core29 min studyLesson 194 in Physics

Match sound at a material interface

Derive pressure-amplitude coefficients and check energy conservation at a lossless normal-incidence boundary.

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What you’ll understand

  • Enforce pressure and normal-velocity continuity.
  • Distinguish pressure reflection from reflected energy fraction.
  • Check reflected and transmitted energy coefficients in lossless fluids.

A pressure pulse that meets a new fluid generally produces both a reflected and a transmitted disturbance. The interface does not choose these amplitudes independently: matching conditions connect them. We will solve the simplest problem, normal incidence between two stationary lossless fluids, and use power flux to catch a common mistake: interpreting a transmitted pressure amplitude greater than the incident amplitude as the creation of energy.

Choose a single sign convention

Let the incident wave travel to the right from medium one toward medium two. Define all normal particle velocities using the same rightward coordinate. At a massless interface with no surface tension contribution to the linear normal stress, total pressure perturbation is continuous. Normal particle velocity is also continuous when the interface moves with both fluids. The reflected wave travels left, so its velocity amplitude enters with the opposite sign from its pressure amplitude. This sign is the key algebraic detail.

Denote positive real characteristic impedances by Z₁ and Z₂. At the boundary, incident plus reflected pressure equals transmitted pressure, while incident minus reflected pressure divided by Z₁ equals transmitted pressure divided by Z₂. Solving these two linear equations gives a pressure reflection coefficient r and a pressure transmission coefficient t. They refer to amplitudes at the interface and assume planar waves, normal incidence, no dissipation, and no additional interface dynamics or mode conversion.

r=PrPi=Z2−Z1Z2+Z1,t=PtPi=2Z2Z1+Z2=1+rr=\frac{P_r}{P_i}=\frac{Z_2-Z_1}{Z_2+Z_1},\qquad t=\frac{P_t}{P_i}=\frac{2Z_2}{Z_1+Z_2}=1+r
The pressure coefficients use one incident wave, one reflected wave, and one transmitted wave at a stationary equilibrium interface between lossless fluids.↔ Scroll wide equations horizontally.

Pressure amplitude is not energy fraction

Each progressive wave carries mean intensity proportional to squared pressure amplitude divided by its medium’s impedance. The reflected energy fraction is therefore r², but the transmitted fraction is t²Z₁/Z₂. The impedance ratio cannot be omitted. A high-impedance second medium can support a larger transmitted pressure with a smaller transmitted velocity. Because energy flux is their product, a large pressure alone does not indicate a large transported power or a violation of conservation.

Insert the amplitude expressions and simplify. Reflection plus transmission equals one for this lossless ideal interface. This check catches reversed impedance ratios, forgotten velocity signs, and the mistaken rule that transmission must equal one minus the pressure reflection coefficient. When the second impedance is much smaller, the reflected pressure changes sign. When the impedances match, reflection disappears even if density and sound speed individually differ. Matching their product is what this normal-incidence model requires.

R=r2,T=Z1Z2t2=4Z1Z2(Z1+Z2)2,R+T=1R=r^2,\qquad T=\frac{Z_1}{Z_2}t^2=\frac{4Z_1Z_2}{(Z_1+Z_2)^2},\qquad R+T=1
R and T are nonnegative fractions of incident normal energy flux. The equality assumes no absorption, interface energy accumulation on average, or additional outgoing channels.↔ Scroll wide equations horizontally.

Know which idealization you are testing

The rigid-wall limit is approached as the second impedance becomes arbitrarily large: reflected pressure nearly equals incident pressure, while total normal velocity at the boundary tends to zero. The pressure-release limit gives a reflected pressure of nearly opposite sign. Real boundaries can have frequency-dependent complex impedance, damping, compliance, or mass. Oblique incidence also changes the matching problem and may involve refraction. The simple two-real-impedance formulas should not be reused without checking those changes in assumptions.

Use the waves lab only to visualize addition of incident-size and reflected-size pressure signals at a chosen boundary point. Predict the sum when the amplitude ratio is one and phase is either zero or 180°. Its components both propagate in the same direction, so the spatial display does not represent an incident/reflected pair. It cannot compute interface motion or transmitted flux. The useful connection is the algebra of adding boundary pressures; the physical matching and energy accounting remain calculations in this lesson.

Pi+Pr=Pt,Pi−PrZ1=PtZ2P_i+P_r=P_t,\qquad \frac{P_i-P_r}{Z_1}=\frac{P_t}{Z_2}
Using one global positive normal direction produces the minus sign for the left-moving reflected wave’s particle velocity.↔ Scroll wide equations horizontally.

A transmitted pressure larger than one

Take Z₂=3Z₁ and incident peak pressure 2 Pa. Find reflected and transmitted pressure amplitudes and the fractions of incident energy reflected and transmitted.

  1. The pressure reflection coefficient is (3−1)/(3+1)=1/2. Reflected peak pressure is therefore 1 Pa and has no pressure phase reversal.

  2. Pressure transmission is 1+r=3/2, so transmitted peak pressure is 3 Pa. This exceeds incident pressure, but medium two has three times the impedance.

  3. Reflected energy fraction is (1/2)²=1/4. Transmitted fraction is (Z₁/Z₂)(3/2)²=(1/3)(9/4)=3/4, and the two fractions add to one.

The result

Pressures are 1 Pa reflected and 3 Pa transmitted; energy fractions are 25% and 75%. No energy is created.

Explore the model

Interactive lab

Interactive lab · waves

Add pressure signals at the boundary

Vary amplitude ratio and phase difference to compare boundary-pressure addition for same-sign and reversed-sign contributions. Predict equal-amplitude addition at 0° and cancellation at 180°. Both displayed waves travel the same way.

Add pressure signals at the boundaryTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab illustrates a local superposition rule, not reflection geometry, impedance matching, or transmission power.

Check your understanding

With Z₂=3Z₁, the transmitted pressure coefficient is 1.5. What is the transmitted energy fraction?

Try it yourself

Now take Z₂=Z₁/3. Find r, t, R, and T. Explain which feature changes relative to the worked example and which energy fractions stay the same.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A lower second impedance reverses the sign of r, but energy reflection uses its square.

Compare with the solution

The pressure coefficients are r=−1/2 and t=1/2. Reflection is 1/4 and transmission is (Z₁/Z₂)t²=3/4. Reflected pressure now has a π phase reversal, and transmitted pressure is smaller. The energy fractions remain 25% and 75%, as the symmetric impedance expression predicts.

Take it with you

  • Boundary conditions determine amplitudes before power fractions are computed.
  • A left-moving wave reverses the pressure-to-signed-velocity relation.
  • Lossless energy transmission includes the ratio of characteristic impedances.

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