelated to experience

What not to write

“At $2$ m instead of $1$ m, the intensity is divided by four, so the level goes from $80$ to $20$ dB.”

What to write

“The intensity is indeed divided by four, but the LEVEL loses $10\log 4=6$ dB: it goes from $80$ to $74$ dB.”

The same power crosses a surface four times larger at

2

I

is divided by

4

and the level loses

6

dB, from

80

74

dB (green), never down to

20

dB (red).

Why: The level is a compressed scale: it turns a factor into an addition. A drop of $60$ dB would correspond to an intensity divided by a million, and so to a distance a thousand times greater.

3. Treating a closed pipe like an open pipe

the whole question, 8 points

What not to write

“The $0.500$ m pipe sounds at $f_{1}=\dfrac{v}{2L}=343$ Hz, and its harmonics are $686$ and $1029$ Hz.”

What to write

“Closed at one end: node at the closed end, antinode at the open end, so $L=\dfrac{\lambda}{4}$ and $f_{1}=\dfrac{v}{4L}=171.5$ Hz. Its harmonics are $3f_{1}=514.5$ Hz and $5f_{1}=857.5$ Hz: the ODD ones only.”

A closed pipe of

0.500

m sounds only at the odd ranks,

171.5

514.5

and

857.5

Hz (green);

343

686

and

1029

Hz (red) are the frequencies of the OPEN pipe.

Why: A closed end imposes a displacement node, an open end an antinode: between the two, there can only be a quarter of a wavelength, then three quarters, then five quarters. Even harmonics are geometrically impossible.

4. Applying the moving-source formula to a moving observer

3 points, and a gap that grows with speed

What not to write

“Whether the ambulance or I am moving, it is the same: $f'=f\dfrac{v}{v-v_{s}}$ .”

What to write

“The two cases are NOT symmetric. Moving source: $f'=f\dfrac{v}{v-v_{s}}$ , the wavelength itself is compressed. Moving observer: $f'=f\dfrac{v+v_{o}}{v}$ , the wavelength is unchanged but more wavefronts are crossed.”

30

m/s, the two formulas differ by only

4.2

Hz; at

250

m/s, the moving source gives

1844

Hz and the moving observer

864

Hz: two mechanisms, two curves.

Why: The air is the absolute reference for sound, unlike for light. At $30$ m/s, the difference between the two formulas remains small, but it becomes huge as the speed approaches that of sound.

5. Getting the sign wrong in the Doppler effect

3 points, and a conclusion that contradicts everyday experience

What not to write

“The ambulance is approaching, so we divide by $v+v_{s}$ : the frequency heard is lower.”

What to write

“On APPROACH, the frequency heard is HIGHER: $f'=f\dfrac{v}{v-v_{s}}$ , with a smaller denominator. The sound is lower only AFTER the ambulance has passed.”

Siren of

500

Hz at

30

m/s: about

548

Hz on approach,

460

Hz after passing (green). The red curve, low pitch first, is the wrong sign.

Why: The check is free: everyone has heard a siren go by. If the result announces a lower sound on approach, the sign is wrong, and no calculation is needed to see it.

6. Taking the beat frequency for a sum or an average

3 points, and the principle of tuning missed

What not to write

“Two tuning forks of $440$ and $444$ Hz: we hear a beat at $884$ Hz.”

What to write

“We hear an AVERAGE pitch of $442$ Hz whose intensity pulses $4$ times per second: $f_{beat}=|f_{1}-f_{2}|=4$ Hz.”

The intensity pulses four times per second,

|444-440|=4

Hz, around a pitch of

442

Hz (green); no note appears at

884

Hz (red).

Why: Beats are a slow variation of the AMPLITUDE, not a new note. It is precisely because this pulsation gets slower and slower that an instrument is tuned until it disappears.

7. Spacing the nodes one wavelength apart

the whole question, and a factor of 2 on the measured wavelength

What not to write

“The nodes are $\lambda$ apart, like two successive crests.”

What to write

“The nodes of a standing wave are $\dfrac{\lambda}{2}$ apart, and so are the antinodes. The distance from a node to the neighbouring antinode is $\dfrac{\lambda}{4}$ .”

Two neighbouring nodes are

\dfrac{\lambda}{2}

apart (green), a node and the neighbouring antinode

\dfrac{\lambda}{4}

; the red arrow, one wavelength long, skips over the circled node.

Why: $\sin(kx)$ vanishes twice per spatial period, not once. This is also what makes it possible to measure a wavelength by locating two successive nodes, then doubling the gap.

8. Believing the supersonic boom happens only once

3 points on the discussion question

What not to write

“The plane crosses the sound barrier, there is a boom, and then it is over.”

What to write

“The shock cone, of angle $\sin\theta=\dfrac{1}{M}$ , ACCOMPANIES the plane permanently and sweeps the ground all along the supersonic flight. Each observer hears the boom at the moment the cone passes over them.”

The cone follows the plane (grey at

t_{1}

, navy at

t_{2}

) and its trace sweeps the ground: everyone hears the boom in turn (green). The single boom at the passage of Mach 1 (red) does not exist.

Why: The sound barrier is not a barrier crossed once: it is a regime, the one in which the plane goes faster than its own waves, which then pile up on a conical surface.

Which method to choose

Which pipe, which series of harmonics

The state of BOTH ends, before any formula

For

0.500

m: the open pipe sounds at

343

686

and

1029

Hz, every rank; the closed pipe, with a node at the closed end, at

171.5

514.5

and

857.5

Hz, odd ranks only.

If both ends are open, or the string is fixed at both ends → $f_{n}=n\dfrac{v}{2L}$ , all the integers

Example: $0.500$ m open: $343$ , $686$ , $1029$ Hz
If one end closed, one end open → $f_{n}=n\dfrac{v}{4L}$ with $n$ odd only

Example: $0.500$ m closed: $171.5$ , $514.5$ , $857.5$ Hz
If you are asked for the wavelength of the fundamental → read it off the drawing: $2L$ if the two ends are alike, $4L$ if they differ

Example: only one quarter wavelength fits in a closed pipe
If you are given two CONSECUTIVE natural frequencies → their difference is $\dfrac{v}{2L}$ for an open pipe, and $\dfrac{v}{2L}$ for a closed one too, whose odd harmonics are spaced $2f_{1}$ apart

Example: lines $343$ Hz apart for a $0.500$ m closed pipe
If the temperature changes → recalculate $v=331+0.6\,T$ : all the natural frequencies change, the length of the pipe does not

Example: from $20$ to $30\,^{\circ}$ C, $f_{1}$ goes from $171.5$ to $174.5$ Hz
If you are asked for the position of the nodes → nodes every $\dfrac{\lambda}{2}$ starting from a closed end, antinodes midway

Example: a node, then an antinode a quarter wavelength further on

A ten-second drawing, two lines for the pipe and a sine curve placed according to the ends, answers half of these branches without any formula. It is the most rewarding move in the chapter.

Doppler effect: who moves, and in which direction

What is moving relative to the AIR

30

m/s for

500

Hz: any closing in rises above

500

Hz (blue), any moving apart falls below (orange); solid dots, the source moves; open circles, the observer moves.

If the SOURCE approaches the stationary observer → $f'=f\dfrac{v}{v-v_{s}}$ , higher frequency

Example: a siren at $30$ m/s takes $500$ Hz to $548$ Hz
If the SOURCE moves away → $f'=f\dfrac{v}{v+v_{s}}$ , lower frequency

Example: this is what you hear after it has passed
If the OBSERVER approaches the stationary source → $f'=f\dfrac{v+v_{o}}{v}$ , higher, but less than in the first case

Example: $500$ Hz becomes $544$ Hz at $30$ m/s
If the OBSERVER moves away → $f'=f\dfrac{v-v_{o}}{v}$

Example: at $v_{o}=v$ , you hear nothing at all
If BOTH are moving → combine the two effects in a single fraction, numerator for the observer, denominator for the source

Example: $f'=f\dfrac{v+v_{o}}{v-v_{s}}$ if they are closing in on each other
If the source exceeds the speed of sound → no more Doppler effect but a shock wave, $\sin\theta=\dfrac{1}{M}$

Example: at Mach 2, a cone of half-angle $30^{\circ}$

The systematic check is qualitative: any closing in raises the frequency, any moving apart lowers it. Setting up the fraction, calculating, then checking that the direction obtained is that one takes five seconds and catches every sign error.

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.

A decibel calculation, written up for the marker

When to use it: The question gives a level and asks for another level, at another distance or with several sources.

1 Go back to INTENSITY on the very first line: $I=I_{0}\times 10^{\beta/10}$ , in W/m $^{2}$ , because it is the only quantity that adds and follows the geometry.
2 Do all the physics on the intensities: adding the sources, dividing by the square of the ratio of the distances.
3 Return to the level at the very end: $\beta=10\log\dfrac{I}{I_{0}}$ .
4 Check with the two rules of thumb, $+3$ dB for a doubled intensity and $-6$ dB for a doubled distance.

Concluding sentence

“At $1.00$ m, $\beta=80.0$ dB gives $I=10^{-4}$ W/m $^{2}$ . At $4.00$ m, the intensity is divided by $16$ , that is $6.25\times 10^{-6}$ W/m $^{2}$ , hence $\beta=68.0$ dB. This agrees with two doublings of the distance, so $-12$ dB.”

The trap: Manipulating the decibels directly, dividing them by four or adding them. No arithmetic operation makes sense on a logarithmic scale, except adding or subtracting a fixed number of decibels.

Marking: Typically 2 marks for converting to intensity, 2 for the physics, 1 for returning to decibels, 1 for the check.

Finding the natural frequencies of a pipe

When to use it: The question gives a length, a temperature or a speed of sound, and asks for the fundamental or the harmonics.

1 Calculate the speed of sound if the temperature is given: $v\approx 331+0.6\,T$ , before anything else.
2 Draw the pipe and set the conditions at the ends: a NODE at each closed end, an ANTINODE at each open end.
3 Read the wavelength of the fundamental off the drawing, $2L$ or $4L$ , and deduce $f_{1}=\dfrac{v}{\lambda_{1}}$ .
4 List the allowed harmonics, all the integers or the odd ones only, and JUSTIFY them by the conditions at the ends.
5 Conclude by naming the frequencies with their rank, and not only their value.

Concluding sentence

“Since the pipe is closed at one end, it has a node at the closed end and an antinode at the open end: it therefore fits a quarter wavelength, $\lambda_{1}=4L=2.00$ m and $f_{1}=171.5$ Hz. Only the odd harmonics exist: $f_{3}=514.5$ Hz and $f_{5}=857.5$ Hz.”

The trap: Calling the second frequency of a closed pipe the “second harmonic”. It corresponds to $n=3$ : it is called the third harmonic, and this is the numbering the marker expects.

Marking: 1 mark for the speed, 2 for the conditions at the ends, 2 for the fundamental, 2 for the justified series of harmonics.

Check before you hand in

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

No operation on decibels

Reread the paper and check that no decibel has been added, multiplied or divided. Only fixed differences, $+3$ or $-6$ , can be added to a level.

A $60+60=120$ dB: two living-room speakers at the level of a jet engine.
The direction of the Doppler shift compared with everyday experience

Check that closing in gives a higher frequency and moving apart a lower one. This check needs no calculation and catches every sign error.

A siren that approaches and is heard lower: the denominator was increased instead of decreased.
The parity of the harmonics

On a closed pipe, check that all the ranks announced are odd and that the ratio of two successive lines is $\dfrac{3}{1}$ , then $\dfrac{5}{3}$ .

A line announced at $343$ Hz for a closed pipe whose fundamental is $171.5$ Hz: that is rank $2$ , it does not exist.
The order of magnitude of an audible frequency

The ear hears from $20$ Hz to $20$ kHz. An organ pipe sounds between a few tens and a few thousand hertz: a result outside this range signals an error in length or in units.

A fundamental at $0.17$ Hz: the length was taken in metres instead of centimetres, or the reverse.
The wavelength compared with the length of the pipe

Check that $\lambda_{1}$ is indeed two or four times the length of the pipe. A wavelength shorter than the pipe cannot be that of the fundamental.

A $\lambda_{1}=0.25$ m for a $0.500$ m pipe: that is a high harmonic, not the fundamental.

The typical problem, taken apart

The organ pipe, from the harmonic to the beat

An organ pipe of length $L=0.500$ m is closed at one end and open at the other. The room temperature is $20\,^{\circ}$ C, where the speed of sound is $343$ m/s.

Determine its fundamental and its next two harmonics, then the frequency of the beats heard with a tuning fork of $168.0$ Hz. Finally, say what happens to this beat when the room warms up to $30\,^{\circ}$ C.

The situation: a

0.500

m pipe closed at one end and open at the other, a room at

20

°C and a tuning fork of

168.0

Hz; nothing has been placed in the pipe yet.

Step 1

Conditions at the ends: displacement node at the CLOSED end, antinode at the OPEN end. The pipe therefore fits a quarter wavelength: $L=\dfrac{\lambda_{1}}{4}$ , so $\lambda_{1}=4L=2.00$ m.

Why

The drawing gives the answer, not the formula: between a node and the nearest antinode, there is exactly a quarter wavelength. Writing this sentence earns the justification marks that the formula alone does not.

Step 2

Fundamental: $f_{1}=\dfrac{v}{\lambda_{1}}=\dfrac{343}{2.00}=171.5$ Hz.

Why

Check the order of magnitude immediately: a few tens to a few hundred hertz for a half-metre pipe, so this is indeed an audible low note.

Step 3

Allowed harmonics: only the ODD ranks, so $f_{3}=3f_{1}=514.5$ Hz and $f_{5}=5f_{1}=857.5$ Hz. Rank $2$ , at $343$ Hz, does not exist.

Why

An even rank would put an antinode at the closed end, which is geometrically impossible. It is also what gives the closed pipe its particular timbre, that of the clarinet.

Step 4

Comparison with the open pipe of the same length: $f_{1}^{open}=\dfrac{v}{2L}=343$ Hz, exactly DOUBLE. The closed pipe sounds an octave lower.

Why

This ratio of $2$ is the quickest check in the chapter: if the closed pipe does not sound an octave below the open pipe of the same length, one of the two formulas has been misapplied.

Step 5

Beats with the tuning fork: $f_{beat}=|171.5-168.0|=3.5$ Hz, that is three and a half beats per second, around an average pitch of $169.75$ Hz.

Why

The difference, not the sum: it is the pulsation of the intensity that the ear perceives. A tuner listens precisely for this pulsation and adjusts until it disappears.

Step 6

At $30\,^{\circ}$ C: $v=331+0.6(30)=349$ m/s, so $f_{1}=\dfrac{349}{2.00}=174.5$ Hz, and the beat becomes $|174.5-168.0|=6.5$ Hz.

Why

The length of the pipe has not changed, but the speed of sound has: it carries the whole variation. That is why an organ goes out of tune when the hall fills up and warms up.

Step 7

Check: the frequency has increased by $1.75\,\%$ , exactly like the speed of sound, which went from $343$ to $349$ m/s.

Why

$f_{1}=\dfrac{v}{4L}$ is proportional to $v$ : the two relative changes must be identical. It is a check independent of the beat calculation.

The conclusion, written out

“The closed pipe fits a quarter wavelength, so $\lambda_{1}=2.00$ m and $f_{1}=171.5$ Hz, followed by the odd harmonics only, $514.5$ Hz and $857.5$ Hz. With a tuning fork of $168.0$ Hz, we hear $3.5$ beats per second. At $30\,^{\circ}$ C, the speed of sound rises to $349$ m/s, the fundamental to $174.5$ Hz and the beats to $6.5$ per second.”

The classic mistake on this problem: Treating the pipe as open and announcing $343$ Hz. The result is exactly an octave too high, and the beats with the tuning fork become absurd, $175$ per second, which no ear perceives as a beat.

Learn by heart

• $v\approx 331+0.6\,T$ , that is $343$ m/s at $20\,^{\circ}$ C, and always $\lambda=\dfrac{v}{f}$ .
• $\beta=10\log\dfrac{I}{I_{0}}$ with $I_{0}=10^{-12}$ W/m $^{2}$ : decibels do not add, intensities do.
• Doubling the intensity: $+3$ dB. Doubling the distance: $-6$ dB.
• Nodes spaced $\dfrac{\lambda}{2}$ apart, node to neighbouring antinode $\dfrac{\lambda}{4}$ .
• Two alike ends: $f_{n}=n\dfrac{v}{2L}$ , every rank. One closed end: $f_{n}=n\dfrac{v}{4L}$ , ODD ranks.
• A closed pipe sounds an OCTAVE lower than the open pipe of the same length.
• Doppler, moving source: $f'=f\dfrac{v}{v\mp v_{s}}$ ; moving observer: $f'=f\dfrac{v\pm v_{o}}{v}$ . Approach: higher pitch.
• Beats: $f_{beat}=|f_{1}-f_{2}|$ , around the average pitch.
• Shock wave: $\sin\theta=\dfrac{1}{M}$ , and the cone sweeps the ground permanently.

Frequently asked questions

How do you add two sound levels in decibels?

You never add them directly. You have to go back to the intensities, add those, then convert back to decibels. Two sources of sixty decibels therefore give sixty-three decibels, not one hundred and twenty: doubling the intensity adds only three decibels, because the scale is logarithmic.

By how much does the sound level drop when the distance doubles?

By six decibels. The intensity falls off as the square of the distance, so it is divided by four, which corresponds to ten times the logarithm of four, about six decibels. The level is never divided by two, it simply loses a fixed amount at each doubling.

Why does a closed pipe have only odd harmonics?

Because a closed end imposes a displacement node and an open end an antinode. Between a node and an antinode, there can only be a quarter of a wavelength, then three quarters, then five quarters. The even ranks would put an antinode on the closed end, which is impossible.

Why does a closed pipe sound lower than an open pipe of the same length?

Because it fits only a quarter of a wavelength whereas the open pipe fits a half. Its fundamental wavelength is therefore twice as long, and its frequency twice as low, an octave below. This is why the lowest organ pipes can be stopped to save space.

What are beats and at what frequency do they occur?

They are slow variations in intensity, produced by two sounds of nearby frequencies. Their frequency is the difference of the two frequencies, and the pitch heard is their average. Two tuning forks of four hundred forty and four hundred forty-four hertz give four beats per second.

Is the Doppler effect the same if the source or the observer moves?

No, the two formulas are different. When the source moves, it really does compress the wavefronts ahead of it. When the observer moves, the wavefronts are unchanged but more of them are crossed. The air is the reference, which makes the two situations physically distinct, unlike the case of light.

Practise it

Corrected exercises: Sound and standing waves, 203-SN3-RE

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

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Simple harmonic motion and mechanical waves Next sheet The Doppler effect

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys3-sound-standing-waves. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: sound and standing waves (203-SN3-RE)

The essentials

Intensity and decibels: a scale that does not add

Standing waves: everything is decided at the ends

Vibrating string: the natural modes

Doppler, beats and shock waves

Two sources in phase: the path difference decides

The mistakes that cost marks

1. Adding decibels

2. Dividing the sound level when the distance doubles

3. Treating a closed pipe like an open pipe

4. Applying the moving-source formula to a moving observer

5. Getting the sign wrong in the Doppler effect

6. Taking the beat frequency for a sum or an average

7. Spacing the nodes one wavelength apart

8. Believing the supersonic boom happens only once

Which method to choose

Which pipe, which series of harmonics

Doppler effect: who moves, and in which direction

How the answer is expected to be written

A decibel calculation, written up for the marker

Finding the natural frequencies of a pipe

Check before you hand in

The typical problem, taken apart

The organ pipe, from the harmonic to the beat

Learn by heart

Frequently asked questions

How do you add two sound levels in decibels?

By how much does the sound level drop when the distance doubles?

Why does a closed pipe have only odd harmonics?

Why does a closed pipe sound lower than an open pipe of the same length?

What are beats and at what frequency do they occur?

Is the Doppler effect the same if the source or the observer moves?

Corrected exercises: Sound and standing waves, 203-SN3-RE

See also

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