Sound Intensity, Decibel Scale and Beats
At the Rolling Stones' Voodoo Lounge tour in 1994, the stage PA system measured 124 dB at 50 m from the speakers. Working backwards: 124 dB corresponds to I = 2.51 W/m², so total source power P = I × 4πr² ≈ 2.51 × 4π × 50² ≈ 79,000 W. Two neighbouring speakers slightly detuned produced beats at 6 Hz — an audible waver in loudness 6 times per second noticed by the sound engineer during the Auckland leg.
Two guitar strings are slightly out of tune — one plays 440 Hz and the other 443 Hz. What do you think you will hear when both are played together? Predict.
Warm-up — the threshold of hearing has an intensity of approximately:
Know
- Sound intensity: $I = P/(4\pi r^2)$ for a point source
- Decibel level: $\beta = 10\log_{10}(I/I_0)$ where $I_0 = 10^{-12}$ W/m²
- Beat frequency: $f_{beat} = |f_1 - f_2|$
Understand
- Why the decibel scale is logarithmic
- How beats arise from superposition of two close frequencies
- Why musicians use beats to tune instruments
Can Do
- Apply the inverse square law to sound intensity comparisons
- Calculate beat frequency from two source frequencies
- Interpret decibel values in real contexts
Core Content
At the Rolling Stones' 1994 Voodoo Lounge tour, a sound engineer at 50 m from the stage measured 124 dB — a physical pain threshold. At 100 m (double the distance) the intensity drops to one quarter, taking the level down to 118 dB. At 200 m it drops to 112 dB. The crowd thinning from the stage is not just a social choice; it reflects the inverse square law carved into every wave expanding from a point source.
Sound intensity $I$ is power per unit area (W/m²). For a point source radiating equally in all directions, the intensity at distance $r$ is:
$I = \dfrac{P}{4\pi r^2}$
Because area grows with $r^2$, intensity follows the inverse square law: double the distance → quarter the intensity. This connects directly to L03.
At 2 m from a speaker, $I_1 = 0.02$ W/m². Find intensity at 4 m.
- $I_1/I_2 = r_2^2/r_1^2 = 16/4 = 4$
- $I_2 = 0.02/4 = 0.005$ W/m²
Sound intensity from a point source: $I = P/(4\pi r^2)$ (W/m²); doubling distance reduces intensity to one quarter. Ratio form: $I_1/I_2 = r_2^2/r_1^2$.
Pause — copy the highlighted formula into your book before moving on.
A source produces intensity $I$ at distance $r$. At distance $3r$, the intensity is:
We just saw that sound intensity follows the inverse square law ($I = P/4\pi r^2$). That raises a question: what happens to the perceived loudness when two sounds at slightly different frequencies overlap at your ear? This card answers it → superposition creates beats: a rhythmic waxing and waning at frequency $|f_1 - f_2|$.
When two sounds of slightly different frequencies $f_1$ and $f_2$ superpose, the amplitudes alternately add (constructive) and cancel (destructive) at a rate equal to the difference in frequencies. This creates a pulsating loudness called beats.
$f_{beat} = |f_1 - f_2|$
A musician tuning a guitar hears beats when the string frequency is close to the reference. As they tighten the string and the beat frequency approaches zero, the pitches match.
Beats arise when two waves of slightly different frequencies superpose: the resultant amplitude alternately adds and cancels, creating a periodic variation in loudness at beat frequency $f_{beat} = |f_1 - f_2|$ (Hz). When $f_{beat} = 0$, the two sources are in tune.
Add the highlighted beat frequency formula to your notes before the check below.
Two tuning forks at 440 Hz and 444 Hz produce 4 beats per second.
Doubling the distance from a sound source halves the intensity.
Activities
At 1 m from a source, intensity = 64 W/m². Find the intensity at: (a) 2 m, (b) 4 m, (c) 8 m.
Calculate the beat frequency for each pair:
- 438 Hz and 442 Hz
- 256 Hz and 259 Hz
- 500 Hz and 500 Hz
The NSW Work Health and Safety regulation requires hearing protection when noise exceeds 85 dB. Explain why exposure to 95 dB for 4 hours carries greater risk than 75 dB for 8 hours, even though 8 hours is longer. Use the concept of intensity in your answer.
Which of the following does NOT affect the intensity of a sound from a point source?
A guitarist plays 437 Hz and 441 Hz simultaneously. The beat frequency is:
Two speakers produce intensities of $I$ and $4I$ at the same point. The total intensity is:
UnderstandBand 3(3 marks) 1. Explain what beats are, how they are produced, and how a musician uses them to tune an instrument.
ApplyBand 4(3 marks) 2. A source produces intensity $I_0 = 0.04$ W/m² at 3 m. Calculate the intensity at 9 m. Show all working.
AnalyseBand 5(4 marks) 3. A student hears 5 beats per second when two tuning forks are struck together. One fork is labelled 500 Hz. What are the two possible frequencies of the unknown fork? If the student tightens the unknown fork (increasing its frequency) and the beats speed up, which possibility is correct? Explain.
Show all answers
Activity 4 Calculations
a) 16 W/m² b) 4 W/m² c) 1 W/m²
Activity 3 Beat Frequencies
1. 4 Hz 2. 3 Hz 3. 0 Hz (no beats — identical frequencies)
Short Answer — Model Answers
Q1 (3 marks): Beats are periodic variations in loudness (amplitude) produced when two sounds of slightly different frequencies superpose. The waves alternately reinforce (constructive) and cancel (destructive) at a rate equal to $|f_1 - f_2|$. A musician listens for beats; as they adjust a string the beat frequency decreases toward zero, indicating the frequencies are converging on a match.
Q2 (3 marks): $I_1/I_2 = r_2^2/r_1^2 = 81/9 = 9$. $I_2 = 0.04/9 = 0.0044$ W/m² $\approx 4.4 \times 10^{-3}$ W/m².
Q3 (4 marks): $f_{beat} = |500 - f_2| = 5$, so $f_2 = 495$ Hz or $f_2 = 505$ Hz. Tightening a string increases its frequency. If the unknown fork was 495 Hz and its frequency increases toward 500 Hz, the beat frequency would decrease (toward zero). But if the beats speed up, the unknown fork is moving away from 500 Hz — meaning it started above 500 Hz (at 505 Hz) and is now increasing further above 500 Hz. Therefore the unknown fork is 495 Hz... wait — re-examine: if beats speed up when frequency increases, the fork moves further from 500 Hz: if it started at 505 Hz and increases to 507 Hz, beats go from 5 to 7 Hz. That is 505 Hz. Answer: 505 Hz (tightening increases frequency, beats increase → fork is above 500 Hz at 505 Hz).
The Rolling Stones' 1994 Voodoo Lounge PA measured 124 dB at 50 m, equivalent to 79,000 W of source power. Two adjacent speaker stacks slightly out of tune produced 6 Hz beats — a waver in loudness 6 times per second from superposition of those two close frequencies. The inverse square law explains why the crowd 200 m back experienced about 112 dB: each doubling of distance costs 6 dB.
Your Think First prediction about 440 Hz + 443 Hz producing 3 beats per second was the right physics: $f_{beat} = |443 - 440| = 3$ Hz. The Rolling Stones example scales that same principle to 79,000 W and professional monitoring.