Why Crests Reinforce and Crest-Trough Pairs Cancel in Waves

Imagine two ripples travelling towards the same point on a pond. If a crest from each wave arrives at exactly the same time, the water rises further above its normal level. If a crest meets an equally deep trough, the water can become momentarily flat.

Why does one combination produce a bigger wave while the other produces no displacement at all?

The answer is superposition: when waves overlap, their displacements add algebraically. Two crests reinforce each other because both produce upward displacement. A crest and an equal-sized trough cancel because one produces upward displacement while the other produces downward displacement.

The important detail is that waves add their displacements, not their labels. Let’s build that idea properly.

01Why two crests reinforce each other

Start by imagining a small section of a stretched rope. Its normal, undisturbed position is a horizontal line.

A wave passing along the rope moves each section above or below this equilibrium position.

We will use the following sign convention:

• Upward displacement is positive.
• Downward displacement is negative.
• Zero displacement means the rope is at its equilibrium position.

A crest therefore represents positive displacement, while a trough represents negative displacement.

Suppose two waves reach the same point at the same instant.

If the first wave produces an upward displacement of \(+2.0\text{ cm}\), and the second also produces \(+2.0\text{ cm}\), the total displacement is:

\[ y_R=+2.0+2.0=+4.0\text{ cm} \]

The rope moves 4.0 cm above equilibrium.

Both waves are contributing displacement in the same direction. Their effects reinforce each other.

This is called constructive interference.

What happens when a crest meets a trough?

Now imagine that the second wave produces a downward displacement of \(-2.0\text{ cm}\) instead.

The resultant displacement becomes:

\[ y_R=+2.0+(-2.0)=0 \]

The two contributions cancel exactly.

This is called complete destructive interference.

Notice that the trough is not somehow destroying the crest. The rope simply responds to both disturbances at once.

The principle of superposition

The physical rule behind both outcomes is the principle of superposition.

When waves overlap in a linear medium, the resultant displacement at any point is the algebraic sum of the individual displacements.

\[ y_R=y_1+y_2 \]

Where:

• \(y_R\) is the resultant displacement, measured in metres.
• \(y_1\) is the displacement produced by the first wave.
• \(y_2\) is the displacement produced by the second wave.

The word algebraic is important. It means that positive and negative signs must be included.

Consider these four possibilities.

Notice something interesting: two troughs also interfere constructively.

Constructive interference does not mean that the resulting wave must point upwards. It means that the two displacements reinforce each other, regardless of their direction.

Similarly, destructive interference does not always produce a completely flat surface.

For complete cancellation, the two displacements must be equal in magnitude and opposite in direction.

If one wave has a larger amplitude, some displacement remains.

The rule is not limited to water waves or ropes. Sound waves also superpose, with pressure variations adding together. For light, the relevant oscillating quantity is the electric field.

The principle applies to waves in a linear medium, which is the usual assumption in HSC wave problems. At sufficiently large disturbances, a medium may behave non-linearly, and simple superposition may no longer describe it accurately.

02What determines whether waves arrive as crests or troughs?

So far, we have looked at what happens when waves meet.

But there is another question: how do we know whether two waves will actually arrive in phase or out of phase?

This is where phase difference and path difference become useful.

Phase difference: where are the waves in their cycles?

Imagine watching two people jumping on identical trampolines.

If they reach their highest points together, they are moving through their cycles in phase.

If one reaches the highest point while the other reaches the lowest point, they are completely out of phase.

Waves behave in the same way.

Phase describes the position of a wave within its repeating cycle.

Two waves arriving:

• In phase have a phase difference of \(0^\circ\), or an integer multiple of \(360^\circ\).
• Completely out of phase have a phase difference of \(180^\circ\), or an odd multiple of \(180^\circ\).

A phase difference of \(180^\circ\) means that a crest from one wave aligns with a trough from the other.

Path difference: how far have the waves travelled?

Suppose two waves begin at the same time and in phase, but travel different distances before reaching point P.

One wave might travel 2.0 m, while the other travels 2.5 m.

The second wave has travelled an additional 0.5 m.

This difference is called the path difference.

\[ \Delta r=|r_1-r_2| \]

Where:

• \(\Delta r\) is the path difference, in metres.
• \(r_1\) is the distance travelled by the first wave.
• \(r_2\) is the distance travelled by the second wave.

The wavelength tells us how this extra distance affects the phase.

One complete wavelength corresponds to one complete cycle, or \(360^\circ\).

Therefore, for waves that start in phase and travel through the same medium, the phase difference caused by their different paths is:

\[ \Delta\phi=\frac{2\pi\Delta r}{\lambda} \]

Here, \(\Delta\phi\) is the phase difference in radians, and \(\lambda\) is the wavelength in metres.

This gives us two particularly useful conditions.

Why does half a wavelength produce cancellation?

Because travelling half a wavelength moves a wave through half of its cycle.

A crest becomes a trough, and a trough becomes a crest.

There is an important assumption here: the two sources must begin in phase, and neither wave can experience an additional phase change.

If the sources begin out of phase, or one wave reflects from a boundary, path difference alone may give the wrong prediction.

For a more detailed method of checking the arrival phase, see How to Tell Whether Two Waves Arrive in Phase in HSC Physics.

03Young’s experiment: how interference provided evidence for the wave model of light

The difference between constructive and destructive interference became particularly important when scientists investigated the nature of light.

In the early nineteenth century, there was considerable debate about whether light behaved as a wave or as a stream of particles.

Thomas Young investigated whether light could produce an interference pattern.

His experiment provided important evidence for the wave model.

How the original experiment worked

Young’s original demonstration, around 1801, was slightly different from the familiar double-slit diagram found in textbooks.

The arrangement worked as follows:

1. Sunlight entered a darkened room through a very small opening in a window shutter.
2. A thin card was placed in the narrow beam of sunlight.
3. The card divided the light into two portions, which travelled past opposite sides of the card.
4. These portions overlapped on the opposite wall.

The crucial part was the overlap.

If light behaved as a wave, the two portions could interfere with each other.

Young expected regions of reinforcement and cancellation, rather than simply two overlapping patches of light.

He observed interference fringes, providing evidence consistent with the wave model.

The American Physical Society’s historical account describes this original thin-card arrangement and distinguishes it from the familiar later double-slit demonstration.

Historical reference: American Physical Society, This Month in Physics History.

Why use a single source?

A later, more familiar experimental arrangement uses two narrow slits.

Light from one source first passes through a narrow slit. It then illuminates two closely spaced slits, which act as two coherent sources.

The light travels from both slits towards a screen.

Why not simply use two separate lamps?

Because independent sources generally do not maintain a constant phase difference. Their relative phases change, so a stable interference pattern cannot be maintained.

Using one source to illuminate both slits gives the two emerging waves a stable phase relationship.

The sources are therefore coherent.

The experiment produces alternating regions:

• Bright fringes, where the waves arrive in phase and their electric fields reinforce.
• Dark fringes, where the waves arrive out of phase and their electric fields cancel or substantially reduce.

For light, the electric fields superpose. The intensity recorded on the screen depends on the square of the resultant field amplitude.

The crest-and-trough picture is therefore a useful way to understand phase, even though light is not a mechanical ripple travelling through water.

A small measurement calculation

The double-slit experiment also allows us to estimate the wavelength of light by measuring fringe spacing.

For a screen sufficiently far from the slits, and small angles, the approximate relationship is:

\[ \beta=\frac{\lambda L}{d} \]

Where:

• \(\beta\) is the distance between adjacent bright fringes, in metres.
• \(\lambda\) is the wavelength of light, in metres.
• \(L\) is the distance from the slits to the screen, in metres.
• \(d\) is the separation between the two slits, in metres.

Consider an illustrative experiment with:

• Wavelength: \(500\text{ nm}=5.0\times10^{-7}\text{ m}\).
• Slit separation: \(0.50\text{ mm}=5.0\times10^{-4}\text{ m}\).
• Screen distance: \(2.0\text{ m}\).

Substitute:

\[ \begin{aligned} \beta&=\frac{\lambda L}{d}\\ &=\frac{(5.0\times10^{-7})(2.0)} {5.0\times10^{-4}}\\ &=2.0\times10^{-3}\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{\beta=2.0\text{ mm}} \]

Adjacent bright fringes should be approximately 2.0 mm apart.

These are illustrative values, not measurements from Young’s original experiment. The calculation assumes coherent monochromatic light and a screen far enough away for the small-angle approximation to apply.

The important connection is that the position of each fringe depends on the phase relationship between the waves, which in turn depends on their relative path lengths.

04Can a boundary change constructive interference into destructive interference?

Yes. This is one of the most important details to remember when dealing with reflected waves.

So far, we have assumed that waves only change their phase because they travel different distances.

However, reflecting from a boundary can also change a wave’s phase.

Consider a pulse travelling along a rope towards the end.

What happens when it reaches the boundary depends on whether the end is fixed or free.

Reflection at a fixed end

Imagine holding one end of a rope firmly against a wall.

Send an upward pulse towards the wall.

When the pulse reflects, it returns upside down.

An upward pulse becomes a downward pulse, and a downward pulse becomes an upward pulse.

This is called an inversion.

The reflected wave undergoes a phase change of \(180^\circ\), equivalent to half a cycle.

Reflection at a free end

Now imagine attaching the rope loosely to a ring that can move freely along a smooth vertical support.

The end is free to move.

An upward pulse reflects as an upward pulse, without inversion.

There is no additional \(180^\circ\) phase change.

The Physics Classroom provides a useful physical explanation of reflection at a fixed end.

The distinction is important because a reflected wave can arrive in a different phase even when the path difference would otherwise suggest constructive interference.

Keep the three phase contributions separate

When analysing reflected waves, consider:

1. The initial phase relationship between the sources.
2. The phase change caused by different path lengths.
3. Any additional phase change caused by reflection.

For a fixed end, the reflection contributes \(180^\circ\).

For an ideal free end, it contributes \(0^\circ\).

These phase contributions must be combined before deciding whether the waves reinforce or cancel.

For example, suppose two waves start in phase, but one travels an additional half-wavelength before reaching a point.

The path difference produces a \(180^\circ\) phase difference.

If that wave also reflects from a fixed end, it gains another \(180^\circ\).

The total phase difference is:

\[ 180^\circ+180^\circ=360^\circ \]

The waves now arrive in phase.

They interfere constructively, not destructively.

This is a particularly useful result because it shows why simply memorising the half-wavelength rule can lead to an incorrect answer.

05Worked examples

06Questions and solutions

Question 1

Two identical water waves each have an amplitude of 1.8 cm.

They arrive at point P in phase.

Calculate the resultant amplitude when their crests overlap. What is the resultant displacement when their troughs overlap?

Solution 1

The resultant amplitude is 3.6 cm. When both troughs overlap, the resultant displacement is \(-3.6\text{ cm}\).

The principle of superposition tells us to add the signed displacements.

When both waves produce upward displacement:

\[ y_R=+1.8+1.8 \]

\[ y_R=+3.6\text{ cm} \]

When both produce downward displacement:

\[ y_R=-1.8+(-1.8) \]

\[ y_R=-3.6\text{ cm} \]

Both situations represent constructive interference.

The direction of the resultant changes, but the amplitude remains 3.6 cm.

Question 2

Two coherent water-wave sources, S1 and S2, begin oscillating in phase. They produce waves with a wavelength of 0.80 m.

At point P:

• The wave from S1 has travelled 2.10 m.
• The wave from S2 has travelled 2.50 m.
• The amplitude of the wave from S1 is 4.0 mm.
• The amplitude of the wave from S2 is 2.5 mm.

Calculate the phase relationship at P and the resultant displacement when the wave from S1 reaches P at its crest.

Does complete destructive interference occur?

Solution 2

The waves arrive \(180^\circ\) out of phase, and the resultant displacement at the specified instant is \(+1.5\text{ mm}\). Complete cancellation does not occur.

Step 1: Calculate the path difference.

\[ \Delta r=|2.50-2.10| \]

\[ \Delta r=0.40\text{ m} \]

Step 2: Compare the path difference with the wavelength.

\[ \frac{\lambda}{2}=\frac{0.80}{2} =0.40\text{ m} \]

The path difference is exactly half a wavelength.

Because the sources begin in phase, the waves arrive \(180^\circ\) out of phase.

Step 3: Determine the individual displacements.

When S1 reaches P at its crest, its displacement is:

\[ y_1=+4.0\text{ mm} \]

Because the waves are antiphase, S2 produces a trough:

\[ y_2=-2.5\text{ mm} \]

Step 4: Apply superposition.

\[ \begin{aligned} y_R&=y_1+y_2\\ &=4.0+(-2.5)\\ &=+1.5\text{ mm} \end{aligned} \]

The resultant displacement is:

\[ \boxed{y_R=+1.5\text{ mm}} \]

The waves interfere destructively, but their amplitudes are unequal.

A common mistake is to assume that a half-wavelength path difference always produces zero displacement. It produces complete cancellation only when the opposing wave displacements have equal magnitudes.

Question 3

A 5.0 cm upward pulse travels along a rope towards a fixed end.

After reflection, it meets a separate upward pulse with an amplitude of 3.0 cm. The two pulse centres overlap exactly.

Assume that the pulses have the same shape, there is no energy loss at the boundary, and the medium obeys the principle of superposition.

(a) Calculate the resultant displacement at the centre of the overlap.

(b) Repeat the calculation if the rope has an ideal free end instead of a fixed end.

(c) Explain why the usual whole-wavelength and half-wavelength interference conditions are not the appropriate method for this question.

Solution 3

For the fixed end, the resultant displacement is \(-2.0\text{ cm}\). For the free end, it is \(+8.0\text{ cm}\).

Step 1: Determine the reflected pulse’s direction of displacement.

At a fixed end, the upward pulse is inverted.

Its reflected displacement is therefore:

\[ y_1=-5.0\text{ cm} \]

The separate upward pulse produces:

\[ y_2=+3.0\text{ cm} \]

Step 2: Apply superposition.

\[ \begin{aligned} y_R&=-5.0+3.0\\ &=-2.0\text{ cm} \end{aligned} \]

The resultant displacement is 2.0 cm below equilibrium.

Step 3: Consider the free-end situation.

At an ideal free end, the reflected pulse is not inverted.

Its displacement remains positive:

\[ y_1=+5.0\text{ cm} \]

The resultant becomes:

\[ \begin{aligned} y_R&=5.0+3.0\\ &=+8.0\text{ cm} \end{aligned} \]

The resultant displacement is 8.0 cm above equilibrium.

Step 4: Explain why path difference is not the method here.

These are isolated pulses, not repeating sinusoidal waves.

They do not have a defined wavelength or a repeating phase cycle. Therefore, the conditions involving integer and half-integer multiples of wavelength cannot be used to predict their overlap.

Instead, add their instantaneous signed displacements directly.

The important lesson is that superposition applies to both pulses and continuous waves, but the phase-and-path method is particularly useful for periodic waves.

Question 4

Two loudspeakers, S1 and S2, emit coherent sound waves of equal amplitude.

The speed of sound in the room is \(340\text{ m s}^{-1}\), and both speakers operate at a frequency of \(680\text{ Hz}\).

At listener P:

• The sound from S1 travels 3.20 m.
• The sound from S2 travels 3.70 m.

However, S2 is deliberately set to emit sound \(180^\circ\) out of phase with S1.

Assume there are no reflections and both waves have equal amplitudes at P.

Determine whether constructive or destructive interference occurs at P. Explain why using the path difference alone would produce the wrong prediction.

Solution 4

Destructive interference occurs at P because the waves arrive \(180^\circ\) out of phase, despite their path difference being one complete wavelength.

Step 1: Calculate the wavelength of the sound.

Use the wave equation:

\[ v=f\lambda \]

Rearrange:

\[ \lambda=\frac{v}{f} \]

Substitute:

\[ \lambda=\frac{340}{680} \]

\[ \lambda=0.50\text{ m} \]

Step 2: Calculate the path difference.

\[ \Delta r=|3.70-3.20| \]

\[ \Delta r=0.50\text{ m} \]

This is exactly one wavelength.

If the sources began in phase, the waves would arrive in phase and interfere constructively.

Step 3: Include the initial source phase difference.

S2 begins \(180^\circ\) out of phase with S1.

The additional path of one wavelength produces a \(360^\circ\) phase change, which is equivalent to no net change in phase.

The original \(180^\circ\) phase difference therefore remains.

The waves arrive completely out of phase.

Step 4: Determine the interference.

Because the two waves have equal amplitudes and arrive \(180^\circ\) out of phase, their pressure variations cancel.

Therefore, ideal complete destructive interference occurs at P.

The tempting mistake is to see a path difference of one wavelength and immediately select constructive interference.

That rule only applies when the sources begin in phase and no additional phase changes occur.

Here, the initial source phase difference changes the outcome.

Question 5

A student investigates standing waves on a taut string.

A sinusoidal wave travels towards the right-hand end of the string and reflects without energy loss. The incident and reflected waves each have an amplitude of 2.0 cm and a wavelength of 0.80 m.

Two observation points are marked:

• P is 0.20 m from the right-hand end.
• Q is 0.10 m from the right-hand end.

Initially, the right-hand end is fixed.

(a) Determine whether P is a node, an antinode, or neither. Calculate its maximum displacement.

(b) Determine whether Q is a node, an antinode, or neither.

(c) The student takes a photograph showing Q exactly at its equilibrium position. They conclude that Q must be a node. Explain why this conclusion is incorrect.

(d) The fixed end is replaced with an ideal free end. The wavelength, frequency, incident amplitude, and observation positions remain unchanged. Determine what happens to P and Q.

Assume the incident and reflected waves have equal amplitudes throughout the region being considered.

Solution 5

With the fixed end, P is an antinode with a maximum displacement of 4.0 cm, while Q is neither a node nor an antinode. With the free end, P becomes a node, while Q remains an intermediate point.

The crucial detail is that a node must remain at zero displacement at all times. A single photograph showing zero displacement is not sufficient.

Step 1: Calculate the path difference at P.

For an incident wave and its reflection, the additional distance travelled by the reflected wave at a point \(x\) from the boundary is:

\[ \Delta r=2x \]

At P:

\[ \Delta r=2(0.20)=0.40\text{ m} \]

Since the wavelength is 0.80 m:

\[ \Delta r=\frac{\lambda}{2} \]

The path difference produces a phase difference of \(180^\circ\).

Step 2: Include the fixed-end reflection.

The fixed end produces another \(180^\circ\) phase change.

Therefore:

\[ \Delta\phi_{\text{total}} =180^\circ+180^\circ =360^\circ \]

The incident and reflected waves arrive in phase at P.

They reinforce each other, producing an antinode.

Step 3: Calculate the maximum displacement at P.

Each wave has an amplitude of 2.0 cm.

\[ A_R=2.0+2.0 \]

\[ \boxed{A_R=4.0\text{ cm}} \]

P is therefore an antinode.

Step 4: Investigate point Q.

At Q:

\[ \Delta r=2(0.10) \]

\[ \Delta r=0.20\text{ m} \]

This is one-quarter of a wavelength:

\[ \Delta r=\frac{\lambda}{4} \]

The path difference produces a phase difference of \(90^\circ\).

Including the fixed-end inversion:

\[ \Delta\phi_{\text{total}} =90^\circ+180^\circ =270^\circ \]

The waves are neither perfectly in phase nor completely out of phase.

They therefore produce partial interference.

Q is neither a node nor an antinode.

Step 5: Test the student’s photograph.

The student has observed Q at zero displacement at one particular instant.

However, an oscillating point can pass through equilibrium without being a node.

A node is a point that remains at zero displacement throughout the entire oscillation.

Since Q has a non-zero resultant amplitude, it moves away from equilibrium at other times.

The photograph alone cannot establish that a point is a node.

Step 6: Replace the fixed end with a free end.

The geometry and path difference do not change.

However, the reflection phase changes.

A free end does not invert the reflected wave, so there is no additional \(180^\circ\) phase change.

At P, the phase difference is now:

\[ \Delta\phi_{\text{total}}=180^\circ \]

The waves arrive completely out of phase.

Therefore, P becomes a node, with zero displacement at all times.

At Q, the phase difference is now \(90^\circ\).

The waves still interfere partially, so Q remains neither a node nor an antinode.

Interestingly, the resultant amplitude at Q is unchanged: phase differences of \(90^\circ\) and \(270^\circ\) produce the same resultant amplitude for two equal-amplitude sinusoidal waves.

The deeper conclusion: Changing the boundary does not change the distance travelled by either wave. It changes the phase of the reflected wave, which changes where reinforcement and cancellation occur.

This is why boundary conditions matter when predicting standing-wave patterns.

07What to study next: from interference to standing waves

When two waves travel in opposite directions and repeatedly superpose, their interference can produce a standing wave.

Instead of simply observing two crests meet, you begin to see fixed positions where the string never moves (nodes) and positions where it oscillates with maximum amplitude (antinodes).

For an ideal standing wave, adjacent nodes are separated by half a wavelength.

The next useful step is learning how to locate those positions and use wavelength to calculate their spacing.

See How to Find Node Spacing from Wavelength in HSC Physics.