Questions: Acoustic Pressure and Amplitude in Sound Waves

Question 1

A sound engineer doubles the pressure amplitude of a speaker output. What happens to the acoustic intensity?

• A. It doubles — intensity is proportional to pressure amplitude
• B. It quadruples — intensity is proportional to the square of pressure amplitude
• C. It increases by √2 — intensity scales with the RMS pressure
• D. It stays the same — intensity depends on frequency, not amplitude

Answer: (B)

Acoustic intensity is proportional to the square of pressure amplitude: I ∝ P_amplitude². Doubling the amplitude multiplies the intensity by 2² = 4. This squared relationship is the same as for all wave types — the 'double amplitude, quadruple intensity' rule — and is precisely why the decibel scale is logarithmic: the ear handles an enormous range of intensities.

Question 2

Acoustic impedance (ρv_sound) links pressure amplitude to particle velocity amplitude. What does a high acoustic impedance mean physically?

• A. Sound travels faster in that medium, so particles vibrate at higher frequency
• B. A large pressure swing is required to drive a given particle velocity in that medium
• C. The medium absorbs more sound energy, reducing amplitude over distance
• D. The medium has lower density, so particles respond more strongly to pressure

Answer: (B)

The relation P_amplitude = ρ × v_sound × v_p shows that high acoustic impedance (large ρv_sound) means more pressure is needed to achieve the same particle velocity — analogous to high electrical resistance requiring more voltage to drive the same current. This is why sound transmits poorly from air (low impedance) into water (high impedance): the mismatch at the boundary causes most energy to be reflected.

Question 3

Acoustic pressure at a point in a medium equals the total air pressure at that point.

• T. True
• F. False

Answer: False

Acoustic pressure is the *deviation* from the undisturbed ambient pressure — not the total pressure. As a wave passes, the local pressure oscillates above and below atmospheric pressure (~101,325 Pa), and the acoustic pressure captures only this oscillating departure. A loud sound might have a pressure amplitude of a few pascals; a whisper around 0.02 Pa. These are tiny fractions of total air pressure, which is why sound barely disturbs the medium even at high volumes.

Question 4

Doubling the pressure amplitude of a sound wave quadruples its acoustic intensity.

• T. True
• F. False

Answer: True

True. The quadratic relationship I ∝ P_amplitude² is fundamental and applies to all wave types. This squared proportionality arises because intensity measures power per unit area — power goes as the square of the driving amplitude. The same relationship exists for water waves, electromagnetic waves, and strings: to transfer four times the energy, you need twice the amplitude.

Question 5

Why can the human ear detect pressure amplitudes as small as 0.00002 Pa, even though atmospheric pressure is roughly 101,325 Pa?

Model answer: The ear responds to changes in pressure (acoustic pressure), not to the absolute ambient pressure. Even though the ambient pressure is huge, the oscillating departures from that baseline — the acoustic pressure — are what the ear's hair cells detect. The auditory system is exquisitely tuned to measure differential pressure across the eardrum, which amplifies tiny vibrations mechanically before they reach the cochlea.

This is a conceptual shift from thinking about total pressure to pressure variation. The ear evolved to detect predators and communication signals, which involve tiny, rapid pressure fluctuations. The mechanism (tympanic membrane deflection → ossicle amplification → cochlear fluid motion → hair cell bending) extracts enormous sensitivity from very small pressure differences, not from the ambient pressure itself.