Why Noise-Cancelling Headphones Work Better at Low Frequencies

On a train, the steady rumble of the engine can almost disappear when you switch on noise-cancelling headphones, yet a nearby keyboard click or sharp announcement still gets through. If the headphones can produce a sound wave that cancels unwanted noise, why can’t they cancel every sound equally well?

Noise-cancelling headphones generally work better on low-frequency sounds because their long wavelengths make the pressure easier to match across the small space around your ear. Higher-frequency sounds change phase much more rapidly with position and time, so a tiny error in microphone position, processing delay, or speaker output can spoil the cancellation. The headphones’ physical seal also blocks some sounds without electronic cancellation.

The key is that the headphones must produce the right pressure disturbance at your eardrum. Producing an inverted waveform somewhere else isn’t enough.

01Sound cancellation starts with air moving back and forth

Imagine a row of air particles between a loudspeaker and your ear.

When the loudspeaker diaphragm moves forwards, it compresses the neighbouring air. When it moves backwards, the air spreads out slightly. These regions of compression and rarefaction travel away from the speaker.

The individual air particles don’t travel all the way from the loudspeaker to your ear. They oscillate backwards and forwards around their equilibrium positions, transferring the disturbance through the air.

This makes sound in air a longitudinal mechanical wave: the particles oscillate parallel to the direction in which the wave travels.

At one position, a passing sound wave produces a changing air pressure.

We’ll use \(p\) to mean the pressure disturbance relative to the normal atmospheric pressure:

• \(p>0\): the air pressure is temporarily higher than normal.
• \(p<0\): the air pressure is temporarily lower than normal.
• \(p=0\): the pressure is momentarily equal to normal atmospheric pressure.

For example, suppose unwanted noise produces a pressure disturbance of \(+0.30\ \mathrm{Pa}\) at your eardrum.

If the headphones simultaneously produce \(-0.30\ \mathrm{Pa}\) at that same position, the disturbances add to zero.

This is destructive interference.

But there’s an important condition. To cancel a continuous sound completely, the unwanted sound and the anti-noise must have equal pressure amplitudes and remain \(180^\circ\) out of phase at the position being protected.

For two sinusoidal waves, this means that every compression from one wave arrives with an equal rarefaction from the other.

The general rule is superposition, covered in more detail in Sound Interference and Superposition for HSC Physics.

Here, the harder problem is maintaining that interference while sound travels through a real headphone.

02Build an experiment to test whether cancellation actually works

Consider a simplified experiment using an over-ear headphone, a test loudspeaker, and an artificial ear.

We want to answer a specific question:

Does reducing the pressure at the headphone’s microphone necessarily reduce the pressure at the eardrum?

The first apparatus: an outside microphone and an internal speaker

Arrange the apparatus in this order.

1. Place a test loudspeaker in front of the headphone. It produces a steady tone of known frequency.
2. Fit the headphone over an artificial ear, keeping the ear cushion sealed against the test fixture.
3. Attach a microphone to the outside of the earcup. It detects the incoming noise before that sound has travelled through the headphone.
4. Connect this microphone to an electronic controller, which calculates an anti-noise signal.
5. Send the controller’s output to the headphone driver, the small loudspeaker inside the earcup.
6. Place a separate, small probe microphone near the artificial eardrum. Connect it to a recording instrument to measure the resulting sound pressure.

The outside microphone and headphone driver form a feedforward active noise cancellation system.

The probe microphone is independent of the control system. Its job is to tell us whether the pressure at the intended listening position has actually decreased.

What should the experiment show?

Suppose the test speaker produces a continuous 120 Hz tone.

The outside microphone detects it, the controller estimates the sound that will reach the ear, and the headphone driver produces an opposing disturbance.

If the controller predicts the arriving pressure perfectly, the probe microphone at the eardrum should record zero resultant pressure disturbance from that tone.

Now replace the 120 Hz tone with a 2400 Hz tone.

Would the same cancellation method still work perfectly?

In an ideal mathematical model, yes. Two equal-amplitude waves that are exactly out of phase at the ear cancel regardless of frequency.

But achieving those conditions becomes more difficult at higher frequencies.

An illustrative result

The following values are invented to demonstrate a plausible experimental pattern. They are not measurements from a particular headphone.

Assume the outside reference microphone measures the same pressure amplitude at both frequencies. For every comparison, keep the sound source, headphone fit, and microphone positions unchanged.

At 120 Hz, switching on active noise cancellation reduces the pressure amplitude from 0.80 Pa to 0.08 Pa.

At 2400 Hz, it decreases only slightly.

Notice something less obvious: even with ANC switched off, the high-frequency pressure at the ear is smaller.

The earcup and cushion can provide passive sound isolation. Some high-frequency sound is reflected, absorbed, or otherwise prevented from entering the enclosed space.

So these results support two different conclusions:

The active electronic system is much more effective at 120 Hz in this example.

The headphone’s physical construction is already reducing the 2400 Hz sound before ANC is switched on.

A smaller ANC-on measurement does not automatically mean better active cancellation. You must compare ANC on and off under the same conditions.

The percentage reductions in this table refer to pressure amplitude, not sound intensity or perceived loudness.

03Why wavelength makes a few centimetres important

A low-frequency sound wave has a long wavelength. A high-frequency sound wave has a short one.

For sound travelling through air,

\[ v=f\lambda \]

where:

• \(v\) is the wave speed in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(f\) is the frequency in hertz (\(\mathrm{Hz}\)).
• \(\lambda\) is the wavelength in metres (\(\mathrm{m}\)).

For our calculations, assume sound travels at \(340\ \mathrm{m\,s^{-1}}\).

Think about the distance between a microphone inside an earcup and your eardrum. It might be only a few centimetres.

At a low frequency, that distance is a tiny fraction of a wavelength. At a high frequency, it may be a substantial fraction.

This matters because the phase of a travelling wave changes as it moves.

04Why a small timing error can ruin cancellation

Position isn’t the only problem. The electronics also need time to process the noise and drive the loudspeaker.

Consider the path through a feedforward ANC system:

The outside microphone detects the disturbance. The controller estimates how it will reach the ear. The driver moves, producing another pressure wave. That new disturbance then travels through the air towards the eardrum.

Every stage introduces a response time.

Imagine the controller produces the correct opposing waveform, but it reaches the ear slightly later than intended.

For a slow oscillation, that delay might represent only a small part of the wave’s period.

For a rapid oscillation, the same delay could represent half a cycle.

The period \(T\) is the time taken to complete one oscillation:

\[ T=\frac{1}{f} \]

For an uncompensated timing error \(\tau\), the magnitude of the resulting phase error is

\[ \delta=360^\circ f\tau \]

Here, \(f\) is in hertz, \(\tau\) is in seconds, and \(\delta\) is in degrees.

The equation explains why the same timing error becomes more serious as frequency increases.

There’s an important qualification, though.

A perfectly steady sinusoidal tone repeats. If a controller knows its frequency and acoustic path accurately, it can adjust the phase to compensate for delay.

A sudden click is different. Its waveform changes rapidly and contains a range of frequencies. A correction that matches one component may not match the others, and a feedback system cannot respond to a completely unexpected disturbance before detecting it.

This is why stable low-frequency hums are usually easier targets than sharp, unpredictable sounds.

It isn’t because high-frequency interference is physically impossible. It’s because achieving the required pressure and phase becomes harder.

05The earcup can create standing waves

So far, we’ve mainly pictured travelling sound waves moving through the air.

But what happens when sound encounters the headphone driver, earcup walls, cushion, or eardrum?

Some of it reflects.

An incoming wave can overlap with a reflected wave travelling in the opposite direction. At suitable frequencies and boundary conditions, this produces a standing-wave pattern.

Unlike a travelling wave, a standing wave has fixed positions where the oscillation amplitude is small or zero, called nodes, and positions where it is largest, called antinodes.

For sound in air, you must distinguish pressure oscillations from particle displacement oscillations.

They don’t have their nodes in the same places.

Pressure nodes and displacement nodes

Imagine a narrow air-filled tube with one end rigidly closed and the other open.

At the rigidly closed end, air particles cannot move backwards and forwards through the boundary.

That makes the closed end a particle displacement node.

For an ideal standing-wave resonance, the same location is a pressure antinode.

At the open end, the air pressure disturbance is approximately zero because the opening is exposed to the surrounding atmosphere.

That makes the open end approximately a pressure node and a particle displacement antinode.

These conditions are particularly useful when studying resonance in air columns.

The ear canal provides a related example. When uncovered, it can be approximated as a tube open at its entrance and closed at the eardrum.

The approximation is imperfect because the eardrum moves and the opening isn’t an ideal pressure-release boundary. Covering the ear with a headphone changes the acoustic conditions at the entrance even further.

Why a quiet microphone may not mean a quiet eardrum

Suppose a microphone is placed close to a pressure node of a standing wave.

It could record a very small pressure amplitude.

A few centimetres away, the same standing wave could have a pressure antinode with a much larger amplitude.

That means a controller can make the pressure at its internal microphone extremely small without guaranteeing the same reduction at the eardrum.

This is a spatial measurement problem, not a failure of the superposition principle.

And it becomes more significant when the wavelength is comparable to the dimensions inside the headphone.

At sufficiently low frequencies, the wavelength is so long that large spatial differences are less likely across a small earcup.

06Revise the experiment: measure the remaining noise inside the headphone

The original experimental arrangement used an outside microphone to predict what sound would reach the ear.

Its weakness is that the outside microphone cannot directly observe the sound after it has passed through the earcup, reflected from surfaces, and combined with the driver’s output.

A revised design adds an internal microphone.

This is the idea behind feedback ANC.

The internal microphone detects the remaining pressure disturbance inside the headphone. The controller uses that information to adjust its output.

Systems combining external feedforward and internal feedback microphones are called hybrid ANC systems.

The extra internal microphone helps, but it doesn’t make the problem disappear.

The feedback microphone hears a mixture of external noise and sound produced by the headphone itself. The controller must account for the playback signal and the acoustic response of the driver.

It also needs to remain stable despite processing delays.

And the internal microphone still isn’t necessarily at the eardrum.

How to tell what is actually causing a poor result

A stronger experiment would keep the external source level and headphone fit fixed, and record pressure at the artificial eardrum with ANC switched off and on.

It would also record pressure at the internal feedback microphone.

Testing several frequencies helps reveal whether the difference changes with wavelength.

Moving the independent probe microphone around the cavity can reveal spatial variations. A pressure minimum at one position and a maximum nearby suggest that wave interference and reflections are affecting the measurement.

Finally, repeat the measurements after deliberately changing the earcup seal.

A leak changes both the amount of external noise entering the cavity and the acoustic path experienced by the anti-noise. It may therefore change passive isolation and active cancellation at the same time.

The diagnostic distinction is:

The main experimental rule is simple: measure the pressure where you want cancellation, compare against the correct baseline, and change one experimental condition at a time.

07Questions and solutions

The following questions build from wavelength calculations to the harder problem of using microphones to infer pressure at a different location.

All numerical setups are original, idealised scenarios rather than data from commercial headphones.

Question 1

A headphone is tested with a 100 Hz motor hum and a 2000 Hz electronic tone.

The distance from an internal microphone to the eardrum is 4.0 cm.

Assume the speed of sound in air is \(340\ \mathrm{m\,s^{-1}}\).

a) Calculate the wavelength of each sound.

b) Calculate the distance as a fraction of a wavelength for each frequency.

c) Explain which sound is more likely to have a similar pressure phase at both positions.

Solution 1

The 100 Hz sound is more likely to have a similar phase at the microphone and eardrum because its wavelength is much longer than their separation.

For part (a), use

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

At 100 Hz,

\[ \begin{aligned} \lambda &=\frac{340}{100}\ \mathrm{m}\\ &=3.40\ \mathrm{m} \end{aligned} \]

At 2000 Hz,

\[ \begin{aligned} \lambda &=\frac{340}{2000}\ \mathrm{m}\\ &=0.170\ \mathrm{m} \end{aligned} \]

For part (b), the separation is \(d=0.040\ \mathrm{m}\).

At 100 Hz,

\[ \frac{d}{\lambda} =\frac{0.040}{3.40} =0.0118 \]

The separation is about 1.18% of a wavelength.

At 2000 Hz,

\[ \frac{d}{\lambda} =\frac{0.040}{0.170} =0.235 \]

The separation is about 23.5% of a wavelength.

For part (c), the phase changes much less across the separation at 100 Hz.

For a single travelling wave, these fractions correspond to approximately \(4.2^\circ\) and \(84.7^\circ\), respectively.

This makes the low-frequency microphone reading a more reliable indication of the phase at the eardrum, assuming otherwise similar acoustic conditions.

The tempting shortcut is to say that lower frequency automatically means greater cancellation. Wavelength explains why matching is easier, but actual cancellation still depends on amplitude, phase, and the headphone’s acoustic response.

Question 2

A simplified ANC controller produces an anti-noise waveform with the correct amplitude and intended opposite phase.

However, an uncompensated timing error means the anti-noise arrives 0.20 ms later than the ideal cancellation signal.

The unwanted sound has pressure amplitude \(0.30\ \mathrm{Pa}\) at the eardrum. Assume the anti-noise has the same amplitude, with reflections neglected.

The controller is tested separately at 125 Hz and 2500 Hz.

a) Calculate the period of each tone.

b) Calculate the phase error at each frequency.

c) Determine the resultant pressure amplitude at 2500 Hz.

d) An engineer suggests adding another 0.20 ms of delay. Explain why this could restore ideal cancellation for a perfectly steady 2500 Hz tone, yet still be a poor general solution.

Solution 2

The timing error causes only a \(9^\circ\) phase mismatch at 125 Hz, but it reverses the intended cancellation at 2500 Hz. At 2500 Hz, the resulting pressure amplitude doubles to \(0.60\ \mathrm{Pa}\).

For part (a), use

\[ T=\frac{1}{f} \]

At 125 Hz,

\[ \begin{aligned} T &=\frac{1}{125}\ \mathrm{s}\\ &=0.0080\ \mathrm{s}\\ &=8.0\ \mathrm{ms} \end{aligned} \]

At 2500 Hz,

\[ \begin{aligned} T &=\frac{1}{2500}\ \mathrm{s}\\ &=0.00040\ \mathrm{s}\\ &=0.40\ \mathrm{ms} \end{aligned} \]

For part (b), a complete period represents \(360^\circ\).

The phase error is

\[ \delta=360^\circ\frac{\tau}{T} \]

where \(\tau=0.20\ \mathrm{ms}\).

At 125 Hz,

\[ \begin{aligned} \delta &=360^\circ\frac{0.20}{8.0}\\ &=9^\circ \end{aligned} \]

At 2500 Hz,

\[ \begin{aligned} \delta &=360^\circ\frac{0.20}{0.40}\\ &=180^\circ \end{aligned} \]

For part (c), the anti-noise was intended to be \(180^\circ\) out of phase with the unwanted sound.

But the additional \(180^\circ\) error at 2500 Hz changes that relationship to \(360^\circ\), equivalent to \(0^\circ\).

The two pressure oscillations now arrive in phase.

Therefore, their amplitudes add:

\[ \begin{aligned} A_{\text{net}} &=0.30+0.30\\ &=0.60\ \mathrm{Pa} \end{aligned} \]

Here, \(A_{\text{net}}\) is the resultant pressure amplitude.

Instead of cancelling the sound, the controller doubles its pressure amplitude in this ideal model.

For part (d), adding another 0.20 ms makes the total error 0.40 ms.

At 2500 Hz, that is exactly one complete period. A perfectly repeating sinusoidal wave would then be back in the correct phase relationship for cancellation.

But at 125 Hz, the total timing error would become

\[ \begin{aligned} \delta &=360^\circ\frac{0.40}{8.0}\\ &=18^\circ \end{aligned} \]

That is worse than the original \(9^\circ\) mismatch.

More importantly, an extra delay that happens to align one steady tone does not align an arbitrary mixture of frequencies or an unpredictable transient.

The subtle point is that delay does not always make cancellation worse at every frequency. Phase repeats every \(360^\circ\). A useful controller needs the correct frequency-dependent phase response, not merely the smallest or largest possible delay.

Question 3

Consider an idealised straight air path inside an acoustic test fixture.

The unwanted sound travels towards the ear position, while the anti-noise travels in the opposite direction.

Both are continuous 2000 Hz waves, with equal pressure amplitudes of \(0.20\ \mathrm{Pa}\).

At a monitoring microphone, the two pressure oscillations are exactly \(180^\circ\) out of phase.

The probe representing the eardrum position is 4.25 cm farther along the path of the unwanted sound.

Assume the waves travel in opposite directions without reflection, spreading, or attenuation, and \(v=340\ \mathrm{m\,s^{-1}}\).

a) Calculate the wavelength.

b) Determine the relative phase of the two waves at the eardrum position.

c) Calculate the resultant pressure amplitude at that position.

d) Explain why perfect cancellation at the monitoring microphone does not contradict the result at the eardrum.

Solution 3

The two waves cancel at the monitoring microphone but reinforce each other at the eardrum position, producing a resultant pressure amplitude of \(0.40\ \mathrm{Pa}\).

For part (a),

\[ \begin{aligned} \lambda &=\frac{340}{2000}\ \mathrm{m}\\ &=0.170\ \mathrm{m} \end{aligned} \]

For part (b), the separation is

\[ d=0.0425\ \mathrm{m} \]

Therefore,

\[ \frac{d}{\lambda} =\frac{0.0425}{0.170} =\frac14 \]

The eardrum probe is one-quarter wavelength from the monitoring microphone.

For each travelling wave, that corresponds to a phase change of

\[ 360^\circ\times\frac14=90^\circ \]

But the waves travel in opposite directions.

Moving from the microphone towards the eardrum changes the phase of one wave by \(90^\circ\) in one sense, and the phase of the other wave by \(90^\circ\) in the opposite sense.

Their relative phase therefore changes by \(180^\circ\).

They were initially \(180^\circ\) out of phase at the monitoring microphone.

At the eardrum, they are now in phase.

For part (c), constructive interference gives

\[ \begin{aligned} A_{\text{net}} &=0.20+0.20\\ &=0.40\ \mathrm{Pa} \end{aligned} \]

This is twice the pressure amplitude of the unwanted sound alone.

For part (d), superposition applies separately at every position.

At the microphone, the two pressure disturbances sum to zero. At the eardrum, their changed phase relationship makes them reinforce.

The tempting assumption is that two waves which are opposite at one location stay opposite everywhere.

That is false when their propagation directions differ.

This example also explains why a controller calibrated at one microphone position may behave poorly elsewhere. A measurement near the eardrum is needed to establish the actual effect at the ear.

The result depends on the idealised travelling-wave assumptions. Real headphone cavities contain reflections, which can create even more complicated spatial patterns.

Question 4

A researcher tests a headphone using a continuous 2200 Hz tone and an artificial ear.

The external loudspeaker produces the same pressure amplitude throughout the experiment.

The researcher measures pressure amplitude at two positions:

• An internal feedback microphone close to the headphone driver.
• An independent probe microphone near the artificial eardrum.

The researcher first tests a good cushion seal, then introduces a small gap in the cushion.

The following results are illustrative, not real experimental measurements.

a) Determine whether active cancellation improves the pressure amplitude at the eardrum for each cushion condition. Give the percentage change.

b) Explain why the feedback microphone gives a misleading impression of the headphone’s performance.

c) One student claims that introducing the gap improves cancellation. Another claims the gap makes the headphone perform worse. Explain what each student may be noticing.

d) Describe a controlled follow-up investigation that could distinguish passive isolation from spatial mismatch in the active cancellation system.

Solution 4

With a good seal, ANC increases the pressure amplitude at the eardrum by 20%. With the small gap, ANC reduces it by approximately 11.4%, but the final pressure remains higher than with the good seal.

For part (a), compare ANC on and off at the same eardrum probe for each cushion condition.

For the good seal, the pressure changes from 0.35 Pa to 0.42 Pa.

The percentage increase is

\[ \begin{aligned} \frac{0.42-0.35}{0.35}\times100\% =20\% \end{aligned} \]

The active system makes the sound pressure amplitude larger at the ear.

For the small gap, the pressure changes from 0.70 Pa to 0.62 Pa.

The percentage reduction is

\[ \begin{aligned} \frac{0.70-0.62}{0.70}\times100\% \approx11.4\% \end{aligned} \]

Here, active cancellation provides a small improvement.

For part (b), the feedback microphone shows a 90% reduction in pressure amplitude in both cases.

With a good seal,

\[ \frac{0.20-0.02}{0.20}\times100\% =90\% \]

With a gap,

\[ \frac{0.48-0.048}{0.48}\times100\% =90\% \]

Those results show that the controller is reducing pressure at its feedback microphone.

They do not show that it is reducing pressure at the eardrum.

At 2200 Hz, different positions inside the cavity may have significantly different pressure amplitudes and phases. Reflections and standing-wave effects can make the microphone’s local measurement unrepresentative of the ear position.

For part (c), the first student may be looking at the improvement from ANC off to ANC on.

With the gap, switching on ANC produces an 11.4% reduction at the eardrum. With the good seal, switching it on produces a 20% increase.

In that limited comparison, the active system provides a more favourable change with the gap.

The second student may be looking at the final sound pressure.

With ANC on, the good seal gives 0.42 Pa, while the gap gives 0.62 Pa.

The smaller absolute pressure amplitude occurs with the good seal.

Both observations are valid, but they measure different aspects of performance. The gap changes how much noise enters the headphone as well as the acoustic response the controller must compensate for.

Neither observation proves that a gap intrinsically improves an ANC algorithm.

For part (d), keep the external source level, test frequency, and microphone calibration fixed.

First, measure the eardrum-position pressure with no headphone, with the headphone fitted and ANC off, and with ANC on. Repeat the full set for each cushion condition.

Comparing no headphone with ANC off helps isolate the contribution of passive isolation.

Comparing ANC off with ANC on, without changing the fit, isolates the additional effect of the active system under that condition.

Next, keep the cushion condition fixed and measure both the amplitude and phase of the residual sound at several positions between the driver and the artificial eardrum.

A strong position-dependent change would support a spatial mismatch explanation.

A frequency sweep could also reveal narrow regions of poor cancellation associated with the cavity’s acoustic response.

The measurements would not automatically identify one unique cause. Timing errors, reflections, driver response, and leakage could interact.

The essential experimental improvement is controlling the baseline and measuring both where the controller listens and where cancellation is intended.

Question 5

An engineer develops an experimental headphone cavity that is deliberately simplified into a uniform air-filled tube of length \(L=5.00\ \mathrm{cm}\).

The headphone-driver end is at \(x=0\), and the eardrum end is at \(x=L\).

For this model, both ends are treated as perfectly rigid boundaries. The tube is lossless, and the speed of sound is \(340\ \mathrm{m\,s^{-1}}\).

A high-frequency disturbance excites the second non-zero standing-wave mode of the closed-closed air column.

Two internal error microphones are placed at

\[ x=\frac{L}{4} \quad\text{and}\quad x=\frac{3L}{4} \]

Both report zero pressure disturbance.

A third diagnostic microphone at the midpoint, \(x=L/2\), records

\[ p_{\text{mid}}(t) =0.40\cos(2\pi ft)\ \mathrm{Pa} \]

The pressure is measured relative to atmospheric pressure, and the sign of this oscillation is calibrated.

a) Calculate the frequency of the standing wave.

b) Explain why the first two microphones report zero while the midpoint microphone does not.

c) Determine the pressure disturbance at the eardrum, including its phase relative to the midpoint.

d) The engineer assumes the midpoint microphone represents the eardrum pressure. The controller therefore commands an anti-noise pressure at the eardrum equal to the negative of the midpoint signal.

Assume, for this part, that an ideal actuator can deliver exactly that additional local pressure, and use superposition to determine the resultant pressure amplitude at the eardrum.

e) Explain why this result is a particularly serious challenge for a microphone-based controller. Suggest a better measurement or calibration strategy, and explain why the same spatial failure is less likely at 100 Hz.

Solution 5

The controller doubles the pressure amplitude at the eardrum instead of cancelling it. The two silent microphones sit at standing-wave pressure nodes, and the midpoint microphone is \(180^\circ\) out of phase with the eardrum in the specified mode.

The result comes from the spatial structure of the standing wave.

For part (a), a closed-closed tube has pressure antinodes at both rigid boundaries.

In the second non-zero mode, the tube length contains two half-wavelengths:

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

Therefore,

\[ \lambda=L=0.0500\ \mathrm{m} \]

Using \(v=f\lambda\),

\[ \begin{aligned} f &=\frac{340}{0.0500}\\ &=6800\ \mathrm{Hz} \end{aligned} \]

The standing wave has a frequency of 6800 Hz.

For part (b), the second mode has pressure antinodes at

\[ x=0,\quad \frac{L}{2},\quad L \]

Its pressure nodes occur at

\[ x=\frac{L}{4}, \quad x=\frac{3L}{4} \]

The two error microphones are positioned exactly at the pressure nodes.

Their zero readings are therefore consistent with a strong standing wave elsewhere in the cavity.

They are not evidence that the disturbance has disappeared.

For part (c), we need the phase, not just the pressure amplitude.

For this ideal second mode, a suitable pressure expression is

\[ p(x,t)=C\cos(kx)\cos(2\pi ft) \]

where \(C\) is a signed pressure coefficient in pascals, \(k=2\pi/L\) is the wave number in radians per metre, and \(x\) is measured from the left boundary.

At the midpoint,

\[ kx=\frac{2\pi}{L}\frac{L}{2} =\pi \]

Since \(\cos\pi=-1\),

\[ p_{\text{mid}}(t) =-C\cos(2\pi ft) \]

The measured signal is \(+0.40\cos(2\pi ft)\ \mathrm{Pa}\), so

\[ C=-0.40\ \mathrm{Pa} \]

At the eardrum, \(x=L\).

Therefore,

\[ kL=2\pi \]

and \(\cos(2\pi)=1\).

The pressure disturbance at the eardrum is

\[ p_{\text{ear}}(t) =-0.40\cos(2\pi ft)\ \mathrm{Pa} \]

It has the same amplitude as the midpoint pressure, but the opposite phase.

For part (d), the controller mistakenly assumes the eardrum pressure equals the midpoint pressure.

It commands the negative of the midpoint reading:

\[ p_{\text{anti}}(t) =-0.40\cos(2\pi ft)\ \mathrm{Pa} \]

But the actual unwanted pressure at the eardrum is already

\[ p_{\text{ear}}(t) =-0.40\cos(2\pi ft)\ \mathrm{Pa} \]

The two disturbances have the same sign at every instant.

By superposition,

\[ \begin{aligned} p_{\text{net}}(t) &=p_{\text{ear}}(t)+p_{\text{anti}}(t)\\ &=-0.80\cos(2\pi ft)\ \mathrm{Pa} \end{aligned} \]

The resultant pressure amplitude is

\[ A_{\text{net}}=0.80\ \mathrm{Pa} \]

This is twice the original \(0.40\ \mathrm{Pa}\) amplitude.

To cancel the unwanted pressure at the eardrum, the ideal local anti-noise would instead need to be

\[ p_{\text{anti}}(t) =+0.40\cos(2\pi ft)\ \mathrm{Pa} \]

Surprisingly, that required signal is in phase with the midpoint microphone reading.

Simply inverting the microphone signal gives exactly the wrong result.

For part (e), the controller faces two separate measurement problems.

First, two microphones can both read zero even when a strong standing wave exists. Their positions may coincide with pressure nodes.

Second, a microphone with a non-zero reading can still have the opposite pressure phase from the eardrum.

Adding microphones doesn’t automatically solve the problem if their locations are poorly chosen or their readings are interpreted without accounting for the acoustic geometry.

A better experimental strategy is to place an independent probe near the eardrum and measure the amplitude and phase relationship between that position and the internal microphones over a range of frequencies.

The controller can then use the measured acoustic transfer relationship to estimate the pressure at the ear rather than assuming it equals the pressure at the microphone.

In a real device, this requires accounting for changing fit, leakage, reflections, and the response of the headphone driver. The perfectly rigid tube and exact nodes used here are mathematical idealisations.

Why is this particular spatial failure less likely at 100 Hz?

At 100 Hz,

\[ \begin{aligned} \lambda &=\frac{340}{100}\\ &=3.40\ \mathrm{m} \end{aligned} \]

The cavity length is only

\[ \frac{0.0500}{3.40} \approx0.0147 \]

of a wavelength.

For a single travelling wave, that corresponds to only about \(5.3^\circ\) of phase change across the entire cavity.

A 100 Hz sound cannot form the same second standing-wave mode in this ideal rigid-ended 5.00 cm tube. Its wavelength is far too long.

That doesn’t guarantee perfect low-frequency cancellation, but it removes the extreme spatial variation responsible for this example.

The deeper conclusion is that destructive interference must be established at the correct position. Neither an inverted signal nor a nearly zero microphone reading is sufficient evidence on its own.

The next useful experiment is to measure the acoustic transfer function between a microphone and the eardrum: how pressure amplitude and phase change between those positions as frequency changes. That measurement connects the HSC ideas of wave propagation, reflection, resonance, and superposition to the engineering problem that real noise-cancelling headphones must solve.