How to Calculate Path Difference Between Two Wave Sources

Two in-phase wave sources are 0.80 m apart. A point \(P\) lies between them, 0.30 m from source \(S_1\) and therefore 0.50 m from source \(S_2\).

If the wavelength is 0.40 m, will the waves reinforce or cancel when they reach \(P\)?

The path difference is:

\[ \Delta r=|0.30-0.50|=0.20\,\text{m} \]

This is half a wavelength, so the waves arrive 180° out of phase and destructively interfere, assuming equal amplitudes.

The key rule is simple: subtract the two distances from the sources to the point.

\[ \boxed{\text{Path difference}=|r_1-r_2|} \]

The challenge is knowing which distances to use, and understanding what that difference tells you about the waves when they meet.

01Finding path difference between two sources

Case 1: The point lies directly between the sources

Consider two sources, \(S_1\) and \(S_2\), separated by a distance \(d\). Point \(P\) lies somewhere along the straight line joining them.

Let:

• \(d\) = separation between the two sources, in metres
• \(x\) = distance from \(S_1\) to \(P\), in metres
• \(r_1\) = distance from \(S_1\) to \(P\)
• \(r_2\) = distance from \(S_2\) to \(P\)

Because \(P\) lies between the sources:

\[ r_1=x \]

\[ r_2=d-x \]

The path difference is therefore:

\[ \Delta r=|x-(d-x)| \]

Simplifying:

\[ \boxed{\Delta r=|2x-d|} \]

Notice what happens at different positions.

• At the midpoint, both waves travel the same distance. The path difference is zero.
• Closer to \(S_1\), the wave from \(S_2\) travels further. The path difference increases.
• Closer to \(S_2\), the wave from \(S_1\) travels further. The path difference also increases.

For example, if the sources are 1.0 m apart and \(P\) is 0.20 m from \(S_1\):

\[ r_1=0.20\,\text{m} \]

\[ r_2=1.0-0.20=0.80\,\text{m} \]

Therefore:

\[ \Delta r=|0.20-0.80| \]

\[ \boxed{\Delta r=0.60\,\text{m}} \]

A common mistake is to add the distances and get 1.0 m. That is the separation between the sources, not the path difference.

Case 2: The point is between the sources but not on their connecting line

Not every question places \(P\) directly on the line joining the sources.

Imagine two speakers on a floor and a listener standing slightly in front of the line between them. The listener is still between the speakers horizontally, but the two sound waves travel along diagonal paths.

You must calculate the actual straight-line distances from each source to \(P\).

If the sources are positioned at:

\[ S_1=(0,0), \qquad S_2=(d,0) \]

and the observation point is:

\[ P=(u,h) \]

Pythagoras gives:

\[ r_1=\sqrt{u^2+h^2} \]

\[ r_2=\sqrt{(d-u)^2+h^2} \]

Then:

\[ \boxed{\Delta r=|r_1-r_2|} \]

The important distinction is that the two distances no longer add to the source separation.

The general method never changes: calculate both distances travelled, then subtract the smaller from the larger.

02Path difference and phase difference are not the same thing

You now have a distance. But how does that tell you whether the waves reinforce or cancel?

First, distinguish two quantities.

Path difference is the difference between the distances travelled by two waves. It is measured in metres.

Phase difference describes how far apart the waves are in their cycles when they meet. It is measured in radians or degrees.

The wavelength connects them.

A wave travels through one complete cycle every time it moves forward by one wavelength, \(\lambda\).

Therefore, for two sources that begin in phase, the magnitude of the phase difference caused by their different travel distances is:

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

Alternatively, in degrees:

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

Here:

• \(\Delta\phi\) = phase difference
• \(\Delta r\) = path difference, in metres
• \(\lambda\) = wavelength, in metres

For example, if the path difference is half a wavelength:

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

Then:

\[ \Delta\phi= \frac{360^\circ}{2} =180^\circ \]

The waves arrive half a cycle apart.

A crest from one wave meets a trough from the other.

Constructive and destructive interference

For two coherent sources that begin in phase, the path difference determines the type of interference.

Coherent sources have the same frequency and maintain a constant phase relationship.

In equation form:

Constructive interference:

\[ \boxed{\Delta r=m\lambda} \]

Destructive interference:

\[ \boxed{\Delta r=\left(m+\frac12\right)\lambda} \]

Here, \(m=0,1,2,3,\ldots\).

There is one important condition: these rules assume the sources begin in phase.

If the sources begin 180° out of phase, the conditions reverse. A whole-wavelength path difference then produces destructive interference, while a half-wavelength path difference produces constructive interference.

This is why you should always check the source phase relationship before classifying a point.

For a more general phase relationship, let \(\phi_s\) be the initial phase of source 2 relative to source 1, and define the signed path difference as \(r_2-r_1\). The phase difference at \(P\), using a wave phase that decreases with distance travelled, is:

\[ \Delta\phi_P= \phi_s-\frac{2\pi}{\lambda}(r_2-r_1) \]

The total phase difference, rather than the path difference alone, determines the interference.

03Superposition: why the waves reinforce or cancel

When two waves reach the same point, their displacements add. This is called the principle of superposition.

Mathematically:

\[ \boxed{y_{\text{resultant}}=y_1+y_2} \]

Here, \(y_1\) and \(y_2\) are the instantaneous displacements caused by each wave.

Consider two waves with equal amplitudes.

If they arrive in phase, their displacements reinforce each other. The resultant amplitude is twice the amplitude of either individual wave.

If they arrive 180° out of phase, their displacements oppose each other. Their resultant amplitude is zero.

However, destructive interference does not always mean complete cancellation.

Suppose one wave has an amplitude of 4 mm and the other has an amplitude of 3 mm. If they arrive exactly out of phase, the resultant amplitude is:

\[ A_{\text{resultant}}=|4-3| \]

\[ \boxed{A_{\text{resultant}}=1\,\text{mm}} \]

The waves interfere destructively, but the point is not perfectly stationary.

This distinction is important:

• A point of constructive interference has increased resultant amplitude.
• A point of destructive interference has reduced resultant amplitude.
• A perfect node requires equal amplitudes and exactly opposite phases.

04Testing path difference using a ripple tank

How could you check that the path difference actually predicts where constructive and destructive interference occur?

A ripple tank provides a useful experimental model.

Setting up the experiment

A typical arrangement contains:

• A shallow tank of water.
• Two small dippers that touch the water surface at separate positions.
• A mechanical vibrator or signal generator connected to both dippers.
• A light source above the tank.
• A translucent screen beneath the tank to display the wave pattern.
• A ruler or scale for measuring wavelength and distances.

Both dippers are driven by the same signal generator. This ensures that they oscillate at the same frequency and maintain a fixed phase relationship.

What should the experiment show?

Suppose the sources are driven in phase.

As the waves spread outwards, their crests and troughs overlap.

At some locations, crests from both sources arrive together. These locations experience constructive interference.

At other locations, a crest from one source meets a trough from the other. These locations experience destructive interference.

The expected result is a pattern of antinodal and nodal lines.

The position of these lines should agree with the path-difference conditions:

\[ \Delta r=m\lambda \]

for constructive interference, and:

\[ \Delta r=\left(m+\frac12\right)\lambda \]

for destructive interference.

Why the first experimental design can fail

Imagine initially setting up two dippers with separate motors.

You might predict a stable interference pattern because both motors have been set to approximately the same frequency.

However, their actual frequencies may differ slightly. Their relative phase then changes over time.

The nodal and antinodal lines can move rather than remaining in fixed positions.

This does not disprove superposition. It reveals that the sources were not sufficiently controlled.

A better design uses one signal generator to drive both dippers. This maintains a fixed phase relationship and produces a stable pattern.

There is another possible problem: reflections from the tank walls. These introduce additional waves that interfere with the original waves.

Placing sponge absorbers around the edges reduces these reflections and makes the direct two-source pattern easier to investigate.

Worked experimental calculation

Consider the following illustrative data from a ripple-tank investigation. These values are examples, not claimed experimental measurements.

Step 1: Calculate the wavelength.

Ten crest-to-crest gaps represent ten wavelengths.

\[ \lambda=\frac{0.500}{10} \]

\[ \boxed{\lambda=0.0500\,\text{m}} \]

Step 2: Calculate the path difference.

\[ \Delta r=|0.315-0.340| \]

\[ \boxed{\Delta r=0.0250\,\text{m}} \]

Step 3: Compare the path difference with the wavelength.

\[ \frac{\Delta r}{\lambda} = \frac{0.0250}{0.0500} =0.5 \]

Therefore:

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

For in-phase sources, this predicts destructive interference.

Step 4: Calculate the wave speed.

Using \(v=f\lambda\):

\[ v=10.0\times0.0500 \]

\[ \boxed{v=0.500\,\text{m/s}} \]

The predicted result at \(P\) is a node, assuming equal amplitudes and negligible reflections.

If the observed pattern agrees with this prediction, it supports the relationship between path difference and interference.

05Boundary effects: when the simple calculation is not enough

There is one final complication worth understanding.

The path-difference calculation compares the direct waves travelling from \(S_1\) and \(S_2\) to \(P\).

It does not automatically account for waves that reflect off a boundary before reaching the observation point.

A reflected wave is an additional wave arriving at the same point.

The total displacement now includes more contributions:

\[ y_{\text{total}}= y_{1,\text{direct}}+ y_{2,\text{direct}}+ y_{\text{reflected}}+\cdots \]

This means that even if the two direct waves are predicted to cancel, the reflected waves may prevent complete cancellation.

The reflection itself can also introduce a phase change. For example, a pulse reflecting from a fixed end of a string is inverted, whereas a pulse reflecting from a free end is not.

The phase behaviour depends on the boundary conditions and the type of wave. You should not automatically assume every reflection introduces a 180° phase change.

What should you do in an HSC question?

Use the following decision process:

For standard two-source questions, assume the direct waves are being considered unless the question explicitly introduces reflections or other boundary effects.

The key is to keep three ideas separate:

1. Path difference tells you how much further one wave travels.
2. Phase difference tells you how far out of step the waves are when they meet.
3. Superposition determines the resultant displacement when the waves overlap.

06Questions and solutions

Question 1

Two coherent, in-phase point sources \(S_1\) and \(S_2\) are separated by 1.00 m.

Point \(P\) lies directly between the sources, 0.30 m from \(S_1\). The wavelength is 0.20 m.

Calculate the path difference and determine whether constructive or destructive interference occurs at \(P\). Assume equal amplitudes.

Solution 1

The path difference is 0.40 m, so constructive interference occurs at \(P\).

Step 1: Calculate the distance travelled by each wave.

\[ r_1=0.30\,\text{m} \]

\[ r_2=1.00-0.30 =0.70\,\text{m} \]

Step 2: Calculate the path difference.

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

\[ =|0.30-0.70| \]

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

Step 3: Compare with the wavelength.

\[ \frac{\Delta r}{\lambda} = \frac{0.40}{0.20} =2 \]

The path difference is two complete wavelengths.

The waves therefore arrive in phase and reinforce each other.

Since their amplitudes are equal, \(P\) is an antinode.

Question 2

Two coherent, in-phase sources are separated by 0.70 m. Point \(P\) lies above the line joining them, rather than directly on the line.

The straight-line distances from the sources to \(P\) are:

\[ r_1=0.50\,\text{m} \]

\[ r_2=0.52\,\text{m} \]

The wavelength is 0.080 m.

Calculate the path difference and phase difference. Determine whether \(P\) is a node, an antinode, or neither.

Explain why using the horizontal separation between \(P\) and either source would give an incorrect result.

Solution 2

Point \(P\) is neither a perfect node nor an antinode. The waves arrive with a phase difference of approximately 90°.

Step 1: Calculate the path difference.

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

\[ =|0.50-0.52| \]

\[ \boxed{\Delta r=0.020\,\text{m}} \]

Step 2: Calculate the phase difference.

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

Substituting:

\[ \Delta\phi= \frac{360^\circ\times0.020}{0.080} \]

\[ \boxed{\Delta\phi=90^\circ} \]

Step 3: Classify the interference.

Constructive interference requires the waves to arrive in phase.

Destructive interference requires them to arrive 180° out of phase.

Here, the phase difference is 90°, so neither condition is satisfied.

The waves undergo partial interference.

The important geometric detail is that \(P\) is off-axis. The horizontal distances alone do not represent the actual distances travelled by the waves.

You must use the straight-line paths from each source to \(P\).

Question 3

Two coherent sources produce waves of equal amplitude. However, the sources begin 180° out of phase.

A point \(P\) lies directly between them. The distances travelled by the waves are:

\[ r_1=0.24\,\text{m} \]

\[ r_2=0.39\,\text{m} \]

The wavelength is 0.10 m.

Determine whether constructive or destructive interference occurs at \(P\).

Explain why a student who uses the standard in-phase interference conditions would reach the wrong conclusion.

Solution 3

Constructive interference occurs at \(P\), even though the path difference is 1.5 wavelengths.

Step 1: Calculate the path difference.

\[ \Delta r=|0.24-0.39| \]

\[ \boxed{\Delta r=0.15\,\text{m}} \]

Step 2: Express the path difference in wavelengths.

\[ \frac{\Delta r}{\lambda} = \frac{0.15}{0.10} =1.5 \]

Therefore:

\[ \Delta r=\frac{3\lambda}{2} \]

If the sources began in phase, this would produce destructive interference.

But they do not.

Step 3: Include the initial source phase difference.

The sources begin 180° out of phase.

The additional path difference of 1.5 wavelengths introduces a further phase shift of:

\[ 360^\circ\times1.5=540^\circ \]

Using the signed phase convention, the total phase difference is:

\[ \Delta\phi_P=180^\circ-540^\circ \]

\[ =-360^\circ \]

A phase difference of \(-360^\circ\) is equivalent to 0°.

The waves therefore arrive in phase.

Conclusion: Constructive interference occurs, and \(P\) is an antinode because the amplitudes are equal.

The tempting mistake is to apply the half-wavelength destructive condition without checking how the sources initially oscillate.

The source phase difference must be included before deciding the interference type.

Question 4

A ripple-tank experiment uses two dippers operating at approximately the same frequency.

At a particular point \(P\), the measured direct path lengths are:

\[ r_1=0.42\,\text{m} \]

\[ r_2=0.47\,\text{m} \]

The nominal wavelength is 0.10 m.

The two dippers are driven by separate motors that are not phase-locked. Their frequencies drift slightly relative to each other. The tank also has hard, reflective edges.

(a) Calculate the direct path difference.

(b) If the sources were coherent and in phase, what interference would be predicted at \(P\)?

(c) Explain why the actual experiment may fail to produce a stable node.

(d) Suggest two modifications to improve the experimental design, explaining what each modification controls.

Solution 4

The direct path difference is 0.050 m, which predicts destructive interference for coherent, in-phase sources. However, the actual experimental arrangement does not provide the conditions needed for a stable two-source interference pattern.

Step 1: Calculate the path difference.

\[ \Delta r=|0.42-0.47| \]

\[ \boxed{\Delta r=0.050\,\text{m}} \]

Step 2: Compare with the wavelength.

\[ \frac{\Delta r}{\lambda} = \frac{0.050}{0.10} =0.5 \]

Therefore:

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

For coherent, in-phase sources, this would produce destructive interference.

Step 3: Explain why the actual experiment may not show a stable node.

There are two separate problems.

First, the dippers are not phase-locked.

Their frequencies drift relative to each other, so their phase difference changes over time. A location that experiences destructive interference at one instant may experience partial or constructive interference later.

Second, the hard tank boundaries reflect waves.

These reflected waves introduce additional contributions to the resultant displacement. Consequently, the observed pattern may differ from the pattern predicted using only the two direct paths.

Step 4: Improve the experiment.

Two modifications address these problems.

• Connect both dippers to the same signal generator. This ensures a common frequency and a fixed phase relationship, producing stable interference.
• Place sponge wave absorbers around the tank edges. These reduce reflected waves and make the direct two-source interference pattern easier to observe.

The important conclusion is that a mismatch between prediction and observation does not automatically mean the path-difference equation is incorrect. The experimental assumptions must also be satisfied.

Question 5

Two coherent point sources, \(S_1\) and \(S_2\), are separated by 0.80 m in a uniform medium.

Point \(P\) lies directly between them, 0.35 m from \(S_1\).

The sources produce equal-amplitude waves. Their relative source phase remains constant when the frequency changes, but it is initially unknown.

The wave speed remains constant at 0.80 m/s.

The tank edges are fitted with absorbers, so reflected waves can be neglected.

The following observations are made:

• At 4.0 Hz, point \(P\) is a clear node.
• At 8.0 Hz, the same point \(P\) is a clear antinode.

(a) Calculate the distances travelled by the two waves to reach \(P\), and determine the path difference.

(b) Determine the wavelength at each frequency.

(c) Use both observations to determine whether the sources are in phase or out of phase. Explain why the two observations must be considered together.

(d) At 8.0 Hz, a second point \(Q\) is placed 0.025 m from \(P\), moving towards \(S_2\). Determine whether constructive or destructive interference occurs at \(Q\).

(e) A student argues that moving \(P\) by 0.025 m should change the path difference by only 0.025 m. Explain the error in this reasoning.

(f) Would the conclusion about the source phase relationship necessarily be reliable if the tank had strongly reflective walls instead of absorbers? Explain.

Solution 5

The sources must be in phase, and point \(Q\) is a node at 8.0 Hz.

This problem combines source geometry, wavelength, phase difference, and superposition. The important detail is that changing frequency changes the wavelength, even though the physical distances between the sources and observation points remain unchanged.

Step 1: Calculate the distances travelled by the waves to \(P\).

The distance from \(S_1\) is given:

\[ r_1=0.35\,\text{m} \]

The sources are separated by 0.80 m, so:

\[ r_2=0.80-0.35 \]

\[ r_2=0.45\,\text{m} \]

Step 2: Calculate the path difference.

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

\[ =|0.35-0.45| \]

\[ \boxed{\Delta r=0.10\,\text{m}} \]

The physical path difference remains 0.10 m throughout the experiment because neither source nor \(P\) moves.

Step 3: Calculate the wavelength at each frequency.

Use the wave equation:

\[ v=f\lambda \]

Rearranging:

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

At 4.0 Hz:

\[ \lambda_1=\frac{0.80}{4.0} \]

\[ \boxed{\lambda_1=0.20\,\text{m}} \]

At 8.0 Hz:

\[ \lambda_2=\frac{0.80}{8.0} \]

\[ \boxed{\lambda_2=0.10\,\text{m}} \]

Doubling the frequency halves the wavelength because the wave speed is constant.

Step 4: Determine the source phase relationship using the first observation.

At 4.0 Hz:

\[ \frac{\Delta r}{\lambda_1} = \frac{0.10}{0.20} =0.5 \]

Therefore:

\[ \Delta r=\frac{\lambda_1}{2} \]

A half-wavelength path difference introduces a 180° phase shift between the waves.

The observation is a node, meaning the waves arrive destructively.

This is consistent with sources that begin in phase.

If the sources began 180° out of phase, the additional 180° phase shift from travelling different distances would make the waves arrive in phase instead.

Therefore, the first observation tells us that the sources begin in phase.

Step 5: Check the second observation.

At 8.0 Hz:

\[ \frac{\Delta r}{\lambda_2} = \frac{0.10}{0.10} =1 \]

The path difference is now one complete wavelength.

For sources that begin in phase, this produces constructive interference.

This agrees with the observed antinode.

Both observations therefore support the same conclusion: the sources are in phase.

Notice the important point: the physical path difference did not change, but the wavelength did. Therefore, the phase difference between the arriving waves changed.

Step 6: Calculate the path difference at \(Q\).

Point \(Q\) is 0.025 m closer to \(S_2\) than \(P\).

Therefore, its distance from \(S_1\) increases:

\[ r_{1,Q}=0.35+0.025 \]

\[ r_{1,Q}=0.375\,\text{m} \]

Its distance from \(S_2\) decreases:

\[ r_{2,Q}=0.45-0.025 \]

\[ r_{2,Q}=0.425\,\text{m} \]

The new path difference is:

\[ \Delta r_Q=|0.375-0.425| \]

\[ \boxed{\Delta r_Q=0.050\,\text{m}} \]

At 8.0 Hz, the wavelength is 0.10 m.

Therefore:

\[ \frac{\Delta r_Q}{\lambda_2} = \frac{0.050}{0.10} =0.5 \]

The path difference is half a wavelength.

Since the sources are in phase, destructive interference occurs.

\[ \boxed{\text{Q is a node.}} \]

Step 7: Explain the student’s mistake.

The student assumes that moving the observation point by 0.025 m changes the path difference by only 0.025 m.

But both paths change.

As \(Q\) moves towards \(S_2\):

• The distance from \(S_1\) increases by 0.025 m.
• The distance from \(S_2\) decreases by 0.025 m.

The difference between the two distances therefore changes by:

\[ 0.025+0.025=0.050\,\text{m} \]

The path difference changes by twice the physical movement.

This is why a relatively small movement of the observation point can change the interference pattern significantly.

Step 8: Consider the effect of reflective boundaries.

The conclusion would not necessarily be reliable if strong reflections were present.

The observations could be influenced by additional reflected waves, which may change the resultant amplitude and the positions of apparent nodes and antinodes.

The direct two-source calculation would no longer describe every contribution to the wave at \(P\).

Absorbing boundaries are therefore important for making the experimental inference reliable.

The deeper lesson is that interference depends on the total phase difference at the observation point, not simply on the physical separation of the sources or the distance moved by the observation point.

07Connecting path difference to interference patterns

Once you can calculate the path difference to a single point, the next step is understanding how that difference changes across an entire region.

Instead of calculating one point at a time, you can identify all positions where the path difference is:

\[ 0,\lambda,2\lambda,3\lambda,\ldots \]

These positions form antinodal lines for in-phase sources.

Similarly, positions satisfying:

\[ \frac{\lambda}{2},\frac{3\lambda}{2}, \frac{5\lambda}{2},\ldots \]

form nodal lines.

This leads to the broader study of two-source interference patterns, including the curved nodal and antinodal lines observed in ripple tanks.

Being able to move confidently between source geometry, path difference, phase difference, and superposition gives you the foundation for analysing those patterns in more complex HSC Physics questions.