Connect compression, displacement and sound pressure

Compression compares neighbors

Suppose two neighboring parcels begin a short distance apart. If the right parcel moves less far right than the left parcel, their separation decreases: that region is compressed. If the right parcel moves farther right, their separation increases: the region expands. A parcel’s displacement alone does not tell you which case applies; compare how displacement changes with position.

In the displacement graph, a negative spatial slope indicates compression in this small-amplitude model. A positive slope indicates expansion or rarefaction. At a displacement maximum, the local slope is zero, so the first-order compression is zero there even though the parcel is farthest from its own equilibrium position.

Interpret pressure as a perturbation

The plotted quantity is pressure change Δp relative to ambient pressure. Positive Δp means compression above the ambient value; negative Δp means rarefaction below it. Negative Δp does not mean negative absolute pressure. For example, if an independently specified ambient pressure were 100,000 Pa, a perturbation of −0.5 Pa would give 99,999.5 Pa absolute pressure.

At 200 Hz and cycle fraction zero, the displacement slope at x = 0 is positive and pressure change is approximately −0.512708 Pa. The reference parcel’s displacement is zero at that instant. At half a period the slope has reversed and pressure change is +0.512708 Pa, again at zero displacement. At a quarter period, displacement is −1 micrometre and pressure change is zero.

Thus displacement and pressure extremes occur at different positions and times. Their sinusoidal variation differs by a quarter cycle. A zero in the displacement graph can coincide with a pressure extreme; a displacement extreme can coincide with zero pressure change. Check both plots rather than treating their vertical values as interchangeable.

Connect the quantities without hiding the assumptions

For a rightward traveling plane wave in a uniform fluid under linear acoustics, Δp = ρ c u, where ρ is equilibrium mass density in kg/m³, c is sound speed in m/s and u is the local longitudinal parcel velocity in m/s. In this model ρ = 1.2 kg/m³ and c = 340 m/s. Their product is 408 kg/(m²·s); multiplying by u gives pascals.

At cycle zero and 200 Hz, u is about −0.001256637 m/s. Multiplying by 408 gives −0.512708 Pa. Pressure and parcel velocity have the same sign and are in phase for this single rightward traveling wave. The same simple sign relation must not be applied unchanged to a leftward wave or a standing-wave mixture. It is also not a statement that parcels move at 340 m/s.

Optional: connect pressure to the displacement gradient

Let ξ, read “xi,” be longitudinal displacement in metres. In the small-amplitude approximation, Δp = −ρc² times the spatial derivative of ξ. The factor ρc² has pressure units. For ξ = A sin(kx − omega t), the spatial derivative is Ak cos(kx − omega t), while parcel velocity is −A omega cos(kx − omega t). Since omega/k = c, both expressions give Δp = ρcu. Here k = 2π/λ in radians per metre, omega = 2πf in radians per second, and A = 0.000001 m. The formulas refer to the unexaggerated displacement.

State what stays fixed when frequency changes

The browser keeps displacement amplitude A at 1 micrometre. At higher frequency, each parcel completes the same displacement excursion in less time, increasing its peak speed. Pressure amplitude is therefore ρc(2πf)A in this model: proportional to frequency when ρ, c and A remain fixed.

At 100 Hz the pressure amplitude is about 0.256354 Pa. At 500 Hz it is about 1.281770 Pa, five times larger, while the displacement graph still reaches only ±1 micrometre. The pressure graph retains its ±1.5 Pa scale so this change remains visible. This is not a general rule that a higher-pitched sound must be louder: a different source might hold pressure amplitude fixed, and perception also depends on other factors.

Transfer to a pressure reading

In the same ideal rightward-wave conditions, a local parcel velocity is +0.002 m/s at an instant. Calculate Δp and describe the compression state. Then decide whether this one reading establishes the displacement amplitude.

Check the sound pressure transfer

Δp = 1.2 × 340 × 0.002 = +0.816 Pa, above ambient pressure: a compression. A single instantaneous velocity or pressure reading does not determine the full-cycle amplitude unless the phase or maximum status is also known. Frequency and sufficient time-dependent evidence would be needed to infer displacement amplitude under the stated model.

OpenStax sound waves explains the displacement and pressure descriptions. This investigation does not compute sound intensity, decibels, exposure limits or hearing response; those require additional definitions and assumptions.