How to Tell Whether Two Waves Arrive in Phase in HSC Physics

Two waves reach the same point after travelling 2.400000 m and 2.4000012 m. Their path difference is just \(1.2\times10^{-6}\ \text{m}\). If their wavelength is \(6.0\times10^{-7}\ \text{m}\), do they arrive in phase?

Yes. The second wave has travelled exactly two extra wavelengths, so it has completed two additional oscillation cycles. It arrives at the same stage of its cycle as the first wave.

This gives us the central rule:

If two waves start in phase, travel through the same medium, and experience no different boundary phase shifts, they arrive in phase when their path difference is a whole number of wavelengths.

But there is a catch. The sources might not start in phase, and one wave might undergo a phase change when it reflects. To answer these questions reliably, you need to account for all three contributions: the starting phase, the path difference, and any boundary effects.

01How Young’s double-slit experiment revealed phase differences

How can we tell whether two waves arrive in phase if we cannot directly see their oscillations?

Young’s double-slit experiment provided a way to investigate this using light. It demonstrated that two waves travelling different distances can combine to produce alternating regions of reinforcement and cancellation.

How the experiment was arranged

In the familiar double-slit arrangement, light from one source first passes through a narrow slit. It then illuminates two closely spaced slits, labelled \(S_1\) and \(S_2\).

A screen is positioned beyond the two slits.

The light travels from each slit towards the screen. At any observation point \(P\), the two waves have travelled different distances:

• \(L_1\): distance from slit \(S_1\) to point \(P\).
• \(L_2\): distance from slit \(S_2\) to point \(P\).

The two slits act as coherent sources. This means they produce waves of the same frequency with a constant phase difference.

What should the experiment show?

At the centre of the screen, both waves travel equal distances.

If the sources begin in phase, the waves arrive in phase. Their crests meet crests, and their troughs meet troughs.

The result is a bright central fringe.

Move away from the centre, and one wave must travel slightly farther than the other. The path difference increases, changing the phase relationship between the waves.

You should therefore predict alternating bright and dark regions:

• Bright fringes occur where the waves arrive in phase and reinforce each other.
• Dark fringes occur where the waves arrive \(180^\circ\) out of phase and oppose each other.

The observed pattern consists of alternating bright and dark fringes, supporting the wave model of light.

Why the experimental design matters

The original experiment used light from a common source to illuminate both slits. This was important because independent light sources would not necessarily maintain a stable phase relationship.

For a cleaner quantitative demonstration, modern experiments commonly use a monochromatic laser. Monochromatic means that the light has one frequency and therefore one wavelength.

White light contains many wavelengths. Each wavelength produces its own interference pattern, making the fringes more complicated.

The underlying observation is the same: the relative distances travelled determine how the waves combine, provided their initial phase relationship is known.

02Path difference is not the same as phase difference

These two quantities are closely related, but they describe different things.

Path difference is the difference between the distances travelled by two waves to reach the same point.

It is measured in metres.

Phase difference describes how far apart the waves are in their oscillation cycles when they reach that point.

It is measured in degrees or radians.

Think of two runners moving around a circular track. If one runner completes an extra full lap, they return to the same position relative to the track. If one completes an extra half lap, they end up on opposite sides.

Waves behave similarly. An extra wavelength represents one complete oscillation, while an extra half-wavelength represents half an oscillation.

The analogy helps explain the repeating cycle, although waves accumulate phase continuously rather than moving around a physical track.

Converting path difference into phase difference

One complete wave cycle corresponds to:

\[ 360^\circ=2\pi\ \text{rad} \]

One wavelength therefore represents a phase change of \(360^\circ\).

Half a wavelength represents:

\[ \frac{1}{2}\times360^\circ=180^\circ \]

A quarter wavelength represents:

\[ \frac{1}{4}\times360^\circ=90^\circ \]

For two waves travelling through the same medium, the phase difference produced by their path difference is:

\[ \boxed{\Delta\phi=\frac{360^\circ\Delta L}{\lambda}} \]

Or, in radians:

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

Where:

• \(\Delta\phi\) is the magnitude of the phase difference.
• \(\Delta L\) is the path difference, in metres.
• \(\lambda\) is the wavelength in that medium, in metres.

The equation tells you how many degrees of oscillation separate the waves because of their different travel distances.

What different path differences mean

Assume two waves begin in phase and travel through the same medium without any additional phase changes.

Notice something important: a phase difference of \(360^\circ\) is equivalent to \(0^\circ\).

The waves have completed different numbers of cycles, but they have returned to the same stage of oscillation.

This is why the condition for two initially in-phase waves to arrive in phase is:

\[ \boxed{\Delta L=n\lambda} \]

Here, \(n=0,1,2,3,\ldots\) is an integer.

For opposite-phase arrival, the condition is:

\[ \boxed{\Delta L=\left(n+\frac12\right)\lambda} \]

These are the two most useful conditions for interference questions.

03A reliable method for checking whether two waves arrive in phase

When you see a question involving two waves and their path lengths, follow this process.

Step 1: Identify the initial phase relationship.

Did the sources begin in phase, or was one source already ahead or behind?

If the question says the sources are coherent and in phase, their initial phase difference is zero.

Step 2: Find the path difference.

Subtract the shorter path length from the longer path length:

\[ \Delta L=L_{\text{long}}-L_{\text{short}} \]

Make sure both distances use the same units.

Step 3: Identify the wavelength.

Use the wavelength in the medium through which the waves travel.

If frequency and wave speed are given, calculate it using:

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

Step 4: Calculate the phase difference caused by the path difference.

\[ \Delta\phi_{\text{path}} =\frac{360^\circ\Delta L}{\lambda} \]

Step 5: Account for any additional phase changes.

For example, reflection from a fixed end of a string produces a \(180^\circ\) phase reversal in transverse displacement.

If only one wave experiences this reversal, it must be included.

Step 6: Reduce the final phase difference to a value between \(0^\circ\) and \(360^\circ\).

Then classify the result:

• \(0^\circ\) or \(360^\circ\): in phase.
• \(180^\circ\): opposite phase.
• Any other value: neither completely in phase nor exactly opposite in phase.

This method is more reliable than memorising a single path-difference rule because it also works when the sources begin out of phase or when reflections are involved.

05What if the sources do not begin in phase?

So far, we have assumed that both sources start their oscillations together.

That assumption is not always valid.

Imagine two speakers producing sound at the same frequency. One speaker might begin its oscillation a quarter of a cycle ahead of the other.

Even if both waves travel exactly the same distance, they will still arrive out of phase.

The path difference changes the phase relationship, but it does not erase the original phase difference.

For this reason, it is useful to distinguish three contributions:

1. Initial phase difference at the sources.
2. Phase difference accumulated through travelling different distances.
3. Additional phase changes caused by reflections or other boundaries.

For two waves travelling through the same medium, we can combine these using:

\[ \boxed{ \Delta\phi_{\text{arrival}} = \Delta\phi_0 - 360^\circ\frac{\Delta L}{\lambda} + \Delta\phi_b } \]

Here:

• \(\Delta\phi_{\text{arrival}}\) is the final phase difference between waves B and A.
• \(\Delta\phi_0\) is the initial phase difference, defined as the phase of B minus the phase of A.
• \(\Delta L=L_B-L_A\) is the signed path difference. A positive value means B travels farther.
• \(\lambda\) is the wavelength.
• \(\Delta\phi_b\) is the difference between the phase changes experienced at boundaries.

The negative sign means that a wave travelling farther falls behind in phase.

A positive initial phase difference means that wave B starts ahead.

The final answer is interpreted modulo \(360^\circ\).

06Boundary effects: When path difference is not enough

There is another important complication: waves can change phase when they reflect.

Consider a transverse wave travelling along a string towards a fixed end.

At the fixed end, the string cannot move. An upward pulse must therefore reflect as a downward pulse to satisfy the boundary condition.

This is an inversion, equivalent to a \(180^\circ\) phase change for a sinusoidal transverse wave.

An ideal free end behaves differently. It allows the string to move, so the reflected pulse is not inverted.

A contrasting case: Equal path difference, different phase

Suppose two initially in-phase waves travel along separate paths to the same point.

The first travels directly.

The second travels an extra wavelength, but also reflects once from a fixed end.

What would you predict?

Looking only at the path difference, you might say that the waves arrive in phase because:

\[ \Delta L=\lambda \]

However, the extra wavelength contributes a \(360^\circ\) phase lag, while the fixed-end reflection contributes another \(180^\circ\) phase change.

The total phase difference is:

\[ \Delta\phi=360^\circ+180^\circ =540^\circ \]

Reducing this modulo \(360^\circ\):

\[ \Delta\phi=180^\circ \]

The waves arrive opposite in phase.

This is the important distinction: a whole-wavelength path difference guarantees in-phase arrival only when the initial phase difference and any additional boundary phase changes are also accounted for.

Applying this to incident and reflected waves

There is an interesting example involving a continuous wave on a string.

A wave travels towards a fixed end and reflects. At a point \(P\), located a distance \(x\) from the fixed end, the reflected wave has travelled an additional distance of:

\[ \Delta L=2x \]

The factor of two appears because the reflected wave travels from \(P\) to the boundary and then back to \(P\).

The total phase difference includes both this extra travel distance and the \(180^\circ\) phase reversal at the fixed end.

At the fixed end itself, \(x=0\).

The path difference is zero, but the reflection contributes \(180^\circ\). The waves are opposite in phase and cancel, producing a displacement node.

Now consider a point one-quarter wavelength from the fixed end:

\[ x=\frac{\lambda}{4} \]

The additional path is:

\[ \Delta L=2\left(\frac{\lambda}{4}\right) =\frac{\lambda}{2} \]

This contributes another \(180^\circ\) phase difference.

Including the fixed-end inversion:

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

The waves are now in phase and reinforce each other.

This explains why the phase relationship between incident and reflected waves changes with position, even though the reflection occurs at the same boundary.

For a more detailed treatment of the reflection process, see Wave Reflection: Direction, Boundaries and Phase in HSC Physics.

Be careful with sound waves

The reflection phase also depends on which physical quantity you are describing.

For a sound wave reflecting from a rigid wall:

• Air particle displacement behaves like a fixed-end reflection and reverses phase.
• Pressure variations do not reverse phase. They reinforce at the wall, producing a pressure antinode.

So you must identify whether a question is asking about displacement, pressure, or another oscillating quantity before applying a reflection rule.

07How phase difference connects to superposition

Once you know whether two waves arrive in phase, you can predict what happens when they overlap.

The principle of superposition states that the resultant displacement is the algebraic sum of the individual displacements:

\[ y_R=y_1+y_2 \]

Here, \(y_R\) is the resultant displacement, while \(y_1\) and \(y_2\) are the individual wave displacements at the same position and time.

The phase relationship determines whether those displacements reinforce or oppose each other.

For two waves with equal amplitudes \(A\):

• In phase: resultant amplitude is \(2A\).
• \(180^\circ\) out of phase: resultant amplitude is zero.

If the amplitudes are different, destructive interference does not necessarily produce complete cancellation.

For example, waves with amplitudes of \(5\ \text{mm}\) and \(3\ \text{mm}\) that arrive opposite in phase produce a resultant amplitude of \(2\ \text{mm}\).

This distinction is important: phase difference determines how the waves combine, but amplitude determines how much cancellation or reinforcement is possible.

08What if the waves travel through different media?

The usual equation

\[ \Delta\phi=\frac{360^\circ\Delta L}{\lambda} \]

assumes that both waves travel through the same medium, so they have the same wavelength.

If one path travels through multiple media, you cannot simply subtract the total physical distances and divide by one wavelength.

Instead, calculate how many wavelengths fit into each section of each path.

For one route:

\[ N=\frac{L_1}{\lambda_1} +\frac{L_2}{\lambda_2} +\cdots \]

Here, each \(L_j\) is the distance travelled in a particular medium, and \(\lambda_j\) is the wavelength in that medium.

Each section contributes its own number of oscillation cycles.

You can then compare the total cycle counts for the two paths and include any initial or boundary phase differences.

This is a more general version of the same method. In ordinary single-medium questions, it reduces to comparing the path difference with one wavelength.

Also remember that coherent waves must have the same frequency and a stable phase relationship. If the frequencies differ, their relative phase changes continuously, so a fixed path difference does not guarantee a stable interference pattern.

09Questions and solutions

Question 1

Two speakers produce coherent sound waves in phase. The sound frequency is \(500\ \text{Hz}\), and the speed of sound is \(340\ \text{m s}^{-1}\).

At point \(P\), the distances travelled by the two waves are \(2.40\ \text{m}\) and \(3.08\ \text{m}\).

Determine whether the waves arrive in phase or opposite in phase.

Solution 1

The waves arrive in phase and therefore interfere constructively.

First calculate the wavelength:

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

Substituting:

\[ \lambda=\frac{340}{500} =0.68\ \text{m} \]

Next, calculate the path difference:

\[ \Delta L=3.08-2.40 \]

\[ \Delta L=0.68\ \text{m} \]

Therefore:

\[ \frac{\Delta L}{\lambda} =\frac{0.68}{0.68} =1 \]

The path difference is exactly one wavelength.

The phase difference is:

\[ \Delta\phi=360^\circ\times1 =360^\circ \]

Since the sources initially emit in phase, the waves arrive in phase.

The tempting mistake is to assume that travelling different distances automatically produces different phases. What matters is how many complete cycles fit into the extra distance.

Question 2

Two coherent water-wave sources, A and B, produce waves with a frequency of \(10\ \text{Hz}\). The wave speed is \(2.0\ \text{m s}^{-1}\).

Source B begins \(90^\circ\) ahead of source A.

At point \(P\), wave B has travelled \(0.15\ \text{m}\) farther than wave A.

The amplitudes of the waves at \(P\) are \(4.0\ \text{mm}\) and \(3.0\ \text{mm}\).

Determine:

• The final phase difference.
• Whether the waves interfere constructively or destructively.
• The resultant amplitude.

Solution 2

The waves arrive \(180^\circ\) out of phase, so they interfere destructively. The resultant amplitude is \(1.0\ \text{mm}\).

First, calculate the wavelength:

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

\[ \lambda=\frac{2.0}{10} =0.20\ \text{m} \]

Next, calculate the phase lag caused by the path difference:

\[ \Delta\phi_{\text{path}} =\frac{360^\circ(0.15)}{0.20} \]

\[ \Delta\phi_{\text{path}}=270^\circ \]

Wave B therefore falls \(270^\circ\) behind during propagation.

However, it started \(90^\circ\) ahead.

The final phase difference is:

\[ \Delta\phi_{\text{arrival}} =90^\circ-270^\circ \]

\[ =-180^\circ \]

This is equivalent to \(180^\circ\), so the waves arrive opposite in phase.

Because the waves have unequal amplitudes, cancellation is incomplete.

The resultant amplitude is:

\[ A_R=|A_1-A_2| \]

\[ A_R=|4.0-3.0| \]

\[ \boxed{A_R=1.0\ \text{mm}} \]

The important point is that the initial phase difference must be included before deciding how the waves interfere. The path difference alone would have given an incomplete answer.

Question 3

Two coherent wave sources, A and B, begin in phase and produce waves of the same frequency.

The waves travel along different paths to point \(P\).

• Path A travels \(0.80\ \text{m}\) through medium X, where the wavelength is \(0.20\ \text{m}\).
• Path B travels \(0.40\ \text{m}\) through medium X, then \(0.225\ \text{m}\) through medium Y, where the wavelength is \(0.15\ \text{m}\).

Assume that transmission at the interfaces introduces no additional relative phase shift.

Determine whether the waves arrive in phase or opposite in phase.

Explain why subtracting the total physical path lengths and dividing by one wavelength would give an unreliable result.

Solution 3

The waves arrive \(180^\circ\) out of phase because their paths accumulate a difference of half an oscillation cycle.

For path A, calculate the number of wavelengths travelled:

\[ N_A=\frac{0.80}{0.20} \]

\[ N_A=4 \]

Path A accumulates four complete cycles.

For path B, calculate the cycles separately in each medium:

\[ N_B=\frac{0.40}{0.20} +\frac{0.225}{0.15} \]

\[ N_B=2+1.5 \]

\[ N_B=3.5 \]

The difference in accumulated cycles is:

\[ \Delta N=4-3.5=0.5 \]

Half a cycle corresponds to:

\[ \Delta\phi=0.5\times360^\circ \]

\[ \boxed{\Delta\phi=180^\circ} \]

Therefore, the waves arrive opposite in phase.

The tempting approach is to calculate the total physical lengths:

\[ L_A=0.80\ \text{m} \]

\[ L_B=0.40+0.225=0.625\ \text{m} \]

Then subtract them.

However, the two paths do not experience the same wavelength throughout. A metre travelled in medium X represents a different number of oscillation cycles from a metre travelled in medium Y.

The correct approach is to count cycles separately in each medium before comparing the paths.

Question 4

A continuous sound wave travels towards a rigid wall and reflects.

A student argues:

“The wall is fixed, so the reflected wave must be \(180^\circ\) out of phase with the incident wave. Therefore, the pressure variations should cancel at the wall.”

Explain why this reasoning is incorrect.

Your answer should distinguish between air particle displacement and pressure variation.

Solution 4

The student’s conclusion about pressure is incorrect. At an ideal rigid wall, pressure variations reinforce each other, producing a pressure antinode.

The mistake is applying the fixed-end reflection rule without identifying which physical quantity is oscillating.

At a rigid wall, air particles cannot move through the wall. Their displacement perpendicular to the wall must therefore be zero at the boundary.

The reflected particle-displacement wave reverses phase, allowing the incident and reflected displacements to cancel at the wall.

Pressure behaves differently.

The incident and reflected pressure variations reinforce each other at the rigid boundary. This produces a pressure antinode, where pressure variation has maximum amplitude.

Therefore:

• Particle displacement has a node at the rigid wall.
• Pressure variation has an antinode at the rigid wall.

The fixed-end reflection rule for a string describes transverse displacement. It cannot be applied blindly to every physical quantity in every type of wave.

Question 5

Two coherent transverse waves travel along separate ideal string paths and meet at point \(P\).

Both sources have a frequency of \(10\ \text{Hz}\). Source B begins \(90^\circ\) ahead of source A.

The path travelled by wave B is \(0.15\ \text{m}\) longer than the path travelled by wave A. In addition, wave B reflects once from an ideal fixed end. Wave A does not reflect.

Initially, the wave speed along both paths is \(6.0\ \text{m s}^{-1}\).

The tension is then adjusted, reducing the wave speed along both paths to \(2.0\ \text{m s}^{-1}\). The source frequency, source phase difference, path lengths, and fixed-end boundary condition remain unchanged.

Assume that both waves have equal amplitudes at \(P\) before and after the adjustment.

Determine:

• Whether the waves arrive in phase before the speed adjustment.
• Whether they arrive in phase after the adjustment.
• The resultant amplitude at \(P\) in each case, if each individual wave has amplitude \(A\).

Solution 5

Initially, the waves arrive \(180^\circ\) out of phase, producing complete destructive interference. After the speed adjustment, they arrive in phase, producing constructive interference.

This happens because changing the wave speed changes the wavelength, which changes the phase difference produced by the fixed path difference.

The initial source phase difference and the fixed-end reflection must also be included.

Before the speed adjustment

Step 1: Calculate the initial wavelength.

\[ \lambda_1=\frac{v_1}{f} \]

\[ \lambda_1=\frac{6.0}{10} =0.60\ \text{m} \]

Step 2: Calculate the phase lag from the path difference.

\[ \Delta\phi_{\text{path}} =\frac{360^\circ(0.15)}{0.60} \]

\[ \Delta\phi_{\text{path}}=90^\circ \]

Wave B falls \(90^\circ\) behind because it travels farther.

Step 3: Include the initial phase difference and reflection.

Wave B starts \(90^\circ\) ahead of A.

The fixed-end reflection adds a \(180^\circ\) phase reversal to its transverse displacement.

Therefore:

\[ \begin{aligned} \Delta\phi_{\text{arrival}} &=90^\circ-90^\circ+180^\circ\\ &=180^\circ \end{aligned} \]

The waves arrive opposite in phase.

Because their amplitudes are equal:

\[ A_R=|A-A|=0 \]

There is complete destructive interference.

After the speed adjustment

Step 4: Calculate the new wavelength.

The frequency remains \(10\ \text{Hz}\), so:

\[ \lambda_2=\frac{v_2}{f} \]

\[ \lambda_2=\frac{2.0}{10} =0.20\ \text{m} \]

Step 5: Recalculate the phase difference caused by the same path difference.

\[ \Delta\phi_{\text{path}} =\frac{360^\circ(0.15)}{0.20} \]

\[ \Delta\phi_{\text{path}}=270^\circ \]

The physical path difference has not changed, but it now represents three-quarters of a wavelength rather than one-quarter of a wavelength.

Step 6: Include the initial phase difference and fixed-end reflection again.

\[ \begin{aligned} \Delta\phi_{\text{arrival}} &=90^\circ-270^\circ+180^\circ\\ &=0^\circ \end{aligned} \]

The waves now arrive in phase.

Their equal amplitudes reinforce:

\[ A_R=A+A \]

\[ \boxed{A_R=2A} \]

The tempting answer is to assume that the phase relationship cannot change because the sources and physical path lengths remain the same.

But phase depends on path length relative to wavelength, not path length alone. Changing the wave speed while keeping frequency constant changes the wavelength and therefore changes the accumulated phase.

The fixed-end reflection also remains part of the calculation. Ignoring it would give the wrong result in both situations.

10What to study next: Standing waves and node spacing

The ability to compare wave phases is essential when studying standing waves.

When two waves of the same frequency and amplitude travel in opposite directions, their repeated superposition creates stationary nodes and antinodes.

At a node, the waves always oppose each other. At an antinode, they reinforce each other.

The spacing between neighbouring nodes is half a wavelength:

\[ \boxed{\text{Node spacing}=\frac{\lambda}{2}} \]

This relationship follows directly from the phase differences between the incident and reflected waves.

Once you can track path difference, phase changes, and superposition, standing-wave patterns become a consequence of the physics rather than another set of formulas to memorise.

For the next step, see How to Find Node Spacing from Wavelength in HSC Physics.