How to Find Pipe Length from Successive Resonant Frequencies

A loudspeaker sends sound into a tube, and a microphone detects two strong resonance peaks at 360 Hz and 600 Hz. The tube is closed at one end. Would you use 360 Hz as the fundamental frequency to calculate its length?

Not necessarily. The first resonance you measure might not be the lowest resonance the tube can produce.

If the two frequencies are successive resonances, their difference gives the effective length of the air column:

\[ L_{\mathrm{eff}}=\frac{v}{2\Delta f} \]

Here:

• \(L_{\mathrm{eff}}\) is the effective acoustic length in metres (m).
• \(v\) is the speed of sound in air in metres per second (m s\(^{-1}\)).
• \(\Delta f\) is the difference between the two successive resonant frequencies, measured in hertz (Hz).

For our example, suppose sound travels at 336 m s\(^{-1}\).

\[ \begin{aligned} \Delta f&=600-360\\ &=240\ \mathrm{Hz} \end{aligned} \]

Therefore,

\[ \begin{aligned} L_{\mathrm{eff}}&=\frac{336}{2(240)}\\ &=0.700\ \mathrm{m} \end{aligned} \]

The effective length is 70.0 cm.

This method works for an ideal pipe open at both ends and for one closed at one end. The important condition is that the frequencies must be consecutive allowed resonances, not merely the next two peaks your microphone happens to detect.

To understand why, we need to look at what the air is doing and what the experiment actually measures.

01How the frequency-sweep experiment works

Imagine trying to measure the length of an air column without putting a ruler inside the tube.

You can send sound through it and measure the frequencies at which the air column resonates.

The apparatus has four main parts:

1. A function generator produces a repeating electrical signal at a chosen frequency.
2. A loudspeaker converts that signal into sound and directs it into the open mouth of the tube.
3. Sound travels along the tube, reflects at its ends, and interferes with the waves travelling in the opposite direction.
4. A small microphone measures pressure oscillations inside the tube. Its output is recorded while the generator frequency changes.

For our illustrative experiment, the tube is horizontal, with its open mouth on the left and a rigid stopper closing the right end. The microphone tip is placed inside, away from a pressure node of the modes being measured.

The speaker’s driving amplitude is kept approximately constant during the frequency sweep.

What should the microphone detect?

If the air column could resonate equally well at every frequency, the microphone response would vary without a distinctive sequence of resonances.

But the ends of the tube restrict which standing-wave patterns are possible.

At certain driving frequencies, the reflected and incoming waves reinforce one another, producing a large standing-wave response. These frequencies appear as peaks in the microphone’s pressure-amplitude readings.

Consider the following illustrative results.

These are invented readings for explaining the measurement method, not observations from a particular experiment.

The two peaks suggest that the air column supports strong standing waves at 360 Hz and 600 Hz.

But why does the frequency difference tell us the length?

02What air particles are doing inside a resonating pipe

Sound in air is a longitudinal wave.

That means air particles oscillate backwards and forwards parallel to the direction in which the sound travels.

Picture a Slinky stretched along a table. If you push and pull one end lengthways, regions where the coils bunch together travel along it. Individual coils move around their starting positions rather than travelling the full length of the Slinky.

Air behaves similarly. The speaker creates alternating regions of compression and rarefaction:

• A compression is a region where pressure and density are temporarily higher.
• A rarefaction is a region where pressure and density are temporarily lower.
• Air particles oscillate locally while the sound disturbance carries energy through the tube.

The Slinky is only a model. Real air particles are not connected by metal coils, and sound in a real pipe involves three-dimensional pressure and velocity fields.

When sound reflects from the ends, waves travelling in opposite directions can interfere to form a standing wave.

A standing wave has fixed locations where the oscillation is zero and other locations where it reaches its greatest amplitude.

These are called nodes and antinodes.

The important detail is that air-particle displacement and pressure variation do not have their nodes in the same places.

At a closed end

A rigid stopper prevents air from moving through the boundary.

Therefore:

• Air-particle displacement has a node at the closed end.
• Pressure variation has an antinode there.

The air immediately beside the stopper can be compressed or decompressed, even though it cannot move through the stopper.

At an open end

The air inside the tube communicates with the surrounding atmosphere.

In the ideal pipe model:

• Pressure variation has a node at the open end.
• Air-particle displacement has an antinode there.

The open end is therefore approximately the opposite of a rigid closed end.

Be careful with the language. A pressure node is not a location where the air has zero pressure. It is where the oscillating part of the pressure is zero, so the pressure remains approximately atmospheric.

Those boundary conditions determine which wavelengths can fit.

That is the physical reason a pipe has a particular set of resonant frequencies.

For a broader explanation of how standing waves and natural frequencies produce resonance, see Resonance in Sound Systems: Natural Frequencies Explained.

03Why successive resonances are separated by the same frequency

Let’s compare the two pipe arrangements.

Pipe open at both ends

Both ends must be displacement antinodes.

The simplest standing-wave pattern fits half a wavelength along the effective length of the pipe:

\[ L_{\mathrm{eff}}=\frac{\lambda_1}{2} \]

Here, \(\lambda_1\) is the wavelength of the fundamental, or lowest-frequency, mode.

The next standing-wave pattern fits two half-wavelengths, the next fits three, and so on.

In general,

\[ L_{\mathrm{eff}}=\frac{n\lambda_n}{2} \]

where \(n=1,2,3,\ldots\) is the harmonic number, and \(\lambda_n\) is the wavelength of that mode.

Remember the wave equation:

\[ v=f\lambda \]

Rearranging and applying the allowed wavelengths gives

\[ f_n=\frac{nv}{2L_{\mathrm{eff}}} \]

where \(f_n\) is the frequency of the \(n\)th harmonic.

The first few frequencies are

\[ \frac{v}{2L_{\mathrm{eff}}},\quad \frac{2v}{2L_{\mathrm{eff}}},\quad \frac{3v}{2L_{\mathrm{eff}}} \]

Each new resonance is exactly one fundamental-frequency step above the previous one.

So their separation is

\[ \Delta f=\frac{v}{2L_{\mathrm{eff}}} \]

Pipe closed at one end

Now one end must be a displacement node, while the other must be an antinode.

The shortest allowed standing-wave pattern fits one-quarter of a wavelength:

\[ L_{\mathrm{eff}}=\frac{\lambda_1}{4} \]

The next allowed pattern fits three-quarters of a wavelength, followed by five-quarters.

The allowed wavelengths satisfy

\[ L_{\mathrm{eff}} =\frac{(2n-1)\lambda_n}{4} \]

Here, \(n=1,2,3,\ldots\) counts successive allowed modes.

Using \(v=f\lambda\), the resonant frequencies become

\[ f_n=\frac{(2n-1)v}{4L_{\mathrm{eff}}} \]

This produces the first, third, fifth, seventh, and subsequent odd harmonics.

The first three frequencies are

\[ \frac{v}{4L_{\mathrm{eff}}},\quad \frac{3v}{4L_{\mathrm{eff}}},\quad \frac{5v}{4L_{\mathrm{eff}}} \]

Notice what happens when we subtract two successive frequencies.

For example,

\[ \begin{aligned} \Delta f &=\frac{3v}{4L_{\mathrm{eff}}} -\frac{v}{4L_{\mathrm{eff}}}\\ &=\frac{2v}{4L_{\mathrm{eff}}}\\ &=\frac{v}{2L_{\mathrm{eff}}} \end{aligned} \]

It’s exactly the same spacing as in the open-open pipe.

The reason is geometric: moving from one allowed standing-wave pattern to the next adds another half-wavelength along the air column, even though the closed-open pipe contains only odd harmonics.

Comparing the two sets of resonances

Suppose two ideal air columns have the same effective length of 0.700 m and the speed of sound is 336 m s\(^{-1}\).

Their predicted frequencies are:

These are ideal-model predictions, not measured data.

The boundary conditions change where the frequency sequence starts.

They do not change the separation between successive resonances when the effective lengths and sound speeds are equal.

This also explains the original 360 Hz and 600 Hz readings. They can be the third and fifth harmonics of a closed-open pipe with an unobserved fundamental at 120 Hz.

The central result is therefore

\[ \boxed{\Delta f=\frac{v}{2L_{\mathrm{eff}}}} \]

and, rearranging,

\[ \boxed{L_{\mathrm{eff}}=\frac{v}{2\Delta f}} \]

These relationships assume an approximately uniform air column, a constant sound speed, and consecutive allowed standing-wave modes.

04The experimental blind spot: a microphone can miss a resonance

The calculations above assume we have identified two consecutive allowed modes.

But a microphone does not necessarily detect every allowed mode.

This is the most important limitation of the measurement method.

Imagine a pipe open at both ends, with a small pressure microphone placed exactly halfway along its length.

For the fundamental mode, the centre of the pipe is a pressure antinode. The microphone can detect a strong pressure oscillation there.

For the second harmonic, the centre is a pressure node.

The second harmonic still exists. The air column can still resonate at that frequency. But the ideal pressure oscillation at the microphone’s exact location is zero.

A microphone at that point could miss the resonance completely.

The same happens for the fourth, sixth, and other even harmonics at the centre of an ideal open-open pipe.

Suppose the pipe’s actual resonant frequencies are

\[ 200,\ 400,\ 600,\ 800,\ 1000\ \mathrm{Hz} \]

A pressure microphone at the centre might detect only

\[ 200,\ 600,\ 1000\ \mathrm{Hz} \]

If you calculate the length from the detected 200 Hz and 600 Hz peaks, you will use a frequency difference of 400 Hz.

But the true separation between allowed modes is 200 Hz.

Your calculated length would be half its correct value.

The error comes from treating consecutive detected peaks as consecutive resonances.

How to improve the experiment

The first design cannot distinguish a missing acoustic mode from a mode that produces zero pressure at the microphone.

The simplest improvement is to move the microphone and repeat the frequency sweep.

A microphone positioned away from the previous pressure node may detect the missing resonance. Taking readings at several different positions is more reliable than relying on one microphone location.

Moving the microphone does not create a new resonant frequency. It changes which parts of an existing standing-wave pressure pattern the instrument can detect.

This matters especially when testing higher harmonics.

Why the calculated length may differ from the ruler length

Even when all the modes are identified correctly, the resonance method measures an effective acoustic length.

For a real open end, the pressure node is generally located a small distance beyond the physical opening.

The sound field extends outside the pipe.

This creates an end correction.

A simplified relationship is

\[ L_{\mathrm{eff}}=L_{\mathrm{tube}}+\delta_{\mathrm{total}} \]

where:

• \(L_{\mathrm{tube}}\) is the physical length measured with a ruler.
• \(\delta_{\mathrm{total}}\) is the combined end correction at the open ends.

In an introductory model, a pipe open at both ends has two open-end corrections, while a pipe closed at one end has one.

The corrections depend on the geometry of the openings and their surroundings. They are not a universal fixed distance.

For a standard idealised HSC calculation, you normally neglect end corrections unless the question provides information about them. Then \(L_{\mathrm{eff}}\) is taken to equal the pipe’s physical length.

Sound speed also matters. It changes with air conditions, particularly temperature. A length calculation using the wrong sound speed will be wrong even if the resonance frequencies were measured perfectly.

Before using the formula, check three things:

• Are the measured peaks successive allowed modes, or could an intermediate mode have been missed?
• Were both frequencies measured with the same boundary arrangement and approximately constant sound speed?
• Is the question asking for an ideal pipe length, an effective acoustic length, or a physical length with end correction?

These checks decide whether the simple frequency-difference method is enough.

05Questions and solutions

Question 1

An ideal pipe is open at both ends. Two successive resonant frequencies are measured at 420 Hz and 560 Hz.

The speed of sound in air is 336 m s\(^{-1}\).

Calculate the length of the pipe.

Solution 1

The pipe is 1.20 m long.

For two successive resonances in an ideal pipe,

\[ L=\frac{v}{2\Delta f} \]

First, calculate the frequency separation:

\[ \begin{aligned} \Delta f&=560-420\\ &=140\ \mathrm{Hz} \end{aligned} \]

Substitute \(v=336\ \mathrm{m\,s^{-1}}\).

\[ \begin{aligned} L&=\frac{336}{2(140)}\\ &=\frac{336}{280}\\ &=1.20\ \mathrm{m} \end{aligned} \]

The result also makes sense when checked against the harmonics.

An open-open pipe of this length has fundamental frequency

\[ \begin{aligned} f_1&=\frac{336}{2(1.20)}\\ &=140\ \mathrm{Hz} \end{aligned} \]

The observed frequencies are the third and fourth harmonics:

\[ 420=3(140),\qquad560=4(140) \]

They are successive resonances even though 420 Hz is not the fundamental.

The mistake to avoid is treating the first measured peak as the first harmonic.

Question 2

A pipe is open at both ends and remains at a fixed physical length. End corrections can be neglected.

Two frequency sweeps are performed.

Scan A: The speed of sound is 340 m s\(^{-1}\). Two successive resonance peaks are detected at 408 Hz and 544 Hz.

Scan B: After the air becomes warmer, two successive resonance peaks are detected at 420 Hz and 560 Hz.

Assume the same modes are measured in both scans.

Determine:

• The length of the pipe.
• The speed of sound during Scan B.
• The harmonic numbers of the four observed frequencies.
• Why subtracting 408 Hz from 420 Hz cannot be used to calculate the pipe length.

Solution 2

The pipe is 1.25 m long, and the sound speed during Scan B is 350 m s\(^{-1}\). The two scans detect the same third and fourth harmonics at different sound speeds.

First, use the successive frequencies from Scan A.

\[ \begin{aligned} \Delta f_A&=544-408\\ &=136\ \mathrm{Hz} \end{aligned} \]

Therefore,

\[ \begin{aligned} L&=\frac{v_A}{2\Delta f_A}\\ &=\frac{340}{2(136)}\\ &=\frac{340}{272}\\ &=1.25\ \mathrm{m} \end{aligned} \]

For Scan B, the new frequency separation is

\[ \begin{aligned} \Delta f_B&=560-420\\ &=140\ \mathrm{Hz} \end{aligned} \]

The length hasn’t changed, so rearrange the spacing equation:

\[ v_B=2L\Delta f_B \]

Substitute:

\[ \begin{aligned} v_B&=2(1.25)(140)\\ &=350\ \mathrm{m\,s^{-1}} \end{aligned} \]

The frequencies increase because the speed of sound has increased while the pipe length remains fixed.

We can also identify the harmonics.

For an open-open pipe, the fundamental frequency equals the spacing.

In Scan A, \(f_1=136\ \mathrm{Hz}\):

\[ 408=3(136),\qquad544=4(136) \]

In Scan B, \(f_1=140\ \mathrm{Hz}\):

\[ 420=3(140),\qquad560=4(140) \]

Both scans measure the third and fourth harmonics.

However, the difference \(420-408=12\ \mathrm{Hz}\) compares the same harmonic at two different sound speeds.

It is not the gap between adjacent harmonics at one fixed sound speed.

Using that difference in the pipe-length equation would confuse a change in wave speed with a change in standing-wave mode.

The useful distinction is between changing the frequency because the wavelength pattern changes and changing it because the wave speed changes.

Question 3

A uniform pipe is initially open at the left end and sealed at the right end with a rigid stopper.

Its successive resonant frequencies include 210 Hz and 350 Hz.

The speed of sound is 350 m s\(^{-1}\).

The stopper is then removed, leaving the same pipe open at both ends.

Assume the sound speed remains constant and all end corrections are negligible.

Determine:

• The length of the pipe.
• The lowest three resonant frequencies after the stopper is removed.
• Whether the original 210 Hz resonance will still be an allowed resonance.
• Why the spacing between successive resonances remains unchanged even though the boundary condition changes.

Solution 3

The pipe is 1.25 m long. After removing the stopper, its first three resonances are 140 Hz, 280 Hz, and 420 Hz. The original 210 Hz resonance is no longer allowed.

First, calculate the original resonance spacing:

\[ \begin{aligned} \Delta f&=350-210\\ &=140\ \mathrm{Hz} \end{aligned} \]

Therefore,

\[ \begin{aligned} L&=\frac{350}{2(140)}\\ &=1.25\ \mathrm{m} \end{aligned} \]

Before removing the stopper, the fundamental frequency is

\[ \begin{aligned} f_1&=\frac{350}{4(1.25)}\\ &=70\ \mathrm{Hz} \end{aligned} \]

The closed-open pipe allows odd harmonics:

\[ 70,\ 210,\ 350,\ 490,\ldots \]

The two original peaks are the third and fifth harmonics.

When the stopper is removed, both ends become displacement antinodes.

The new fundamental frequency is

\[ \begin{aligned} f_1&=\frac{350}{2(1.25)}\\ &=140\ \mathrm{Hz} \end{aligned} \]

The open-open pipe allows every positive integer harmonic:

\[ f_n=n(140) \]

So its first three resonances are

\[ 140,\ 280,\ 420\ \mathrm{Hz} \]

The old 210 Hz frequency is not among them. In fact,

\[ \frac{210}{140}=1.5 \]

An open-open pipe requires a whole-number harmonic, not a harmonic number of 1.5.

The tempting prediction is that a fixed-length tube must keep its resonant frequencies when the stopper is removed.

But the allowed modes depend on both the length and the boundary conditions.

The surprising part is that the frequency gap remains 140 Hz.

Originally,

\[ \Delta f=\frac{2v}{4L} \]

After opening both ends,

\[ \Delta f=\frac{v}{2L} \]

These are the same expression.

Both arrangements add one more half-wavelength of standing-wave pattern between successive allowed modes, but their fundamental patterns are different.

Consequently, the resonance spacing alone cannot tell you whether a pipe has one closed end or two open ends.

Question 4

A laboratory investigates a cylindrical tube whose physical length is unknown.

First, both ends are open. The measured separation between successive resonance frequencies is 200.0 Hz.

Then the right end is sealed with a rigid cap that sits flush with the original opening. The measured separation becomes 206.1 Hz.

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

For this question, use a simplified end-correction model:

• Each open end contributes the same positive correction \(\delta\).
• A rigidly closed end contributes no correction.
• Sealing the right end does not change the correction at the remaining open end.
• The physical length stays unchanged.

Calculate the physical length of the tube and the end correction at each open end.

Explain why sealing the tube can change the frequency spacing without physically changing the tube length.

Solution 4

The tube’s physical length is approximately 0.800 m, and each open end contributes an end correction of approximately 0.0252 m, or 2.52 cm.

The two frequency gaps reveal two different effective acoustic lengths.

For the pipe open at both ends,

\[ L_{\mathrm{OO}}=\frac{v}{2\Delta f_{\mathrm{OO}}} \]

Here, the subscript \(\mathrm{OO}\) means open at both ends.

\[ \begin{aligned} L_{\mathrm{OO}}&=\frac{340}{2(200.0)}\\ &=0.8500\ \mathrm{m} \end{aligned} \]

For the pipe with one end closed,

\[ L_{\mathrm{OC}}=\frac{v}{2\Delta f_{\mathrm{OC}}} \]

The subscript \(\mathrm{OC}\) means one end open and one closed.

\[ \begin{aligned} L_{\mathrm{OC}}&=\frac{340}{2(206.1)}\\ &\approx0.82484\ \mathrm{m} \end{aligned} \]

Now apply the simplified end-correction model.

With two open ends,

\[ L_{\mathrm{OO}}=L_{\mathrm{tube}}+2\delta \]

With one open end,

\[ L_{\mathrm{OC}}=L_{\mathrm{tube}}+\delta \]

Subtract the second equation from the first.

The physical length cancels:

\[ \begin{aligned} \delta&=L_{\mathrm{OO}}-L_{\mathrm{OC}}\\ &=0.8500-0.82484\\ &=0.02516\ \mathrm{m} \end{aligned} \]

Therefore,

\[ \delta\approx0.0252\ \mathrm{m} \]

Finally, calculate the physical length:

\[ \begin{aligned} L_{\mathrm{tube}}&=L_{\mathrm{OC}}-\delta\\ &=0.82484-0.02516\\ &=0.79968\ \mathrm{m}\\ &\approx0.800\ \mathrm{m} \end{aligned} \]

The frequency spacing increases after sealing one end because the effective acoustic length decreases from approximately 0.850 m to 0.825 m.

This does not mean the tube physically shrinks.

In the model, the second open-end correction disappears when that end is capped.

Because

\[ \Delta f=\frac{v}{2L_{\mathrm{eff}}} \]

a smaller effective length produces a larger frequency separation.

The important distinction is that the frequency-spacing method measures how long the air column behaves acoustically, not necessarily the distance between its physical ends.

This result depends on the question’s equal-correction assumptions. A more realistic model may need to account for the tube diameter, opening geometry, and how the surrounding apparatus affects the sound field.

Question 5

An ideal cylindrical pipe is open at both ends. Its length is unknown, and the speed of sound is 360 m s\(^{-1}\).

A loudspeaker drives the air column through a continuous frequency sweep from 250 Hz to 1600 Hz.

The pressure microphone is very small and does not disturb the standing waves. Assume:

• Every allowed resonance within the scanned frequency range is excited.
• The microphone detects a resonance whenever the pressure oscillation at its position is non-zero.
• A microphone placed exactly at a pressure node does not detect that mode.
• The measured peak lists are complete for their microphone positions.

Two scans are made without changing the pipe or the sound speed.

Scan A: The microphone is placed halfway along the pipe. It detects peaks at 300 Hz, 900 Hz, and 1500 Hz.

Scan B: The microphone is moved to one-quarter of the pipe length from the left end. It detects peaks at 300 Hz, 600 Hz, 900 Hz, and 1500 Hz.

A student claims that the 300 Hz and 900 Hz peaks in Scan A are successive resonances, so the pipe must be 0.300 m long.

Another student argues that 1200 Hz cannot be an allowed resonance because neither microphone position detects it.

Determine:

• The actual ideal pipe length.
• Every allowed resonance frequency between 250 Hz and 1600 Hz.
• Why 600 Hz is missing from Scan A.
• Why 1200 Hz is missing from both scans, and whether it is actually an allowed resonance.
• Why an alternative model with a 100 Hz fundamental cannot explain all the observations.
• One new microphone position that would allow the 1200 Hz resonance to be detected.

Solution 5

The actual pipe length is 0.600 m, and its allowed resonances in the scanned range are 300 Hz, 600 Hz, 900 Hz, 1200 Hz, and 1500 Hz.

The 1200 Hz resonance is real in the ideal model. It is hidden from both microphone positions because both happen to lie at pressure nodes for that particular mode.

The key is to separate which frequencies the pipe supports from which frequencies the microphone detects.

First, consider the standing-wave pressure patterns in an open-open pipe.

Both ends are pressure nodes.

For the first harmonic, there is a pressure antinode halfway along the tube.

For the second harmonic, the centre becomes a pressure node.

This alternating pattern continues. At the centre:

• Odd harmonics have pressure antinodes.
• Even harmonics have pressure nodes.

Under the question’s assumptions, a central microphone detects the odd harmonics but misses every even harmonic.

That means the spacing between consecutive detected peaks at the centre is twice the actual spacing between allowed modes.

From Scan A,

\[ \begin{aligned} \Delta f_{\mathrm{detected}}&=900-300\\ &=600\ \mathrm{Hz} \end{aligned} \]

Therefore, the spacing between consecutive allowed harmonics is

\[ \begin{aligned} \Delta f&=\frac{600}{2}\\ &=300\ \mathrm{Hz} \end{aligned} \]

Now calculate the length:

\[ \begin{aligned} L&=\frac{v}{2\Delta f}\\ &=\frac{360}{2(300)}\\ &=0.600\ \mathrm{m} \end{aligned} \]

The open-open fundamental frequency is

\[ \begin{aligned} f_1&=\frac{360}{2(0.600)}\\ &=300\ \mathrm{Hz} \end{aligned} \]

Consequently, the complete set of allowed modes within the specified range is:

The table explains both scans.

At 600 Hz, the second harmonic has a pressure node at the midpoint. That is why Scan A misses it.

At one-quarter of the pipe length, however, the second harmonic has a pressure antinode. Scan B can detect it.

The fourth harmonic at 1200 Hz is more interesting.

It fits four half-wavelengths into the pipe. Its pressure nodes occur at both ends and at one-quarter, one-half, and three-quarters of the pipe length.

Both microphone positions therefore coincide with pressure nodes.

This demonstrates why the second student’s argument is incorrect.

The absence of a microphone peak does not necessarily mean the standing wave is forbidden. A pressure microphone can fail to detect a mode even when the air column supports it.

The first student’s length of 0.300 m is also inconsistent with the observations.

At that length, the open-open fundamental would be

\[ \begin{aligned} f_1&=\frac{360}{2(0.300)}\\ &=600\ \mathrm{Hz} \end{aligned} \]

Its allowed frequencies would begin at 600 Hz, 1200 Hz, and 1800 Hz.

That model could not produce an open-open resonance at the observed 300 Hz.

So it fails an independent physical constraint.

What about a longer pipe with a 100 Hz fundamental?

Such a pipe could mathematically support 300 Hz, 900 Hz, and 1500 Hz as its third, ninth, and fifteenth harmonics.

But it would also support a fifth harmonic at 500 Hz.

For an open-open pipe, the fifth harmonic has a pressure antinode at the centre. Under the assumption that every allowed mode is excited and detected away from nodes, the central microphone would detect 500 Hz.

Scan A does not contain that peak.

The 100 Hz fundamental model therefore fails to explain the complete evidence.

Finally, we need a microphone position that can detect the fourth harmonic.

Moving the microphone to one-sixth of the pipe length from the left end would work.

Since \(L=0.600\ \mathrm{m}\),

\[ \begin{aligned} x&=\frac{L}{6}\\ &=\frac{0.600}{6}\\ &=0.100\ \mathrm{m} \end{aligned} \]

Here, \(x\) is the microphone’s distance from the left opening.

The fourth-harmonic pressure pattern does not have a node at \(x=0.100\ \mathrm{m}\), so the pressure microphone can detect the 1200 Hz resonance there.

This is the decisive experimental test.

The first two scans appear to provide independent evidence that 1200 Hz is absent. But both microphone positions happen to sample zero-pressure-oscillation points of the same allowed standing-wave pattern.

Their agreement is therefore misleading.

A third measurement taken at a carefully chosen position can distinguish an absent resonance from one that the original apparatus was unable to detect.

06What to test after calculating the pipe length

The difference between successive resonant frequencies is a useful way to measure the effective length of an air column without knowing the harmonic numbers in advance.

But the frequency gap alone doesn’t identify the pipe’s boundary conditions, and it cannot tell you whether a microphone has missed a mode.

After calculating a length, the next useful step is to test the model against the full resonance pattern.

For an ideal open-open pipe, check whether the frequencies are whole-number multiples of \(v/(2L)\).

For an ideal closed-open pipe, check whether they are odd multiples of \(v/(4L)\).

If the pattern doesn’t fit, reconsider the microphone position, the possibility of missing peaks, the sound speed, and the acoustic end corrections before changing the length calculation.

That takes the experiment beyond finding a number. It lets the measured frequencies reveal the standing-wave structure of the air inside the pipe.