Waves • Topic 3 of 3

Beats & the Doppler Effect

So far we have superposed waves of the same frequency. What if the two frequencies differ slightly? The answer is one of the most useful effects in acoustics: beats.

Beats are the periodic rise and fall in the loudness of sound heard when two notes of slightly different frequencies are sounded together. At some instants the two waves are in phase and reinforce (loud); a moment later they slip out of phase and cancel (soft). One loud–soft cycle is one beat.

  • Beat frequency $=|f_1-f_2|$ — the number of beats heard per second equals the difference of the two frequencies.
  • Beats are clearly audible only when the difference is small (roughly up to 6–7 Hz); beyond that the ear cannot resolve the separate loud–soft cycles.
  • Beats are the basis of tuning: a musician adjusts an instrument against a reference until the beats slow down and vanish, meaning the frequencies match exactly.

For example, two tuning forks of 256 Hz and 260 Hz sounded together produce $|260-256|=4$ beats per second.

The Doppler effect is the apparent change in the frequency (pitch) of a sound when there is relative motion between the source and the observer. You hear it every day: the pitch of an approaching ambulance siren is higher than its true pitch, and it suddenly drops to a lower pitch as the ambulance passes and recedes.

The general formula for the observed frequency $f'$ when the source emits frequency $f$ in still air of sound speed $v$ is:

  • $f'=f\left(\frac{v\pm v_o}{v\mp v_s}\right)$, where $v_o$ is the observer's speed and $v_s$ is the source's speed.
  • Sign convention: choose the signs so that motion of either body towards the other raises $f'$, and motion away lowers it. Use the upper sign in the numerator ($+v_o$) when the observer moves towards the source; use the upper sign in the denominator ($-v_s$) when the source moves towards the observer.

Some standard cases (sound speed $v$):

  • Source approaching, observer at rest: $f'=f\left(\frac{v}{v-v_s}\right)$ — pitch rises.
  • Source receding, observer at rest: $f'=f\left(\frac{v}{v+v_s}\right)$ — pitch falls.
  • Observer approaching, source at rest: $f'=f\left(\frac{v+v_o}{v}\right)$ — pitch rises.

Note that the Doppler effect for sound is not symmetric in $v_o$ and $v_s$ — a moving source and a moving observer at the same speed give slightly different shifts, because sound travels in a medium (air) which provides a frame of reference. Applications include speed-detecting radar guns used by traffic police, weather radar for storms, medical ultrasound to measure blood flow, and sonar for detecting submarines. Astronomers use the Doppler shift of starlight (red shift) as evidence that the universe is expanding.

Doppler effect: wavefronts bunch ahead of a moving source and spread behindv_sbunched: higher pitchspread: lower pitchobserverf' = f (v ± v_o)/(v ∓ v_s)
1
Worked Example
Two tuning forks of frequencies 480 Hz and 484 Hz are sounded together. Find the beat frequency.
Solution
  1. Step 1: Beat frequency $=|f_1-f_2|$.
  2. Step 2: Substitute: $=|484-480|$.
  3. Step 3: Compute: $=4\ \text{Hz}$ (4 beats per second).

Answer: Beat frequency $=4\ \text{Hz}$.

2
Worked Example
A tuning fork of unknown frequency gives 5 beats per second with a 256 Hz fork. On loading the unknown fork with wax, the beat frequency drops to 3 per second. Find the unknown frequency.
Solution
  1. Step 1: The unknown frequency is either $256+5=261$ Hz or $256-5=251$ Hz.
  2. Step 2: Loading with wax lowers the unknown fork's frequency. If it were 261 Hz, lowering it towards 256 would reduce beats — consistent with the drop to 3.
  3. Step 3: If it were 251 Hz, lowering it further would increase the beats (to more than 5), which contradicts the observation. So the unknown frequency is 261 Hz.

Answer: The unknown frequency is $261\ \text{Hz}$.

3
Worked Example
A source of frequency 500 Hz moves towards a stationary observer at 30 m/s. Find the observed frequency. (Speed of sound $=330\ \text{m/s}$.)
Solution
  1. Step 1: Source approaching, observer at rest: $f'=f\left(\frac{v}{v-v_s}\right)$.
  2. Step 2: Substitute: $f'=500\left(\frac{330}{330-30}\right)=500\left(\frac{330}{300}\right)$.
  3. Step 3: Compute: $f'=500\times1.1=550\ \text{Hz}$.

Answer: $f'=550\ \text{Hz}$ (pitch rises).

4
Worked Example
The same 500 Hz source now moves away from the stationary observer at 30 m/s. Find the observed frequency. (Speed of sound $=330\ \text{m/s}$.)
Solution
  1. Step 1: Source receding, observer at rest: $f'=f\left(\frac{v}{v+v_s}\right)$.
  2. Step 2: Substitute: $f'=500\left(\frac{330}{330+30}\right)=500\left(\frac{330}{360}\right)$.
  3. Step 3: Compute: $f'=500\times0.9167\approx458\ \text{Hz}$.

Answer: $f'\approx458\ \text{Hz}$ (pitch falls).

5
Worked Example
An observer moves towards a stationary source of frequency 400 Hz at 33 m/s. Find the observed frequency. (Speed of sound $=330\ \text{m/s}$.)
Solution
  1. Step 1: Observer approaching, source at rest: $f'=f\left(\frac{v+v_o}{v}\right)$.
  2. Step 2: Substitute: $f'=400\left(\frac{330+33}{330}\right)=400\left(\frac{363}{330}\right)$.
  3. Step 3: Compute: $f'=400\times1.1=440\ \text{Hz}$.

Answer: $f'=440\ \text{Hz}$.

6
Worked Example
A car sounding a horn of 600 Hz approaches a stationary listener at 20 m/s. Find the frequency heard. (Speed of sound $=340\ \text{m/s}$.)
Solution
  1. Step 1: Source approaching, observer at rest: $f'=f\left(\frac{v}{v-v_s}\right)$.
  2. Step 2: Substitute: $f'=600\left(\frac{340}{340-20}\right)=600\left(\frac{340}{320}\right)$.
  3. Step 3: Compute: $f'=600\times1.0625=637.5\ \text{Hz}$.

Answer: $f'\approx637.5\ \text{Hz}$.

Key Points

  • Beats arise from superposing two waves of slightly different frequencies; beat frequency $=|f_1-f_2|$.
  • Beats are audible only for small frequency differences (about 6–7 Hz) and are used to tune instruments to a reference.
  • The Doppler effect is the apparent change in pitch due to relative motion of source and observer: $f'=f\left(\frac{v\pm v_o}{v\mp v_s}\right)$.
  • Motion towards raises the observed frequency; motion away lowers it; the effect is not symmetric in $v_o$ and $v_s$ for sound.
  • Applications: radar speed guns, weather radar, medical ultrasound, sonar, and astronomical red shift.
Tap an option to check your answer0 / 4
Q1.The beat frequency produced by two tuning forks of frequencies $f_1$ and $f_2$ is:
Explanation: The number of beats per second equals the magnitude of the difference of the two frequencies, $|f_1-f_2|$.
Q2.When a source of sound moves towards a stationary observer, the observed frequency:
Explanation: Wavefronts bunch up ahead of the approaching source, so the observed frequency (pitch) rises: $f'=f\frac{v}{v-v_s}$.
Q3.Beats can be heard clearly only when the two frequencies differ by:
Explanation: The ear can resolve separate loud–soft cycles only when the difference is small, roughly up to 6–7 Hz.
Q4.The Doppler effect is used in all of the following EXCEPT:
Explanation: Tuning a guitar by ear uses beats, not the Doppler effect; the other three rely on Doppler frequency shifts.