Why Two Speakers Create Quiet Spots in a Room

You can stand between two speakers playing the same note, move a short distance sideways, and suddenly hear the sound become much quieter. Neither speaker has turned down. Nothing has blocked the sound, either.

The explanation is destructive interference. The sound waves from the two speakers travel different distances to reach your ears. At certain positions, one wave arrives half a cycle behind the other. Their pressure variations oppose each other, reducing the resulting sound.

This explains the quiet spots. But there’s another complication: the walls, floor, and ceiling also reflect sound. In a real room, you’re hearing a combination of direct sound and reflected sound, so the quiet spots can be more complicated than a simple two-speaker calculation predicts.

01Why two speakers can cancel each other’s sound

Imagine two speakers producing the same steady musical note. Both emit sound waves at the same frequency and in the same phase.

Before considering the room, make a prediction.

If you stand exactly halfway between the speakers, will you hear a particularly quiet sound?

Actually, you’ll generally hear a louder sound.

Because you’re equally far from both speakers, the sound waves travel equal distances. They arrive in phase, meaning their pressure variations reinforce each other.

The quieter positions occur elsewhere, where the difference in the distances travelled causes the waves to arrive out of phase.

Start with what happens when waves overlap

Sound travels through air as a longitudinal wave. It consists of changing regions of higher and lower pressure.

A compression is a region where air pressure is temporarily higher than normal. A rarefaction is a region where it is temporarily lower.

When sound waves overlap, their pressure variations add. This is called the principle of superposition.

Consider two waves arriving at the same point.

• If both produce a positive pressure variation at the same time, they reinforce each other.
• If both produce a negative pressure variation, they also reinforce each other.
• If one produces a positive variation while the other produces an equal negative variation, they cancel.

The last situation is destructive interference.

For two equal-amplitude waves, the difference is:

Notice that destructive interference does not mean the sound waves have disappeared.

The individual waves continue to exist. Their pressure variations simply add to a smaller resultant at that particular location.

Sound energy is not magically destroyed. Interference redistributes the sound energy through the surrounding space, while absorption at surfaces and in the air gradually converts some of it into other forms of energy.

There is also an important distinction between pressure and intensity. Sound intensity is proportional to the square of the pressure amplitude:

\[ I\propto p_{\max}^{\,2} \]

where \(I\) is sound intensity and \(p_{\max}\) is the maximum pressure variation.

This means even a substantial reduction in pressure amplitude can produce a noticeable reduction in the sound you hear.

02How path difference creates quiet spots

The key to predicting where destructive interference occurs is the difference between the distances travelled by the two waves.

This is called the path difference.

Imagine two speakers, A and B, producing sound waves that reach a listener at point P.

Let:

• \(r_A\) = distance travelled from speaker A to P, in metres
• \(r_B\) = distance travelled from speaker B to P, in metres
• \(\Delta r\) = path difference, in metres

We will use the sign convention:

\[ \Delta r=r_B-r_A \]

A positive path difference means the wave from B has travelled further.

If both speakers emit sound in phase, the wave travelling the longer distance arrives later.

How much later depends on the wavelength.

The wavelength, represented by \(\lambda\), is the distance between two consecutive points in the same phase of a wave.

It is related to wave speed and frequency by:

\[ v=f\lambda \]

Rearranging:

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

where:

• \(v\) = speed of sound, in \(\text{m s}^{-1}\)
• \(f\) = frequency, in hertz (Hz)
• \(\lambda\) = wavelength, in metres (m)

A difference of one complete wavelength means the second wave has travelled one extra cycle. It therefore arrives back in phase.

A difference of half a wavelength means it has travelled an extra half-cycle. It arrives 180° out of phase.

This gives us the two important conditions.

Constructive interference

For speakers that initially emit waves in phase:

\[ |\Delta r|=m\lambda \]

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

The waves arrive in phase and reinforce each other.

Destructive interference

For speakers that initially emit waves in phase:

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

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

The waves arrive 180° out of phase and oppose each other.

These conditions explain why a room can have several quiet spots rather than just one.

As you move around, the distances from the two speakers change. So does the path difference. Whenever the path difference becomes an odd number of half-wavelengths, destructive interference can occur.

For two ideal point sources, the positions satisfying each particular path-difference condition form hyperbolic curves in a two-dimensional horizontal plane. In three dimensions, the corresponding surfaces are hyperboloids.

This is why quiet regions can form curved bands rather than neat, evenly spaced straight lines.

If you want to practise finding the actual distance difference between two sources and an observation point, see How to Calculate Path Difference Between Two Wave Sources.

04An important complication: The speakers might not start in phase

So far, we have assumed that both speakers begin producing their sound waves at exactly the same stage of the oscillation.

This is a useful starting model, but it isn’t always true.

Two speakers can produce waves of the same frequency while having a phase difference at their sources.

For example, one speaker might produce its maximum compression a quarter of a cycle before the other.

This means path difference alone is not enough to determine whether the waves reinforce or cancel.

We must consider two contributions:

1. The phase difference already present when the speakers emit the waves.
2. The additional phase difference created by the different distances travelled.

Using our earlier sign convention, the phase difference at P is:

\[ \Delta\phi_P = \Delta\phi_s-\frac{2\pi\Delta r}{\lambda} \]

where:

• \(\Delta\phi_P\) = phase difference at P, in radians
• \(\Delta\phi_s\) = initial phase difference between speaker B and speaker A, in radians
• \(\Delta r=r_B-r_A\) = signed path difference, in metres
• \(\lambda\) = wavelength, in metres

The negative sign accounts for the delay caused by speaker B travelling further when \(\Delta r\) is positive.

The final phase difference determines the interference:

\[ \Delta\phi_P=2m\pi \]

for constructive interference, and

\[ \Delta\phi_P=(2m+1)\pi \]

for destructive interference, where \(m\) is an integer.

You can also work in degrees rather than radians, using 360° for a complete cycle.

For a more detailed explanation of this calculation, see How to Tell Whether Two Waves Arrive in Phase in HSC Physics.

05Why walls and boundaries create additional quiet spots

Now we can introduce the complication that makes real rooms more interesting.

The two-speaker model considers only the direct sound travelling from each speaker to the listener.

But sound doesn’t simply stop when it reaches a wall.

It reflects.

The sound travelling towards a wall can bounce back into the room. That reflected wave travels along another path and overlaps with the direct sound.

The reflected wave is effectively another contribution to the superposition at your listening position.

What happens when sound reflects?

The phase of a reflected wave depends on the boundary.

For an ideal rigid wall, the reflected sound pressure wave is not inverted at the boundary. Its pressure variation reflects in phase.

This is different from the behaviour of a wave reflecting from an ideal pressure-release boundary, where the pressure reflection is inverted.

There is another important factor: the reflected wave travels a different distance.

Therefore, its phase at the listener depends on both:

• The phase change associated with reflection at the boundary.
• The additional distance travelled.

This is exactly the same underlying reasoning we used for two speakers, except that now we’re comparing direct and reflected waves.

A real room can contain many reflected waves arriving from different directions. The listener experiences their combined effect.

Standing waves and room resonances

Repeated reflections can create an especially interesting pattern called a standing wave.

A standing wave forms when waves of the same frequency and amplitude travel in opposite directions and repeatedly superpose.

Instead of the entire pattern travelling through the room, certain positions remain fixed.

For a sound wave, we distinguish between:

• A pressure antinode: a position where the pressure variation has maximum amplitude.
• A pressure node: a position where the pressure variation has minimum amplitude.

A pressure node can therefore be a particularly quiet position.

Consider a simplified room model with two parallel, rigid walls separated by a distance \(L\).

For a standing-wave mode to form along this dimension:

\[ L=\frac{n\lambda}{2} \]

where \(n=1,2,3,\ldots\) represents the mode number.

Combining this with \(v=f\lambda\):

\[ f_n=\frac{nv}{2L} \]

This equation gives the resonant frequencies for the idealised one-dimensional room.

For the fundamental mode, \(n=1\).

The pressure variation is greatest at the rigid walls and smallest halfway between them.

That means the centre of the room can be a pressure node, even if only one speaker is producing the sound.

This is a useful distinction: a quiet spot does not necessarily require two separate speakers. A direct wave and its reflection can also produce destructive interference.

In a real room, the boundaries are not perfectly rigid, and some sound is absorbed. The standing-wave pattern is therefore less perfect than the ideal model.

Direct interference versus room boundaries

It’s worth separating these two effects when answering HSC questions.

In an actual room, both effects occur together.

This is why calculating the path difference between two speakers can predict a quiet region without perfectly predicting the sound level you will measure there.

06How could you test why the quiet spots occur?

A useful experimental investigation would separate the direct-wave interference from the effects of room reflections.

Imagine setting up two matched speakers, both connected to the same signal generator through an amplifier.

The signal generator produces a steady sinusoidal tone. The two speakers are placed a fixed distance apart, facing into an open measurement area.

A microphone connected to a sound-level meter is then moved through a series of positions. At each position, the sound pressure level is recorded.

What should the experiment show?

Suppose the speakers produce a frequency of \(680\text{ Hz}\).

Using a sound speed of \(340\text{ m s}^{-1}\):

\[ \lambda=\frac{340}{680} =0.50\text{ m} \]

The first destructive interference condition occurs when:

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

Therefore:

\[ \Delta r=0.25\text{ m} \]

The prediction is that the microphone should record a lower sound level at positions where the two direct paths differ by \(0.25\text{ m}\).

Further minima should occur at path differences of \(0.75\text{ m}\), \(1.25\text{ m}\), and so on.

What might go wrong with the original setup?

If this experiment is performed in an ordinary classroom, the microphone also detects reflected sound.

The walls, ceiling, floor, and furniture all affect the measured pressure.

Consequently, the observed minima may not occur exactly where the simple two-speaker model predicts. Some may be deeper, while others may be less noticeable.

The first experiment has not disproved superposition. It has revealed that the model leaves out additional waves.

How could the experiment be improved?

Repeat the measurement in an anechoic chamber, where the walls, floor, and ceiling are designed to absorb sound rather than reflect it.

Also ensure that:

• Both speakers receive the same frequency and maintain a stable phase relationship.
• Their outputs are calibrated to give comparable pressure amplitudes.
• The microphone remains at a consistent height and orientation.
• Background noise is kept low.

This revised setup reduces the contribution from reflected waves.

The measured minima should then agree more closely with the predictions from the direct two-source interference model.

A separate experiment in a normal room can then investigate how changing the room boundaries alters the sound distribution.

The distinction is important: the experiment tests superposition, while the differences between the ideal prediction and the real room measurement reveal the influence of boundaries.

07Questions and solutions

The following questions move from straightforward path-difference calculations to situations where source phase and room reflections must be considered together.

Question 1

Two identical speakers, A and B, produce sound waves in phase at a frequency of \(850\text{ Hz}\).

The speed of sound is \(340\text{ m s}^{-1}\).

At point P, the sound from speaker A travels \(5.10\text{ m}\), while the sound from speaker B travels \(5.30\text{ m}\).

Determine:

a) The wavelength of the sound.

b) The path difference at P.

c) Whether constructive or destructive interference occurs.

Assume the pressure amplitudes of the two waves are equal at P.

Solution 1

The waves undergo complete destructive interference at P in the idealised model, because they arrive half a wavelength out of phase and have equal amplitudes.

Step 1: Calculate the wavelength.

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

Substitute:

\[ \lambda=\frac{340}{850} \]

\[ \boxed{\lambda=0.40\text{ m}} \]

Step 2: Calculate the path difference.

\[ \Delta r=r_B-r_A \]

\[ =5.30-5.10 \]

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

Step 3: Compare the path difference with the wavelength.

\[ \frac{\Delta r}{\lambda} =\frac{0.20}{0.40} =0.5 \]

Therefore:

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

Step 4: Identify the interference.

A half-wavelength path difference produces a 180° phase difference when the speakers initially emit in phase.

The waves therefore arrive in opposite phases.

Since their amplitudes are equal, their pressure variations cancel completely.

P is an ideal pressure node.

Question 2

Two speakers produce sound waves in phase at a frequency of \(500\text{ Hz}\).

The speed of sound is \(340\text{ m s}^{-1}\).

At point Q, the path lengths are:

• Speaker A: \(3.00\text{ m}\)
• Speaker B: \(3.34\text{ m}\)

The speakers have equal pressure amplitudes at Q.

The frequency is then reduced to \(425\text{ Hz}\), without moving either speaker or the listener.

Determine:

a) Why the original position is a point of complete destructive interference.

b) The new wavelength.

c) The new phase difference at Q.

d) Whether Q remains a point of complete destructive interference.

Solution 2

Q is initially a point of complete destructive interference, but it is no longer a point of complete destructive interference after the frequency decreases.

Step 1: Calculate the original wavelength.

\[ \lambda_1=\frac{340}{500} \]

\[ \lambda_1=0.68\text{ m} \]

Step 2: Calculate the path difference.

\[ \Delta r=3.34-3.00 \]

\[ \Delta r=0.34\text{ m} \]

Compare:

\[ \Delta r=\frac{0.68}{2} \]

The path difference is half a wavelength.

Therefore, the waves initially arrive 180° out of phase.

Because their amplitudes are equal, they cancel completely.

Step 3: Calculate the new wavelength.

The speaker positions and sound speed remain unchanged, but the frequency decreases.

\[ \lambda_2=\frac{v}{f_2} \]

\[ =\frac{340}{425} \]

\[ \boxed{\lambda_2=0.80\text{ m}} \]

Step 4: Calculate the new phase difference.

The path difference remains \(0.34\text{ m}\).

However, it now represents a different fraction of a wavelength:

\[ \frac{\Delta r}{\lambda_2} =\frac{0.34}{0.80} =0.425 \]

The phase difference is:

\[ \Delta\phi_P =360^\circ(0.425) \]

\[ \boxed{\Delta\phi_P=153^\circ} \]

Step 5: Determine whether complete cancellation still occurs.

Complete destructive interference requires a phase difference of 180° (or an equivalent odd multiple).

The new phase difference is 153°.

Therefore, the waves are still substantially opposing each other, but they are no longer exactly out of phase.

Q is no longer a point of complete destructive interference.

The tempting mistake is to assume that because the physical path difference has not changed, the interference condition must remain unchanged.

But the wavelength has changed. The path difference must always be compared with the current wavelength.

Question 3

A narrow experimental room is modelled as a one-dimensional space between two parallel, rigid walls separated by \(2.0\text{ m}\).

A single speaker near one end produces a steady \(85\text{ Hz}\) tone.

The speed of sound is \(340\text{ m s}^{-1}\).

Assume that the room supports an ideal fundamental standing-wave mode, with negligible energy losses.

a) Calculate the wavelength.

b) Show that the room length corresponds to the fundamental resonant mode.

c) Explain why a quiet position can occur halfway between the walls, even though only one speaker is operating.

Solution 3

A pressure node occurs halfway between the walls because the incident and reflected sound waves combine to form the fundamental standing-wave pattern.

Step 1: Calculate the wavelength.

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

\[ =\frac{340}{85} \]

\[ \boxed{\lambda=4.0\text{ m}} \]

Step 2: Compare the room length with the wavelength.

The room length is:

\[ L=2.0\text{ m} \]

Therefore:

\[ L=\frac{4.0}{2} \]

\[ L=\frac{\lambda}{2} \]

This satisfies the fundamental standing-wave condition.

Alternatively, calculate the fundamental resonant frequency:

\[ f_1=\frac{v}{2L} \]

\[ =\frac{340}{2(2.0)} \]

\[ f_1=85\text{ Hz} \]

This matches the frequency produced by the speaker.

Step 3: Explain the quiet position.

The sound travels towards the rigid wall and reflects back into the room.

The incident and reflected pressure waves travel in opposite directions and superpose.

At the fundamental mode, the pressure variation is greatest at the rigid walls and smallest halfway between them.

The midpoint is therefore a pressure node.

This produces a quiet position even though only one speaker is operating.

The important point is that destructive interference can occur between a wave and its own reflection. Two separate speakers are not essential.

Question 4

Two speakers, A and B, produce coherent sound waves at \(400\text{ Hz}\).

The speed of sound is \(320\text{ m s}^{-1}\).

Speaker B initially emits its pressure wave 180° out of phase with speaker A.

At point P, the wave from B travels \(0.40\text{ m}\) further than the wave from A.

The two waves have equal pressure amplitudes at P.

Determine whether constructive or destructive interference occurs at P. Explain why using only the path difference would lead to an incorrect conclusion.

Solution 4

The waves undergo constructive interference at P because the initial 180° phase difference is cancelled by the additional 180° phase delay from B’s longer path.

Step 1: Calculate the wavelength.

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

\[ =\frac{320}{400} \]

\[ \boxed{\lambda=0.80\text{ m}} \]

Step 2: Compare the path difference with the wavelength.

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

\[ \frac{\Delta r}{\lambda} =\frac{0.40}{0.80} =0.5 \]

The path difference is half a wavelength.

Therefore, the additional phase difference caused by travelling further is:

\[ 180^\circ \]

Step 3: Include the initial source phase difference.

Using the signed phase convention:

\[ \Delta\phi_P =\Delta\phi_s -360^\circ\frac{\Delta r}{\lambda} \]

Substitute:

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

\[ \boxed{\Delta\phi_P=0^\circ} \]

Step 4: Identify the interference.

The waves arrive in phase.

They therefore undergo constructive interference.

The incorrect approach would be to see a half-wavelength path difference and immediately conclude that the waves must cancel.

That rule assumes the sources initially emit in phase.

Here, speaker B starts out inverted relative to A. Its additional travel distance introduces another phase reversal, bringing the two waves back into phase at P.

Question 5

This final question combines direct interference, reflected sound, and the phase change associated with a boundary.

Two speakers, A and B, produce coherent sound waves in phase at a frequency of \(680\text{ Hz}\).

The speed of sound is \(340\text{ m s}^{-1}\).

At point P, three sound contributions are present:

• The direct wave from A travels \(4.00\text{ m}\).
• The direct wave from B travels \(4.25\text{ m}\).
• A wave from A reflects off a nearby rigid wall before reaching P. Its total travel distance is \(4.25\text{ m}\).

The following idealised conditions apply:

• All three waves have equal pressure amplitudes at P.
• Reflection from the rigid wall does not invert the reflected pressure wave.
• There are no other significant reflections.
• All three waves have the same frequency.

A student considers only the two direct waves from A and B and concludes that P must be completely silent.

Is the student correct?

Determine the actual resultant pressure amplitude at P in terms of the pressure amplitude \(p_0\) of each individual contribution. Explain the role of the reflected wave.

Solution 5

The student is incorrect. The resultant pressure amplitude is \(p_0\), not zero.

The student has considered the two direct waves correctly but has ignored the additional reflected wave.

Step 1: Calculate the wavelength.

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

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

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

Step 2: Compare the two direct paths.

The direct path lengths are:

\[ r_A=4.00\text{ m} \]

\[ r_B=4.25\text{ m} \]

Their path difference is:

\[ \Delta r=4.25-4.00 \]

\[ \Delta r=0.25\text{ m} \]

Therefore:

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

Since the speakers initially emit in phase, their direct waves arrive 180° out of phase.

Choose the direct pressure variation from A to be positive at a particular instant.

The direct wave from B must then have an equal negative pressure variation.

The two direct waves cancel.

Step 3: Consider the reflected wave from A.

The reflected path is \(4.25\text{ m}\), while the direct path from A is \(4.00\text{ m}\).

Therefore, the reflected wave travels an additional:

\[ \Delta r_{\text{reflection}} =4.25-4.00 \]

\[ =0.25\text{ m} \]

This is also half a wavelength.

Because the wall is rigid, the reflected pressure wave is not inverted at the boundary.

The additional half-wavelength path therefore makes the reflected wave arrive 180° out of phase with A’s direct wave.

Step 4: Add all three contributions.

At the chosen instant, let the direct wave from A have pressure variation \(+p_0\).

The direct wave from B has pressure variation:

\[ -p_0 \]

The reflected wave from A also has pressure variation:

\[ -p_0 \]

Therefore:

\[ p_{\text{resultant}} =p_0-p_0-p_0 \]

\[ \boxed{p_{\text{resultant}}=-p_0} \]

The negative sign indicates the direction of the pressure variation at that instant.

The resultant pressure amplitude is:

\[ \boxed{p_{\max}=p_0} \]

The surprising part is that the reflected wave reinforces the direct wave from B.

Although A’s direct wave and B’s direct wave cancel, the reflected wave from A arrives in phase with B’s direct wave.

The final result is therefore equivalent to one wave of amplitude \(p_0\).

This is why the two-speaker calculation alone cannot determine the actual sound level in a room when reflections are significant.

The question is not simply whether two waves cancel. It is whether every significant wave arriving at the listening position has been included in the superposition.

08Where to take this concept next

You now have three separate ideas to keep track of:

1. Path difference tells you how much further one wave travels than another.
2. Phase difference tells you how the waves line up when they reach the same position.
3. Boundary effects introduce reflected waves, which can create additional interference patterns and standing waves.

The next useful connection is standing-wave node spacing.

For a standing wave, adjacent nodes are separated by half a wavelength, while a node and its nearest antinode are separated by a quarter wavelength.

Understanding why those positions occur will help you connect the quiet spots created by two speakers with the fixed nodes and antinodes formed by repeated wave reflection.