How to Calculate the Fundamental Frequency of an Open Air Column

A tube is \(0.80\ \mathrm{m}\) long and open at both ends. A loudspeaker produces sound near one opening. If the speed of sound is \(340\ \mathrm{m\,s^{-1}}\), should the tube’s lowest resonant frequency be \(212.5\ \mathrm{Hz}\) or \(425\ \mathrm{Hz}\)?

You might predict \(425\ \mathrm{Hz}\), assuming one complete wavelength must fit inside the tube. But the correct answer is \(212.5\ \mathrm{Hz}\). The fundamental standing wave only needs half a wavelength to fit between the open ends.

To calculate the fundamental frequency of an air column open at both ends, use:

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

where:

• \(f_1\) is the fundamental frequency in hertz (\(\mathrm{Hz}\)).
• \(v\) is the speed of sound in air in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(L\) is the length of the air column in metres (\(\mathrm{m}\)).

For our \(0.80\ \mathrm{m}\) tube:

\[
\begin{aligned}
f_1&=\frac{340}{2(0.80)}\\
&=212.5\ \mathrm{Hz}
\end{aligned}
\]

This is the lowest non-zero resonant frequency predicted by the ideal open-pipe model.

But why does the denominator contain \(2L\) rather than \(L\)? To understand that, we need to look at what the air actually does.

01Why the fundamental fits half a wavelength

Air particles move differently from the sound wave

When sound travels through a pipe, individual air particles oscillate backwards and forwards along the pipe’s length.

They don’t travel all the way from one opening to the other. Instead, each small region of air moves around its equilibrium position, pushing against neighbouring regions.

A slinky stretched along a table gives a useful picture. Push a few coils forwards, then pull them back. A compression travels along the slinky, even though individual coils only move a short distance.

Air behaves similarly, except its particles aren’t connected by metal springs. Pressure forces transmit the disturbance.

Inside a tube, sound waves reflect from its ends. The incoming and reflected waves can overlap to produce a standing wave.

In a standing wave, some positions have almost no air-particle displacement, while others have large oscillations.

These positions have names:

• A displacement node is a position where the air’s oscillation amplitude is zero in the ideal model.
• A displacement antinode is a position where the air’s oscillation amplitude is greatest.

Now consider the two open ends.

Could the air at an opening remain completely stationary, as though a rigid wall were holding it in place?

No. Air at an open end is relatively free to move in and out. Each open end therefore behaves approximately as a displacement antinode.

The fundamental standing wave must satisfy three conditions:

1. A displacement antinode at the left opening.
2. A displacement antinode at the right opening.
3. The simplest non-zero standing-wave pattern between them.

That pattern has a single displacement node halfway along the tube.

How the boundary conditions give the wavelength

In the fundamental mode, the standing-wave pattern runs from an antinode at one end, through a node at the centre, to an antinode at the other end.

That distance is half a wavelength.

\[
L=\frac{\lambda_1}{2}
\]

Rearranging:

\[
\lambda_1=2L
\]

Here, \(\lambda_1\) is the wavelength of the fundamental in metres.

For a \(0.80\ \mathrm{m}\) open tube:

\[
\begin{aligned}
\lambda_1&=2(0.80\ \mathrm{m})\\
&=1.60\ \mathrm{m}
\end{aligned}
\]

The wavelength is twice the physical length of the ideal air column.

We then use the wave equation:

\[
v=f\lambda
\]

Frequency is wave speed divided by wavelength. For the fundamental:

\[
f_1=\frac{v}{\lambda_1}
\]

Substituting \(\lambda_1=2L\) gives:

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

That’s the reasoning behind the formula, not just a rule to memorise.

Pressure nodes are not displacement nodes

There’s an important detail that can make standing-wave diagrams confusing.

At the centre of the fundamental mode, air particles barely move, yet the pressure variation is greatest.

Imagine a moment when the air on both sides moves towards the centre. Air becomes compressed around the middle, raising its pressure. Half a cycle later, the air moves away from the centre, producing a region of lower pressure.

At the openings, pressure stays approximately equal to atmospheric pressure, so the variation in pressure is small.

That means pressure and displacement have opposite node locations.

A pressure node does not mean zero air pressure. It means approximately zero variation from atmospheric pressure.

02Measuring the fundamental: why the first experiment can mislead you

Suppose we want to determine the fundamental frequency experimentally rather than just calculate it.

We could drive a tube with a loudspeaker and search for the frequencies at which it resonates.

The challenge is knowing whether a missing signal means no resonance exists or whether our detector is simply in the wrong position.

Setting up a resonance experiment

Consider a straight tube with these properties:

Here, \(x\) is distance measured from the left opening along the tube.

Place a loudspeaker immediately outside the left opening without sealing it. Connect the loudspeaker to a signal generator that produces a steady sinusoidal sound wave at an adjustable frequency.

Insert a small pressure microphone into the air column at its midpoint, \(x=0.40\ \mathrm{m}\). Connect the microphone to a readout that displays the amplitude of the pressure variation.

As the signal generator sweeps upwards through different frequencies, the microphone records where pressure oscillations become particularly strong.

What should the experiment show?

For the ideal tube, the expected fundamental is:

\[
f_1=212.5\ \mathrm{Hz}
\]

Higher resonances can occur when additional half-wavelengths fit inside the tube.

Their frequencies are integer multiples of the fundamental:

\[
f_n=nf_1
\]

Here, \(n=1,2,3,\ldots\) is the harmonic number.

So we would predict resonances at \(212.5\ \mathrm{Hz}\), \(425\ \mathrm{Hz}\), \(637.5\ \mathrm{Hz}\), and so on.

Would the microphone at the centre detect all of them equally well?

No. That’s the problem with this first design.

Illustrative results

The following values are invented, physically plausible data for exploring the method. They are not measurements from a reported experiment.

They allow for a small increase in the effective acoustic length of the tube, which we’ll explain shortly.

Why might the \(416\ \mathrm{Hz}\) resonance be missing from the microphone’s readings?

The second harmonic has a pressure node at the centre of an open-open pipe.

Even though the pipe can resonate at that frequency, the local pressure variation is approximately zero where the microphone is positioned.

The microphone isn’t necessarily broken, and the second harmonic isn’t forbidden. We’re measuring at the wrong place.

Revising the experiment

Move the pressure microphone from the midpoint to a point one-quarter of the tube length from the left opening:

\[
x=\frac{0.80}{4}=0.20\ \mathrm{m}
\]

Keep the tube, air conditions, loudspeaker, and frequency sweep unchanged.

At this new position, the second harmonic has a pressure antinode. A clear resonance signal near \(416\ \mathrm{Hz}\) should become detectable in the illustrative model.

That change tells us something important.

The second harmonic existed before we moved the microphone. Changing the measurement position revealed a mode that the original arrangement was unable to detect effectively.

This is a useful experimental distinction: the resonant frequencies belong to the air column, but the strength of a measured signal also depends on where and how we measure it.

04Why the measured frequency can differ from the calculation

In our illustrative experiment, the ideal fundamental was \(212.5\ \mathrm{Hz}\), but the lower frequency of \(208\ \mathrm{Hz}\) was recorded.

The difference isn’t necessarily a mistake.

Our original formula assumes that the displacement antinodes occur exactly at the physical openings.

Real open pipes are slightly more complicated.

Air just outside each opening also participates in the oscillation. As a result, the pipe behaves approximately as though its acoustic length extends a small distance beyond its ends.

This is called end correction.

If each end contributes the same small correction \(e\), then:

\[
L_{\mathrm{eff}}=L+2e
\]

where:

• \(L_{\mathrm{eff}}\) is the effective acoustic length in metres.
• \(L\) is the physical length in metres.
• \(e\) is the approximate end correction at each opening in metres.

The corresponding fundamental frequency becomes:

\[
f_1\approx\frac{v}{2L_{\mathrm{eff}}}
\]

Because the effective length is slightly greater, the real fundamental frequency can be slightly lower than the ideal prediction.

This is an approximation. The size of the correction depends on factors such as the tube diameter, the shape of the openings, and the surrounding conditions.

For a standard idealised HSC calculation, use \(f_1=v/(2L)\) unless the problem provides information requiring a correction.

05Open at both ends or closed at one end?

A common mistake is to use the open-open formula whenever a question mentions a pipe.

But the boundary conditions are what determine the fundamental wavelength.

Imagine inserting a rigid bung into one end of our \(0.68\ \mathrm{m}\) tube.

The air at the bung can no longer move backwards and forwards through the end. That creates a displacement node at the closed boundary.

The open end remains a displacement antinode.

The simplest standing wave connecting a displacement node to an antinode occupies only a quarter of a wavelength.

The closed-open fundamental needs twice the wavelength of an open-open tube of the same length.

For a \(0.68\ \mathrm{m}\) tube with \(v=340\ \mathrm{m\,s^{-1}}\):

• Open at both ends: \(f_1=250\ \mathrm{Hz}\).
• Closed at one end: \(f_1=125\ \mathrm{Hz}\).

Notice the subtle point: closing one end does more than halve the fundamental frequency. It also changes which higher harmonics are allowed.

For an ideal closed-open pipe, only odd multiples of its fundamental are resonances. For an ideal open-open pipe, all positive integer multiples are allowed.

For the full relationship between harmonic number, wavelength, and standing-wave patterns, see Standing waves in open pipes: fundamental and harmonics.

A quick method for deciding what to calculate

The formula gives an allowed frequency. Whether that mode produces a large measured response also depends on how the sound is excited, where it is detected, and how much energy is lost.

06Questions and solutions

Question 1

A straight air column is open at both ends and has a length of \(0.75\ \mathrm{m}\).

The speed of sound is \(330\ \mathrm{m\,s^{-1}}\).

Calculate:

(a) The wavelength of the fundamental.

(b) The fundamental frequency.

Solution 1

The fundamental wavelength is \(1.50\ \mathrm{m}\), and the fundamental frequency is \(220\ \mathrm{Hz}\).

Both ends are open, so a displacement antinode occurs at each end. The fundamental contains half a wavelength.

For part (a):

\[
\begin{aligned}
\lambda_1&=2L\\
&=2(0.75\ \mathrm{m})\\
&=1.50\ \mathrm{m}
\end{aligned}
\]

For part (b):

\[
f_1=\frac{v}{\lambda_1}
\]

\[
\begin{aligned}
f_1&=\frac{330\ \mathrm{m\,s^{-1}}}{1.50\ \mathrm{m}}\\
&=220\ \mathrm{Hz}
\end{aligned}
\]

The tube’s lowest ideal resonance is therefore \(220\ \mathrm{Hz}\).

The tempting mistake is using \(0.75\ \mathrm{m}\) as one complete wavelength. That would give \(440\ \mathrm{Hz}\), which is the second harmonic rather than the fundamental.

Question 2

Two straight tubes, A and B, are open at both ends.

They contain air at the same temperature, have identical opening shapes and diameters, and are sufficiently narrow for the one-dimensional standing-wave model.

Their physical lengths and fundamental frequencies are given below.

These are idealised measurement values.

Assume both tubes have the same speed of sound, \(v\), and the same end correction \(e\) at each opening.

(a) Determine the end correction at each opening.

(b) Determine the speed of sound.

(c) Explain why simply using \(v=2Lf_1\) with the physical lengths would produce two different estimates of the sound speed.

Solution 2

Each opening has an end correction of \(0.010\ \mathrm{m}\), or \(1.0\ \mathrm{cm}\), and the common speed of sound is \(360\ \mathrm{m\,s^{-1}}\).

Both tubes follow the effective-length relation:

\[
v=2f_1(L+2e)
\]

Because the air conditions and opening corrections are assumed identical, both expressions must produce the same speed.

For tube A:

\[
v=2(300)(0.58+2e)
\]

For tube B:

\[
v=2(225)(0.78+2e)
\]

The numerical frequencies are in hertz, and the lengths and \(e\) are in metres.

Equating the expressions and dividing by \(2\):

\[
\begin{aligned}
300(0.58+2e)&=225(0.78+2e)\\
174+600e&=175.5+450e\\
150e&=1.5\\
e&=0.010\ \mathrm{m}
\end{aligned}
\]

So the correction per opening is \(1.0\ \mathrm{cm}\).

Now substitute this into tube A’s equation:

\[
\begin{aligned}
v&=2(300)(0.58+0.020)\\
&=2(300)(0.600)\\
&=360\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

A check using tube B gives:

\[
\begin{aligned}
v&=2(225)(0.78+0.020)\\
&=360\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Both tubes now give the same result.

For part (c), ignoring the end correction would give:

\[
\begin{aligned}
v_A&=2(0.58)(300)=348\ \mathrm{m\,s^{-1}}\\
v_B&=2(0.78)(225)=351\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Those estimates differ even though the speed of sound is assumed to be the same.

The apparent inconsistency comes from using the physical lengths in place of the effective acoustic lengths.

The key constraint is that the same value of \(v\) must explain both tubes. That lets us determine the unknown end correction rather than treating the two frequency measurements independently.

Question 3

A rigid tube is \(0.70\ \mathrm{m}\) long and initially open at both ends. The speed of sound is \(336\ \mathrm{m\,s^{-1}}\).

A loudspeaker produces a constant frequency of \(720\ \mathrm{Hz}\).

The right opening is then sealed completely with a rigid bung. The tube length, air conditions, and loudspeaker frequency remain unchanged.

(a) Determine whether \(720\ \mathrm{Hz}\) is an exact resonance before the bung is inserted. Identify the harmonic.

(b) Determine whether \(720\ \mathrm{Hz}\) is an exact resonance after the bung is inserted.

(c) Find the greatest resonant frequency below \(720\ \mathrm{Hz}\) for the closed-open tube.

(d) Explain why knowing that the fundamental frequency halves is not enough to determine the answer to part (b).

Solution 3

The \(720\ \mathrm{Hz}\) sound is initially an exact third-harmonic resonance, but it is not an exact resonance after the right end is sealed. The greatest allowed closed-open resonance below \(720\ \mathrm{Hz}\) is \(600\ \mathrm{Hz}\).

For part (a), both ends are initially open:

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

\[
\begin{aligned}
f_1&=\frac{336}{2(0.70)}\\
&=240\ \mathrm{Hz}
\end{aligned}
\]

The harmonic number of the driving frequency is:

\[
\begin{aligned}
n&=\frac{720}{240}\\
&=3
\end{aligned}
\]

So \(720\ \mathrm{Hz}\) is the third harmonic.

For part (b), the sealed end must now be a displacement node, while the open end remains an antinode.

The fundamental changes to:

\[
f_1=\frac{v}{4L}
\]

\[
\begin{aligned}
f_1&=\frac{336}{4(0.70)}\\
&=120\ \mathrm{Hz}
\end{aligned}
\]

It might seem that \(720\ \mathrm{Hz}\) is still allowed, since:

\[
\frac{720}{120}=6
\]

But this is the trap.

A closed-open pipe supports only odd harmonics:

\[
f_n=nf_1,\qquad n=1,3,5,7,\ldots
\]

The sixth harmonic is not permitted because it cannot satisfy a displacement node at one end and an antinode at the other.

Therefore \(720\ \mathrm{Hz}\) is not an exact resonance after the bung is inserted.

For part (c), the first few permitted frequencies are:

\[
\begin{aligned}
f_1&=120\ \mathrm{Hz}\\
f_3&=3(120)=360\ \mathrm{Hz}\\
f_5&=5(120)=600\ \mathrm{Hz}\\
f_7&=7(120)=840\ \mathrm{Hz}
\end{aligned}
\]

The greatest resonance below \(720\ \mathrm{Hz}\) is \(600\ \mathrm{Hz}\).

For part (d), halving the fundamental frequency doesn’t preserve the full harmonic sequence.

Closing an end changes both the fundamental wavelength and the permitted standing-wave patterns. You must check the harmonic boundary conditions, not just whether the driving frequency is an integer multiple of the new fundamental.

Question 4

An ideal open-open air column has a fixed physical length of \(1.00\ \mathrm{m}\).

A loudspeaker drives it at a constant frequency of \(1870\ \mathrm{Hz}\).

Initially, the speed of sound is \(340\ \mathrm{m\,s^{-1}}\). The air conditions then change so that the speed of sound becomes \(374\ \mathrm{m\,s^{-1}}\), while the tube length and driving frequency stay the same.

A student argues:

“Increasing the speed of sound increases all the pipe’s resonant frequencies. Since the loudspeaker frequency stays fixed, the pipe must stop resonating.”

(a) Determine whether \(1870\ \mathrm{Hz}\) is an exact resonance before and after the change.

(b) Identify the harmonic number in each state.

(c) Explain why the student’s conclusion is not necessarily correct.

Assume the ideal standing-wave model, with negligible end corrections.

Solution 4

The fixed \(1870\ \mathrm{Hz}\) sound is an exact resonance in both states. It changes from the eleventh harmonic to the tenth harmonic when the speed of sound increases.

For an open-open pipe:

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

Rearranging to determine the harmonic number:

\[
n=\frac{2Lf_n}{v}
\]

Initially:

\[
\begin{aligned}
n_{\mathrm{initial}}
&=\frac{2(1.00)(1870)}{340}\\
&=11
\end{aligned}
\]

The initial frequency is the eleventh harmonic.

After the speed increases:

\[
\begin{aligned}
n_{\mathrm{final}}
&=\frac{2(1.00)(1870)}{374}\\
&=10
\end{aligned}
\]

The final frequency is the tenth harmonic.

Both harmonic numbers are positive integers, so both standing waves satisfy the open-open boundary conditions.

We can verify this using wavelength.

Initially:

\[
\begin{aligned}
\lambda_{\mathrm{initial}}
&=\frac{340}{1870}\\
&=0.1818\ \mathrm{m}
\end{aligned}
\]

After the change:

\[
\begin{aligned}
\lambda_{\mathrm{final}}
&=\frac{374}{1870}\\
&=0.200\ \mathrm{m}
\end{aligned}
\]

Initially, eleven half-wavelengths fit into the tube:

\[
11\left(\frac{0.1818}{2}\right)
\approx1.00\ \mathrm{m}
\]

Afterwards, ten half-wavelengths fit:

\[
10\left(\frac{0.200}{2}\right)
=1.00\ \mathrm{m}
\]

The student’s reasoning correctly identifies that the resonant frequency of each individual harmonic increases with sound speed.

What it overlooks is that the fixed driving frequency may coincide with a different harmonic after the change.

Resonance is not tied to one harmonic number. It requires the driving frequency to match any permitted mode.

The exact match in this question is deliberately chosen. A small arbitrary change in sound speed would usually move the permitted frequencies away from a fixed driving frequency. Real high-order modes may also differ in strength.

Question 5

Two resonators are hidden inside separate enclosures. They contain air with a sound speed of \(340\ \mathrm{m\,s^{-1}}\).

A pressure microphone is installed inside each resonator. A frequency sweep is used to identify sharp resonance peaks, rather than simply detecting any sound from the loudspeaker.

The two possible arrangements are:

Model A: A \(1.00\ \mathrm{m}\) tube open at both ends. Its microphone is located exactly halfway along the tube.

Model B: A \(0.50\ \mathrm{m}\) tube closed at its left end and open at its right end. Its microphone is located immediately beside the closed end.

Ignore end correction, damping, and any disturbance caused by the microphone.

The initial frequency sweep covers \(100\) to \(900\ \mathrm{Hz}\).

Both microphones report strong resonance peaks at:

\[
170,\quad510,\quad850\ \mathrm{Hz}
\]

Neither reports a resonance peak at \(340\ \mathrm{Hz}\) or \(680\ \mathrm{Hz}\).

A technician then moves the microphone in each resonator to one-quarter of that resonator’s physical length from its left end. No other conditions change.

One of the resonators now produces a clear resonance peak at \(340\ \mathrm{Hz}\).

At that same microphone position, however, no resonance peak is detected at \(680\ \mathrm{Hz}\).

(a) Show why the original frequency measurements cannot distinguish between Models A and B.

(b) Identify which model produces the new \(340\ \mathrm{Hz}\) peak. Explain the physical reason.

(c) Is \(680\ \mathrm{Hz}\) an allowed resonance of the identified model? Explain why it is not detected at the quarter-length position.

(d) Give a microphone position, measured from the left end, where a strong pressure response at \(680\ \mathrm{Hz}\) could be detected. Justify your choice using the standing-wave pattern.

(e) Explain why moving the microphone has not created a new resonant frequency.

Solution 5

Both models can produce the three initially observed peaks, but only Model A can explain the appearance of a \(340\ \mathrm{Hz}\) resonance after the microphone moves. The \(680\ \mathrm{Hz}\) resonance also exists in Model A, but its pressure variation is zero at the quarter-length microphone position.

Part (a): Why the original observations are inconclusive

For Model A, both ends are open:

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

\[
\begin{aligned}
f_1&=\frac{340}{2(1.00)}\\
&=170\ \mathrm{Hz}
\end{aligned}
\]

All positive integer harmonics are allowed:

\[
170,\ 340,\ 510,\ 680,\ 850,\ldots
\]

But the microphone is at the exact centre of the tube.

At that position, the pressure variation is zero for the second, fourth, sixth, and other even harmonics.

The microphone therefore detects the odd-numbered modes strongly:

\[
170,\ 510,\ 850,\ldots
\]

It can miss \(340\ \mathrm{Hz}\) and \(680\ \mathrm{Hz}\), even though both are permitted resonances.

Now consider Model B.

Its left end is closed, and its right end is open:

\[
f_1=\frac{v}{4L}
\]

\[
\begin{aligned}
f_1&=\frac{340}{4(0.50)}\\
&=170\ \mathrm{Hz}
\end{aligned}
\]

A closed-open pipe permits only odd harmonics:

\[
170,\ 510,\ 850,\ldots
\]

The microphone is near the closed end, where the pressure variation is large for these modes.

Therefore both models can produce the same initial detected peaks.

The crucial distinction is that Model A permits the even harmonics but hides them at the microphone’s location. Model B does not permit those harmonics at all.

Part (b): Identifying the resonator

The new \(340\ \mathrm{Hz}\) resonance identifies Model A.

For Model A:

\[
\begin{aligned}
f_2&=2f_1\\
&=2(170)\\
&=340\ \mathrm{Hz}
\end{aligned}
\]

The second harmonic is allowed because an open-open pipe can accommodate two half-wavelengths along its length.

At this frequency:

\[
\begin{aligned}
\lambda_2&=\frac{v}{f_2}\\
&=\frac{340}{340}\\
&=1.00\ \mathrm{m}
\end{aligned}
\]

The pressure nodes of this second-harmonic pattern occur at:

\[
x=0,\quad0.50,\quad1.00\ \mathrm{m}
\]

The pressure antinodes occur halfway between adjacent nodes:

\[
x=0.25,\quad0.75\ \mathrm{m}
\]

Moving Model A’s microphone from the midpoint, \(x=0.50\ \mathrm{m}\), to the quarter-length position, \(x=0.25\ \mathrm{m}\), moves it from a pressure node to a pressure antinode.

The \(340\ \mathrm{Hz}\) resonance becomes detectable.

Model B cannot produce a genuine \(340\ \mathrm{Hz}\) resonance in the ideal model because that frequency is an even multiple of its fundamental.

Moving its microphone cannot change the allowed standing-wave boundary conditions.

Part (c): Why \(680\ \mathrm{Hz}\) is still missing

The frequency \(680\ \mathrm{Hz}\) is the fourth harmonic of Model A:

\[
\begin{aligned}
f_4&=4f_1\\
&=4(170)\\
&=680\ \mathrm{Hz}
\end{aligned}
\]

Its wavelength is:

\[
\begin{aligned}
\lambda_4&=\frac{340}{680}\\
&=0.50\ \mathrm{m}
\end{aligned}
\]

Four half-wavelengths fit into the \(1.00\ \mathrm{m}\) tube.

For this fourth harmonic, pressure nodes occur at:

\[
x=0,\ 0.25,\ 0.50,\ 0.75,\ 1.00\ \mathrm{m}
\]

The technician has placed the microphone at \(x=0.25\ \mathrm{m}\), exactly on one of these pressure nodes.

So the fourth harmonic is allowed, but the microphone still misses it.

The quarter-length position is excellent for detecting the second harmonic, yet unsuitable for detecting the fourth.

Part (d): Finding a better microphone position

A suitable microphone position is:

\[
x=0.125\ \mathrm{m}
\]

This is one-eighth of the tube length from its left opening.

For the fourth harmonic, adjacent pressure nodes are separated by:

\[
\frac{\lambda_4}{2}=0.25\ \mathrm{m}
\]

A pressure antinode occurs halfway between adjacent nodes.

Between the nodes at \(x=0\) and \(x=0.25\ \mathrm{m}\), the antinode is at:

\[
\begin{aligned}
x&=\frac{0+0.25}{2}\\
&=0.125\ \mathrm{m}
\end{aligned}
\]

A pressure microphone positioned there can detect a strong fourth-harmonic response when the tube is driven at \(680\ \mathrm{Hz}\).

Other fourth-harmonic pressure antinodes also exist, but one valid position is sufficient.

Part (e): What moving the microphone actually changes

Moving the microphone changes which pressure variations it can measure. It does not change the resonator’s boundary conditions, length, sound speed, or permitted frequencies.

Model A supported both \(340\ \mathrm{Hz}\) and \(680\ \mathrm{Hz}\) before the microphone moved.

The technician merely changed the observation point.

This is the deeper lesson of the experiment: an absent measured peak is not necessarily an absent standing-wave mode.

To establish whether a frequency is truly forbidden, check the boundary conditions first. Then check whether the measurement position lies at a node of the quantity being measured.

That same distinction guides the next useful decision in an air-column experiment. Moving a microphone can reveal an existing mode; sealing an open end changes which modes can exist in the first place.