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Lesson 4 of 8

Waves · Lesson 4 of 8

Superposition, Interference and Reflection

“Add overlapping waves, derive phase-dependent interference, and apply fixed and free boundary conditions with consistent reflection equations.”

Learning Objectives

• Use signed displacement addition to explain pulse overlap and recovery. • Derive the resultant of equal-amplitude waves with a fixed phase difference. • Distinguish constructive and destructive interference without treating cancellation as destruction of energy. • Construct reflected waves that reverse direction and satisfy fixed or free boundary conditions.

What happens when two pulses meet?

Send an upward pulse and a downward pulse towards each other along a rope. Before meeting, each pulse has its own shape. While they overlap, there is only one rope, so a particular element has one net displacement. In the small-amplitude linear model, calculate it by adding the two displacements with their signs at that same position and instant. After overlap, the pulses emerge and continue in their original directions. A temporary complicated shape is not evidence that the pulses permanently changed each other.

Opposite pulses overlap and recover12345Dashed: individual contributions. Dark: their signed sum.At stage 3 displacement cancels; the pulses continue through.
Pulse overlap— Read downward through time. The sum is straight at full overlap but the contributions continue.
Principle of superpositionLaTeX
Apply displacement addition at the same x and t. It assumes a linear medium and sufficiently small disturbances.

If two simultaneous contributions are +3 mm and −2 mm, the net displacement is +1 mm, not 5 mm. Equal opposite pulses can momentarily give zero displacement throughout their overlap. Zero displacement at an instant does not imply zero particle velocity, and it does not establish that all mechanical energy vanished. The moving material may still carry kinetic energy. The waves can reappear after overlap because the instantaneous displacement alone is not the entire dynamical state.

For several right-moving shapes with the same speed, y = F₁(x − vt) + F₂(x − vt) + … is another translating shape. Superposition is more general: opposite propagation directions and different frequencies may also be added, but their sum need not be one rigidly translating wave. At large amplitudes the restoring force can become nonlinear, and this simple addition rule may fail.

Interference of two matching harmonic waves

Now take two continuous harmonic waves travelling in the same direction with equal amplitude a, equal k and equal ω, but with a fixed phase difference φ. Set θ = kx − ωt. Then y₁ = a sin θ and y₂ = a sin(θ + φ). Their maintained phase difference is what makes the interference stable. This comparison is different from pulses that overlap only briefly.

The sum-to-product identityLaTeX
Choose U = θ and V = θ + φ. The average angle is θ + φ/2; the half-difference contributes cos(φ/2), since cosine is even.
Resultant travelling waveLaTeX
The amplitude is the nonnegative magnitude A = 2a|cos(φ/2)|. If the displayed prefactor is negative, absorb its sign into a phase shift of π.

The frequency, wavelength and direction are unchanged because the sine still contains the same kx − ωt. The amplitude changes with the relative phase. At φ = 0, both waves raise or lower an element together and the resultant amplitude is 2a: constructive interference. More generally φ = 2mπ produces the same result. At φ = (2m + 1)π, one displacement is always opposite to the other and the resultant is zero: complete destructive interference for these equal waves.

Fixed phase difference controls reinforcementIn phase: A = 2aPhase difference 2π/3: A = aOpposite phase: A = 0At one instant: light curves are contributions; dark curve is the sum.
Harmonic interference— Spatial snapshots of matching harmonic waves at three fixed phase differences.

Complete cancellation requires equal amplitudes as well as opposite phase. With unequal amplitudes, opposition leaves a residual oscillation equal to the amplitude difference. Also distinguish displacement from intensity: amplitudes add with phase; intensities are not generally added as if they were signed displacements. In the exact idealisation of two equal co-propagating waves everywhere opposite, their total displacement and velocity field can vanish. Describing two separately generated components does not create extra energy in their zero total field; energy accounting must include the sources and interference terms.

Phase can strengthen or weaken the same two waves

Problem
Two equal waves of amplitude 4 mm have the same frequency, wave number and direction. Find the resultant amplitude for phase differences 0, 2π/3 and π.

  1. 1.Use A = 2a|cos(φ/2)|, rather than adding two positive amplitude numbers for every case.
  2. 2.For φ = 0, A = 8 mm. For φ = 2π/3, cos(π/3) = 1/2, so A = 4 mm.
  3. 3.For φ = π, cos(π/2) = 0, so the ideal resultant vanishes. The intermediate result shows that interference need not mean either complete reinforcement or complete cancellation.

Reflection is controlled by what the boundary allows

A pulse arriving at a boundary exerts a force on it. The boundary’s response acts back on the medium and can create a reflected disturbance. At an ideal fixed end of a string, the endpoint cannot move, so the total displacement there must be zero at every time. An incident upward pulse therefore reflects as a downward pulse. At a lossless fixed boundary the reflected pulse has the same magnitude and shape with an inversion, often described as a phase reversal of π for a harmonic wave.

A boundary determines the reflected displacementFixed end: invertedendpoint displacement = 0Free end: not invertedendpoint slope = 0Top: incoming. Middle: overlap. Bottom: reflected. Schematic.
Fixed and free reflection— Ideal lossless boundary schematics; free-end displacement doubles at complete overlap.

At an ideal free end, modelled by a string attached to a freely sliding light ring, the endpoint can move transversely. There is no net transverse force at the endpoint, so the string tangent has zero slope there in the small-slope model. An upward pulse reflects upward without inversion. At full overlap the endpoint displacement can be twice the displacement amplitude of either incident or reflected pulse. This does not mean that a free endpoint is fixed at a nonzero height: it oscillates.

Reflection equations must reverse the direction as well as satisfy the end condition

Choose the boundary at x = 0 and the string on x ≤ 0. An incident wave moving right towards the boundary is yᵢ = a cos(kx − ωt), with k,ω positive. A reflected wave must move left, so its time-dependent phase must involve kx + ωt. Choose its sign by enforcing the boundary condition. The arbitrary phase origin is chosen for clarity; it does not change pulse inversion.

A fixed end, using cosine phasesLaTeX
At x = 0, a cos(ωt) − a cos(ωt) = 0 for all t. The reflected wave moves left and its displacement is inverted.
A free end, using cosine phasesLaTeX
At x = 0, y = 2a cos(ωt), and the slopes of incident and reflected waves cancel. Reflection is not inverted.

As a cross-check, if instead yᵢ = a sin(kx − ωt) in this same coordinate system, the fixed-end reflected wave is +a sin(kx + ωt). At x = 0 the two terms are −a sin(ωt) and +a sin(ωt), so they cancel. The free-end reflected wave is −a sin(kx + ωt). These apparent sign changes arise from the sine phase convention, not from a change of the physical reflection rule. Use the endpoint test to prevent sign errors. Leaving kx − ωt unchanged in a supposed reflected wave fails to reverse the propagation direction.

An interface can both reflect and transmit

A perfectly fixed or perfectly free end is an ideal limiting boundary. At an interface between two real elastic media, some incident energy is generally reflected and some transmitted. For incidence at an angle, the transmitted direction may bend: refraction. The directions obey the reflection law and, in the appropriate isotropic wave model, the refraction relation sin i/sin r = v₁/v₂. Here i and r are measured from the normal, and v₁ and v₂ are the corresponding phase speeds. This geometric relation is stated for context; deriving transmission amplitudes lies outside this chapter.

For sound the phrase “phase reversal” must specify the variable. A rigid closed end is a displacement node but a pressure antinode; an ideal open end is a pressure node but a displacement antinode. A reflected pressure disturbance can therefore have the opposite sign convention to a reflected displacement disturbance at the same end. We will build this distinction into the air-column modes, rather than applying the string’s displacement rule blindly to pressure.

A falling stone and a returning sound

Problem
A stone is dropped from rest from a 300 m tower into a pond. When is the splash heard at the top, taking g = 9.8 m s⁻² and sound speed 340 m s⁻¹? Neglect air resistance.

  1. 1.The stone and the sound make successive journeys by different physical mechanisms. For the fall, h = ½gt² gives t₁ = √(2h/g) = √(600/9.8) = 7.8246 s.
  2. 2.Only after impact is the splash sound produced. Its upward travel takes t₂ = h/v = 300/340 = 0.88235 s.
  3. 3.Add successive times: t = t₁ + t₂ = 8.7070 s ≈ 8.7 s. This is not a 600 m sound round trip. For an actual echo from a reflector a distance d away, the sound would make a two-way trip and d = vΔt/2.

Quick Quiz

Quiz

Quick check

Two contributions at one point are +5 mm and −3 mm. Their resultant is

Explanation

Superposition adds signed displacements: 5 − 3 = 2 mm.

Quick check

Two equal co-propagating waves have phase difference π. The ideal resultant amplitude is

Explanation

Their displacements cancel at every point and time.

Quick check

An upward pulse at an ideal fixed string end reflects

Explanation

The fixed endpoint must have zero net displacement.

Quick check

For yᵢ = a cos(kx − ωt) at a fixed boundary x = 0, which reflected wave works?

Explanation

It travels left and cancels the incident displacement at x = 0.

Quick check

A rope is momentarily straight when opposite pulses overlap. Which conclusion follows?

Explanation

Displacement alone does not specify velocity or the later evolution.

Practice Problems

Practice Problems
  1. Two pulses contribute +6 mm and −4 mm at the same point and instant. Find the net displacement and explain why you must not add their absolute values.
  2. Derive the resultant of a sin(kx − ωt) and a sin(kx − ωt + φ) using a sum-to-product identity. Give the physical amplitude for arbitrary φ.
  3. Two matching waves each have amplitude 3 mm. Calculate resultant amplitudes for phase differences 0, π/2, 2π/3 and π, and identify complete reinforcement and cancellation.
  4. Draw an upward pulse before and after reflection at a fixed endpoint and then a free endpoint. Explain the boundary condition responsible for the different reflected shapes.
  5. Place a boundary at x = 0 with the string on x ≤ 0. For yᵢ = a sin(kx − ωt), construct left-moving reflected waves at a fixed and a free end. Verify displacement or slope at the boundary for all times.
  6. A stone falls from rest through 300 m and its splash is heard above. Calculate the elapsed time for g = 9.8 m s⁻² and sound speed 340 m s⁻¹, explaining why two different time calculations are needed.
  7. Explain why zero displacement during pulse overlap does not necessarily imply zero mechanical energy. Contrast a snapshot of displacement with the full displacement-and-velocity state.

Key Takeaways

Key Takeaways

• Linear superposition adds signed displacements at the same event. • Equal co-propagating waves have resultant amplitude 2a|cos(φ/2)|; the frequency and wavelength remain unchanged. • A fixed end inverts a displacement pulse; a free end reflects it without inversion. • A valid reflected-wave equation must reverse direction and satisfy the boundary condition. Pressure and displacement reflection conventions differ.