3E

Sound waves, beats and the Doppler effect

Sound as a longitudinal pressure wave, pitch and frequency, loudness and intensity, the inverse square law for sound intensity, decibels, beats from two close frequencies, and the Doppler effect for moving sources and observers.

An ambulance races towards you and its siren sounds high and urgent; the moment it passes, the pitch drops. The siren hasn't changed at all. What has changed is how closely the compressions are packed together when they reach your ear. Police speed guns, weather radar at the Bureau of Meteorology and ultrasound scans of blood flow all use the same effect, with light or sound.

Year 9 introduced sound, echoes and the speed of sound. Here we model sound as a pressure wave, connect pitch and loudness to wave properties with numbers, and explain beats and the Doppler effect.

Sound as a longitudinal pressure wave

A vibrating object (a speaker cone, a guitar string, your vocal folds) pushes the air next to it back and forth. Each push squeezes air molecules together into a compression; each pull-back leaves a rarefaction where they are spread out. The molecules oscillate parallel to the direction the sound travels, so sound in air is a longitudinal wave.

Compressions are regions of higher pressure; rarefactions are regions of lower pressure.

Bunched-up molecules mean higher pressure, so a sound wave can also be described as a pressure wave: the pressure rises and falls a tiny amount above and below normal atmospheric pressure. A microphone, and your eardrum, respond to these pressure changes.

One subtle point that HSC questions test: the pressure variation and the displacement of the molecules are a quarter of a wavelength out of step. At the centre of a compression the molecules on each side have moved in towards it, but the molecules at the centre itself are at their rest position (zero displacement) while the pressure is greatest. Where displacement is largest, the pressure is normal.

The speed of sound in dry air is about 331 m/s at 0 °C and increases by about 0.6 m/s for every degree, giving 343 m/s at 20 °C. Many textbooks and exams round this to 340 m/s; use whatever value the question gives.

Pitch and loudness

  • Pitch is how high or low a sound seems. It depends on the frequency: the higher the frequency, the higher the pitch. Young people can hear roughly 20 Hz to 20 000 Hz; the upper limit falls with age. Above 20 kHz is ultrasound; below 20 Hz is infrasound.
  • Loudness is how strong a sound seems. It depends mainly on the amplitude of the pressure variation, which determines the energy the wave carries. Doubling the amplitude carries four times the energy.

On a microphone and oscilloscope trace (a displacement–time graph of the pressure), a higher pitch shows more waves across the screen and a louder sound shows taller waves.

Intensity and the inverse square law

The intensity of a wave is the power it carries per unit area, measured at right angles to the direction of travel:

$I = \dfrac{P}{A}$   (W m$^{-2}$)

A small source sending sound out equally in all directions spreads its power over the surface of an ever-growing sphere of area $4\pi r^2$, so

$I = \dfrac{P}{4\pi r^2} \quad \Rightarrow \quad I \propto \dfrac{1}{r^2} \quad \Rightarrow \quad I_1 r_1^2 = I_2 r_2^2$

At twice the distance the same power covers four times the area, so the intensity is a quarter.

Double the distance and the intensity drops to a quarter; triple it and it drops to a ninth. This assumes a point source, no absorption and no reflections. Indoors, echoes off walls mean the intensity falls more slowly than the law predicts.

Because the ear handles intensities from about $10^{-12}$ W m$^{-2}$ (the threshold of hearing) up to about 1 W m$^{-2}$ (painful), sound levels are given on the logarithmic decibel scale. Each increase of 10 dB is a tenfold increase in intensity; a fall of 6 dB is close to a quarter, which is what you get by doubling your distance from a source.

Worked example: a speaker at a festival

A sound engineer measures an intensity of $2.4 \times 10^{-2}$ W m$^{-2}$ at 5.0 m from a speaker stack. (a) Find the intensity at 20 m. (b) Find the acoustic power of the stack, assuming it radiates equally in all directions.

$I_2 = I_1 \dfrac{r_1^2}{r_2^2} = 2.4 \times 10^{-2} \times \dfrac{5.0^2}{20^2} = 2.4 \times 10^{-2} \times \dfrac{1}{16} = 1.5 \times 10^{-3}$ W m$^{-2}$.
ratio
$P = I \times 4\pi r^2 = 2.4 \times 10^{-2} \times 4\pi(5.0)^2 = 7.5$ W.
power
Four times the distance gives one-sixteenth of the intensity, as the inverse square law requires.
check

Beats

When two sounds of slightly different frequency $f_1$ and $f_2$ play together, their waves drift in and out of step. When compressions coincide they interfere constructively and the sound is loud; a little later a compression of one meets a rarefaction of the other and the sound is soft. The loudness rises and falls regularly. Each rise and fall is a beat, and the number of beats per second is

$f_{\text{beat}} = |f_2 - f_1|$

Adding two waves of slightly different frequency gives a wave whose loudness rises and falls at the beat frequency.

A correction to a common description. Some books say that beats produce "a third frequency that is heard". That is misleading. You do not hear a separate note at $f_2 - f_1$ (4 Hz would be far too low to hear as a pitch anyway). You hear one tone at the average frequency, $\frac{f_1 + f_2}{2}$, whose loudness pulses $|f_2 - f_1|$ times per second. The beat frequency is the frequency of the loudness variation, not of a new sound.

Musicians use beats to tune: play a string together with a reference note and adjust the tension until the beats slow down and disappear. Beats are only heard clearly when the difference is less than about 10 to 15 Hz; above that the ear hears a rough sound or two separate notes.

Worked example: tuning a guitar

A guitarist plays a string with a 440 Hz tuning fork and hears 3 beats per second. Tightening the string slightly makes the beats faster. What was the string's frequency, and what should the guitarist do?

$f_{\text{string}} = 440 \pm 3$, so 437 Hz or 443 Hz.
two possibilities
Tightening raises the frequency. Since the beats got faster, the difference grew, so the string was already above 440 Hz: it was 443 Hz.
use the clue
Loosen the string until the beats disappear.
action

The Doppler effect

The Doppler effect is the change in observed frequency when the source of a wave and the observer move relative to each other. If they are getting closer, the observed frequency is higher; if they are moving apart, it is lower.

Each circle was emitted when the source was at its centre. Ahead of the source the wavefronts are bunched; behind they are spread out.

Moving source. The source emits each compression from a little further forward than the last, so ahead of it the compressions are closer together (shorter wavelength) and behind it they are further apart. The speed of sound in the air is unchanged, so the observer ahead receives more compressions per second: a higher frequency.

Moving observer. The wavelength in the air is unchanged, but an observer moving towards the source meets the compressions faster than they would standing still, so receives more per second.

Both cases combine into one formula, measured with the speeds as positive when the source and observer move towards each other:

$f' = f\,\dfrac{(v_{\text{wave}} + v_{\text{observer}})}{(v_{\text{wave}} - v_{\text{source}})}$

If the observer moves away from the source, $v_{\text{observer}}$ is negative; if the source moves away from the observer, $v_{\text{source}}$ is negative. Check every answer with common sense: approaching means $f' > f$.

Worked example: a passing ambulance

An ambulance siren emits 700 Hz. The ambulance travels at 25 m/s and the speed of sound is 340 m/s. Find the frequency heard by a stationary pedestrian (a) as it approaches and (b) after it has passed.

$f' = 700 \times \dfrac{340 + 0}{340 - 25} = 700 \times \dfrac{340}{315} = 756$ Hz.
approaching
The source moves away, so $v_{\text{source}} = -25$ m/s: $f' = 700 \times \dfrac{340}{340 + 25} = 652$ Hz.
receding
Higher while approaching, lower while receding, and the drop as it passes is about 100 Hz.
check
Worked example: a moving observer

A cyclist rides at 10 m/s towards a stationary car whose horn sounds at 400 Hz. Take $v = 340$ m/s.

$f' = 400 \times \dfrac{340 + 10}{340 - 0} = 400 \times 1.029 = 412$ Hz.
substitute

The Doppler effect applies to all waves. Doppler ultrasound measures the speed of blood in arteries from the frequency shift of sound reflected off moving blood cells. For light, the shift of spectral lines from distant galaxies towards the red end (redshift) showed that the universe is expanding, which you will meet in Year 12.

InteractiveDoppler effect and beats
Try this
  • At the start a 500 Hz source moves at 34 m/s towards the observer: the readouts show 555.6 Hz ahead and 454.5 Hz behind. Check both with the formula.
  • Set the source speed to 0 and the observer speed to 34 m/s. The observer now hears 550.0 Hz: moving observer and moving source at the same speed do not give exactly the same shift.
  • Push the source speed towards 300 m/s and watch the wavefronts pile up in front of it.
  • In Beats mode with forks at 256 Hz and 260 Hz, the readouts show 4.0 beats per second and a heard pitch of 258.0 Hz. Count the loud bulges on the 1-second trace. Then set both forks to the same frequency.
Seven steel tuning forks of different sizes lying side by side
A set of tuning forks. Each vibrates at one precise frequency, stamped on its stem; smaller forks give higher notes. Two forks a few hertz apart, struck together, produce clear beats.Photo: K. Venkataramana, CC0

Test yourself

  1. At 3.0 m from a small source the sound intensity is $8.0 \times 10^{-4}$ W m$^{-2}$. At 6.0 m it is

  2. Two tuning forks of 512 Hz and 508 Hz are sounded together. A listener hears

  3. A train moving at 30 m/s sounds a 600 Hz horn as it approaches a stationary person. Using $v = 340$ m/s, the frequency heard is closest to

  4. In a sound wave, at the centre of a compression

Label the Doppler diagram.

source moving rightahead: wavefronts bunched,shorter λ, higher fbehind: wavefronts spread out, longer λ, lower f

What does the listener hear compared with the source frequency?

The pitch of a sound depends on its , and its loudness depends mainly on its . At three times the distance from a point source the intensity is one of the original. Two notes of 330 Hz and 336 Hz produce beats per second. When a source moves towards a listener, the observed frequency is .

6 marks

A police car with a 900 Hz siren drives at 20 m/s past a pedestrian standing on the footpath. Take the speed of sound as 340 m/s. (a) Calculate the frequency the pedestrian hears as the car approaches and as it moves away. (b) Explain, in terms of wavefronts and wavelength, why the frequency heard is different from 900 Hz. (c) A second, parked police car sounds an identical siren. Explain what the pedestrian hears while the moving car approaches, with a calculation.