MCAT content library / Chemistry & physics / Sound intensity: distinguish level, frequency, and attenuation
A quieter transmitted wave can retain the same frequency. Use energy flow, source timing, and propagation conditions to identify what actually changed.
About 18 minutes, including the worked examples and checks.
Useful foundations
On this page
- Describe the wave before describing how it is heard
- Intensity is a physical rate; decibels describe a ratio
- Worked application: distance and barriers are separate comparisons
- Distinguish spreading from loss along a path
- The medium controls propagation through its mechanical response
- Check your reasoning
- Continue learning
- Sources
Describe the wave before describing how it is heard
A microphone can detect a change in pressure amplitude without a change in the repetition rate of the pressure oscillation. Those changes answer different questions. Frequency helps determine perceived pitch; intensity describes energy transfer per area and is related to perceived loudness. Neither a decibel reading nor a frequency alone completely describes a listener’s experience. AAMC category 4D includes these sound properties, propagation, and attenuation.1
In a fluid, a vibrating source produces alternating compressions and rarefactions. The disturbance propagates while fluid elements oscillate locally along the direction of travel. Sound requires matter to support those disturbances. The wave’s propagation speed is not the speed at which one molecule travels from speaker to microphone.2
Intensity is a physical rate; decibels describe a ratio
Sound intensity I is the time-averaged power crossing an area perpendicular to propagation, divided by that area. Its unit is W/m². Sound intensity level is β =10log10(I/I0)dB. Here use the conventional reference I0 =10−12W/m². The logarithm receives a dimensionless ratio, not a quantity carrying units.3
A level of 0 dB on this scale means I = I0, not zero intensity. A negative level means a smaller positive intensity than the reference. This reference is not an individualized hearing threshold and does not turn the calculation into a hearing test.
For a comparison made with the same reference, subtracting levels gives Δβ =10log10(I2/I1). A tenfold increase adds 10 dB; a hundredfold increase adds 20 dB. Doubling intensity adds approximately 3.0 dB. These increments add because the underlying intensity ratios multiply.
In the same medium for a progressive sound wave, intensity is proportional to squared pressure amplitude. Doubling pressure amplitude therefore quadruples intensity and adds approximately 6.0 dB, rather than 3.0 dB. This is why a pressure-amplitude ratio uses 20log10, while an intensity ratio uses 10log10. Do not apply the same pressure-to-intensity conversion across unlike media without accounting for their properties.3
Worked application: distance and barriers are separate comparisons
The following microphone measurements are original illustrative data. A steady 500 Hz source radiates approximately uniformly into an open region. The no-barrier model neglects absorption, reflections, and background sound, and the listed distances are far enough from the small source for spherical spreading to apply. All measurements use the same calibrated intensity measure.
| Condition | Distance (m) | Intensity (W/m²) | Level (dB) |
|---|---|---|---|
| Open path | 1.0 | 1.00×10−6 | 60.0 |
| Open path | 2.0 | 2.50×10−7 | 54.0 |
| Open path | 4.0 | 6.25×10−8 | 48.0 |
| Barrier X in path | 2.0 | 2.50×10−8 | 44.0 |
| Barrier Y in path | 2.0 | 2.50×10−9 | 34.0 |
With a constant emitted power and spherical spreading, the area of a wavefront is 4πr². Thus I is proportional to 1/r². Doubling distance from 1 to 2 m spreads the same power over four times the area and reduces intensity to one quarter. The level change is 10log10(1/4) ≈−6.0 dB. Doubling distance again produces another approximately 6.0 dB decrease.3
The exact 2 m level is 10log10(2.50×105) ≈53.98 dB, rounded to 54.0. Multiplying that dB number by one quarter would not describe the next distance. Apply the distance relation to intensity first, or add the corresponding logarithmic level change.
Compare each barrier with the no-barrier reading at the same2 m distance. X leaves one tenth of that intensity, a 10 dB decrease. Y leaves one hundredth, a 20 dB decrease. Y’s transmitted intensity is one tenth of X’s; its lower reading does not establish that Y absorbs ten times as much energy. Reflection or redirection can also reduce the energy reaching this microphone.
Distinguish spreading from loss along a path
Geometric spreading reduces local intensity while the ideal total power through a surrounding sphere remains constant. Material attenuation can instead reduce the directed wave as it travels: absorption transfers acoustic energy into internal energy, while scattering redistributes energy into other directions. A transmission measurement by itself does not separate these mechanisms.4
A repeatable barrier experiment holds source output, frequency, microphone location, orientation, and surroundings fixed while changing the barrier. Measuring at 1 m without a barrier and 2 m with a barrier would mix barrier effects with spreading. A measured reduction also needs comparison with the background level if the received signal approaches that background.
For a monochromatic source passing through a stationary linear barrier, a lower transmitted amplitude need not change frequency. The waveform can still repeat 500 times per second while carrying less energy. For a sound containing several frequencies, unequal attenuation across components can change its spectral balance, so a description of the source and measurement still matters.
The medium controls propagation through its mechanical response
Sound speed depends on elastic response and inertia. For a simple fluid, v =√(B/ρ), where B is bulk modulus and ρ is density. Greater stiffness raises speed if density is held fixed; greater density lowers it if stiffness is held fixed. Comparing real substances by density alone ignores changes in stiffness. Sound is commonly faster in liquids and solids than in gases, but that is not an unconditional ranking of every material.5
For an illustrative 500 Hz wave, a specified speed 340 m/s gives wavelength 0.68 m. If it transmits into a stationary medium with speed 1500 m/s while source frequency remains 500 Hz, the wavelength is 3.0 m. The source timing stays the same; the spatial distance occupied by a cycle changes.
Increasing source amplitude in a linear medium does not, by itself, require a new propagation speed or frequency. Increasing frequency does not, by itself, require a larger intensity. Identify the quantity that the measurement supports: travel time informs speed, cycle timing informs frequency, and energy flow informs intensity. Then apply the decibel conversion only to the appropriate ratio.
ACTIVE PRACTICE · SOUND PHYSICS
Separate Geometric Spreading From Barrier Attenuation
The original 500-Hz source produces 1.00 × 10⁻⁶ W/m² at 1.0 m without a barrier: 60.0 dB referenced to 10⁻¹² W/m². Assume ideal far-field spherical spreading without air absorption or reflections. A selected barrier multiplies the local unblocked intensity by a fixed transmission fraction.
Intensity = W/m². Level = dB. Relative to 1 m, no barrier: dB.
Frequency stays 500 Hz in this stationary-source, stationary-barrier model. The barriers’ effects are stipulated transmission fractions, not a separation of absorption from scattering. Decibels describe a logarithm of an intensity ratio and cannot be multiplied as if they were intensities.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
At the same microphone location, source intensity rises from 2.0×10⁻⁷ W/m² to 8.0×10⁻⁷ W/m². What is the approximate increase in sound intensity level?
A. 3 dB
Review this choice. 3 dB corresponds to approximately doubled intensity. The stated intensity increases by a factor of 4.
B. 6 dB
Correct. Δβ=10log₁₀(8/2)=10log₁₀4≈6.0 dB.
C. 12 dB
Review this choice. This applies the 20log rule to an intensity ratio. That factor is for an amplitude ratio whose square sets intensity under the same conditions.
D. 40 dB
Review this choice. Decibels are logarithmic. Multiplying the intensity ratio 4 by 10 omits the logarithm.
The principle to carry forward
Decide whether the ratio is intensity or amplitude before choosing the logarithmic factor.
Check 2
A stationary linear barrier reduces the intensity level of a 500 Hz sinusoidal sound by 20 dB. Measurements before and after the barrier are in the same medium. Each isolates a progressive traveling wave, with reflected waves excluded; the transmitted signal is still sinusoidal at 500 Hz. Which comparison is correct?
A. Intensity halves and pressure amplitude halves.
Review this choice. Halving intensity is approximately −3 dB, not −20 dB. Halving pressure amplitude would quarter intensity.
B. Intensity is one tenth and pressure amplitude is one tenth.
Review this choice. A 20 dB decrease makes intensity one hundredth. A one-tenth pressure amplitude is consistent with that squared relationship.
C. Intensity is one hundredth and pressure amplitude is one tenth.
Correct. 10log(I₂/I₁)=−20 gives I₂/I₁=0.01. In the same medium, the pressure-amplitude ratio is√0.01=0.10; frequency remains the measured 500 Hz.
D. Intensity is one hundredth and frequency is one tenth.
Review this choice. The intensity ratio is correct, but reduced energy flow does not require lower frequency. The transmitted timing is explicitly unchanged.
The principle to carry forward
Attenuation can change amplitude and intensity while leaving a monochromatic wave’s frequency unchanged.
Check 3
Compare two ideal fluid models with the same density. Fluid B has four times fluid A’s bulk modulus. A source drives both at the same frequency. What are the speed and wavelength in B relative to A?
A. Both are one half as large.
Review this choice. Greater stiffness at fixed density raises sound speed. The square-root relationship gives a factor of 2, not 1/2.
B. Both are unchanged.
Review this choice. Equal density alone does not fix speed. The bulk modulus also appears in v=√(B/ρ).
C. Both are twice as large.
Correct. v_B/v_A=√4=2. Since the driving frequency is the same, λ=v/f also doubles.
D. Both are four times as large.
Review this choice. This omits the square root in the speed relation. A fourfold modulus gives a twofold speed.
The principle to carry forward
Separate stiffness and inertia, then use unchanged source frequency to obtain wavelength.
Continue learning
Related lessons
- Waves: separate amplitude, frequency, and speed
- Work, energy, and power: keep the system account consistent
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Sources
1. AAMC: C/P Content Category 4D
2. OpenStax University Physics1: Sound Waves
3. OpenStax University Physics1: Sound Intensity