Wavelength: L = 3λ/2 → λ = 2L / 3

Math: λ = (2 * 1.2) / 3 = 2.4 / 3 = 0.8 m

Frequency: f = v / λ OR f_3 = 3 * (v / 2L)

Math: f = 340 / 0.8 = 425 Hz

λ = 0.8 m

f = 425 Hz

Technical Comparison Key

Closed Ends (Closed Tube)

One Open End, One Closed End

Node Patterns (Closed Tube)

Node (closed) to Antinode (open)

Fundamental Pattern (Closed Tube)

L = 1/4 \(\lambda\)

Harmonic Series (Open Tube)

1, 2, 3, 4, 5... (ALL harmonics)

2. Musical Application

If you cut a pipe in half, the frequency will ____________________ because the wavelength that "fits" is now half as long. This is why smaller instruments play ____________________ pitches.

Annotation Exercise:

In the box below, explain in your own words why a specific pipe cannot play "every" note, but only certain ones.

IV. Open-End Air Columns

In an open-end pipe, the air is free to move at both ends. To satisfy this boundary condition, the standing wave must have a displacement antinode at both ends of the tube.

The simplest way to connect two antinodes is to place a single node in the exact center of the pipe. This pattern represents the Fundamental Frequency (or the 1st Harmonic).

The Harmonic Series

Unlike closed pipes, open pipes are "harmonically complete." They can produce every integer multiple of the fundamental frequency:

• 1st Harmonic: \(f_1\)
• 2nd Harmonic: \(2 \cdot f_1\)
• 3rd Harmonic: \(3 \cdot f_1\)
• ...and so on.

Technical Data

Wavelength Relation

\(\lambda_n = \frac{2L}{n}\)

Frequency Relation

f_n = \frac{nv}{2L}

"Each harmonic adds half a wavelength to the pattern."

Critical Thinking

If an open pipe has a fundamental frequency of 440 Hz (Concert A), what is the frequency of its 4th harmonic?

Answer: ____________________ Hz

Open Pipe Harmonic Visualization

Instructions: Draw the displacement envelopes for the first three harmonics. Label all Nodes (N) and Antinodes (A).

n = 1

Fundamental

Drawing Space

n = 2

2nd Harmonic

Drawing Space

n = 3

3rd Harmonic

Drawing Space

V. Closed-End Air Columns

Closed-end air columns are asymmetrical systems. One end is open, while the other is permanently sealed. This creates a more restrictive set of rules for our standing waves.

The closed end must be a node, and the open end must be an antinode. The simplest pattern that fits this rule is one-quarter of a wavelength (\(\frac{1}{4}\lambda\)).

The "Missing" Harmonics

Because of the node-antinode requirement, closed pipes cannot produce even harmonics. If you try to create the 2nd harmonic, you'd end up with a node at both ends, which is physically impossible for an open opening.

Result: Closed pipes only produce ODD harmonics (n = 1, 3, 5, 7...).

Technical Data

Wavelength Relation

\(\lambda_n = \frac{4L}{n}\)

Frequency Relation

f_n = \frac{nv}{4L}

"Each step adds half a wave, but only hits odd-numbered quarters."

Instrument Spotlight: Clarinet

The clarinet acts as a closed-end pipe. This is why it has a "hollow" or "woody" sound compared to a flute; it is physically missing all the even harmonic overtones!

Closed Pipe Harmonic Visualization

Instructions: Draw the displacement envelopes. Label the Closed End (Node) and the Open End (Antinode).

n = 1

Fundamental

Drawing Space

n = 3

1st Overtone

Drawing Space

n = 5

2nd Overtone

Drawing Space

VI. The Inverse Reality: Pressure Waves

Up to this point, we have visualized harmonics as displacement waves. However, sound can also be described as pressure waves. This is where many students get confused, but the rule is simple:

The 90-Degree Phase Shift

A displacement Antinode is always a pressure Node.
A displacement Node is always a pressure Antinode.

Think about the closed end of a pipe. Molecules can't move (Displacement Node), but they are getting smashed against the wall by the wave (Maximum Pressure Antinode).

Summary Table: Open Pipes

Summary Table: Closed Pipes

VII. Acoustic Mastery Check

1. Boundary Logic: If you blow into a bottle (closed end), why is the fundamental frequency lower than a pipe of the same length that is open at both ends?

2. Harmonic Math: A 0.85m open pipe and a 0.85m closed pipe are in a room where sound travels at 340 m/s. Calculate the fundamental frequency for both.

OPEN PIPE

CLOSED PIPE

3. Pattern Recognition: A tube is resonating at a frequency that is exactly 3 times its fundamental frequency. Is it possible for this tube to be a closed tube? Is it possible for it to be an open tube? Explain.

4. Synthesis: How does the presence or absence of even harmonics affect the "timbre" (quality) of a sound? Why does a clarinet sound different from a flute?

Mastery Question:

In the box below, explain why the length of the pipe is the primary factor in determining the pitch of the instrument.

IV. Open-End Air Columns

In an open-end pipe, the air is free to move at both ends. To satisfy this boundary condition, the standing wave must have a displacement antinode at both ends of the tube.

The simplest way to connect two antinodes is to place a single node in the exact center of the pipe. This pattern represents the Fundamental Frequency (or the 1st Harmonic).

The Harmonic Series

Open pipes support all integer harmonics (1, 2, 3, 4...). Each harmonic is a multiple of the fundamental frequency.

Frequency Relations

Fundamental Frequency (\(f_1\))

f_1 = \frac{v}{2L}

Harmonic Relation

f_n = n \cdot f_1

where n = 1, 2, 3, 4...

Pitch increases linearly with the harmonic number.

Critical Thinking

If an open pipe has a fundamental frequency of 440 Hz (Concert A), what is the frequency of its 3rd harmonic?

Calculation: _____________________________________ Hz

Open Pipe Harmonic Visualization

Instructions: Draw the displacement envelopes for the first three harmonics. Ensure the boundaries match the "Open-Open" rules.

n = 1

Fundamental

L = ½ λ

A - N - A

n = 2

2nd Harmonic

L = 1 λ

A-N-A-N-A

n = 3

3rd Harmonic

L = 1.5 λ

A-N-A-N-A-N-A

V. Closed-End Air Columns

Closed-end air columns are asymmetrical systems. One end is open, while the other is permanently sealed. This creates a unique constraint: the closed end must be a node, and the open end must be an antinode.

The simplest pattern that fits this rule is one-quarter of a wavelength (\(\frac{1}{4}\lambda\)).

The "Missing" Harmonics

Because of the node-antinode requirement, closed pipes only produce odd harmonics (n = 1, 3, 5, 7...).

Crucial Fact: A closed pipe of the same length as an open pipe will have a fundamental frequency exactly HALF as high.

Frequency Relations

Fundamental Frequency (\(f_1\))

f_1 = \frac{v}{4L}

Harmonic Relation

f_n = n \cdot f_1

where n = 1, 3, 5, 7... (ODD ONLY)

Even harmonics (n=2, 4...) do not resonate in closed pipes.

Instrument Spotlight: Clarinet

The clarinet acts as a closed-end pipe. This is why it has a unique "hollow" timbre. Since it lacks even harmonics, the frequency gaps between its overtones are larger than those of a flute.

Closed Pipe Harmonic Visualization

Instructions: Draw the displacement envelopes. The sealed end (left) must be a Node (N).

n = 1

Fundamental

L = ¼ λ

N - A

n = 3

1st Overtone

L = ¾ λ

N - A - N - A

n = 5

2nd Overtone

L = 1¼ λ

N-A-N-A-N-A

VII. Displacement vs. Pressure

Sound is a pressure wave. While we draw displacement waves to show particle movement, we could also draw pressure waves.

The Inverse Rule:

A displacement Antinode corresponds to a pressure Node.
A displacement Node corresponds to a pressure Antinode.

Example: At the closed end of a bottle, molecules are stationary (Displacement Node). However, because they are constantly being smashed against the wall, the pressure variation is at its maximum (Pressure Antinode).

VIII. Acoustic Mastery Check

1. Synthesis: Why does an open pipe play a frequency exactly one octave higher (twice the frequency) than a closed pipe of the same length?

2. Calculation: A organ pipe is 2.4 meters long. If the speed of sound is 344 m/s, calculate the fundamental frequency if the pipe is open at both ends versus closed at one end.

Open Pipe Solution

f_1 = _________ Hz

Closed Pipe Solution

f_1 = _________ Hz

3. Pattern Rule: A student blowing into a pipe observes the 3rd harmonic. They then seal one end of the pipe. Can the 3rd harmonic still resonate at the same frequency? Explain why or why not.

Critical Thinking Box

Explain why a specific pipe cannot play every note, but only certain frequencies in a series.

IV. Open-End Air Columns

Open-end pipes have openings at both ends. To resonate, the wave pattern must show a displacement antinode at both boundaries.

The simplest pattern that meets this requirement has a single node in the center. This is the 1st Harmonic (Fundamental). Every subsequent harmonic adds another half-wavelength (\(\frac{1}{2}\lambda\)) to the tube.

The Integer Series

Open pipes are "harmonically full," meaning they support all integer multiples: 1, 2, 3, 4, 5...

The Physics Formulas

Fundamental Frequency (\(f_1\))

f_1 = \frac{v}{2L}

Harmonic Frequency Relation

f_n = n \cdot f_1

\(n = 1, 2, 3, 4, ...\)

"Wavelength is always \(\lambda_n = \frac{2L}{n}\)"

The Frequency Law

If an open pipe has a fundamental frequency of 250 Hz, what is the frequency of its 4th harmonic?

Calculation:

Hz

Open Pipe Workspace

Instructions: Draw the displacement envelopes for each resonant state. Ensure Antinodes (A) are at both open ends. Label every Node (N) and Antinode (A).

n = 1

1st Harmonic

L = ________ λ

Draw Pattern

n = 2

2nd Harmonic

L = ________ λ

Draw Pattern

n = 3

3rd Harmonic

L = ________ λ

Draw Pattern

V. Closed-End Air Columns

Closed-end air columns are asymmetrical. One end is open, while the other is permanently sealed by a rigid wall.

The constraint is absolute: the closed end must be a Node, and the open end must be an Antinode. The simplest pattern that satisfies this is exactly one-quarter of a wavelength (\(\frac{1}{4}\lambda\)).

The Missing Harmonics

Because of the N-A rule, closed pipes skip all even harmonics. The series proceeds from n=1 directly to n=3, then n=5.

Critical Logic Check:

"A closed pipe of length L plays a frequency twice as deep (one octave lower) as an open pipe of the same length."

The Physics Formulas

Fundamental Frequency (\(f_1\))

f_1 = \frac{v}{4L}

Harmonic Frequency Relation

f_n = n \cdot f_1

\(n = 1, 3, 5, 7, ...\)

"Wavelength is always \(\lambda_n = \frac{4L}{n}\)"

Instrument Profile: The Clarinet

While a flute acts as an open pipe, the clarinet (despite its shape) acts as a closed pipe due to the vibrating reed. This is why clarients produce a "hollow" timbre—they are physically missing all the even harmonic overtones!

Closed Pipe Workspace

Instructions: Draw the displacement envelopes. The sealed end (left) must be a Node (N). The open end (right) must be an Antinode (A).

n = 1

1st Harmonic

L = ________ λ

Draw Pattern

n = 3

3rd Harmonic

L = ________ λ

Draw Pattern

n = 5

5th Harmonic

L = ________ λ

Draw Pattern

VII. The Inverse: Pressure Waves

Sound waves can be described in two ways: the movement of air particles (displacement) or the change in air density (pressure). The "Physics Classroom" diagrams we've drawn are displacement waves.

The 90-Degree Phase Shift

"A point of maximum molecule movement (Displacement Antinode) is a point of constant density (Pressure Node)."

Displacement End

Closed End = Node

Open End = Antinode

Pressure End

Closed End = Antinode

Open End = Node

Reference Summary Table

VIII. Mastery Assessment

1. Conceptual Synthesis: Two pipes are in the same room. Pipe A is open at both ends and Pipe B is open at only one end. If they are the same length, why does Pipe B play a lower note?

Student Response Space...

2. Calculations: A 1.8-meter long pipe is in a lab where \(v_{sound} = 344 \text{ m/s}\). Calculate the frequency of the 3rd Harmonic for both pipe types.

OPEN PIPE

f_3 = ____________ Hz

CLOSED PIPE

f_3 = ____________ Hz

3. Boundary Violation: A student tries to draw the 2nd Harmonic for a closed pipe. Explain, using physics boundary rules, why this harmonic is impossible to sustain.

Student Response Space...

This is why musical instruments play discrete notes. These "allowed" frequencies of vibration are called harmonics.

Thinking Exercise

Consider a flute. When the player covers or uncovers holes, they are effectively changing the length of the resonating tube. Based on the "fit" requirement, explain why this changes the note played.

IV. Open-End Air Columns

In an open-end pipe, the air is free to move at both ends. To satisfy the boundary condition, the standing wave must show a displacement antinode at each end.

Fundamental Frequency (n=1):

The simplest pattern is a tube with an antinode at each end and one node in the center. This pattern represents exactly half of a wavelength (\(\frac{1}{2}\lambda\)).

Integer Harmonic Series

Open pipes are capable of producing a full harmonic series. This means they support all integer multiples of the fundamental frequency: \(n = 1, 2, 3, 4, ...\)

Technical Reference: Open

Fundamental Formula (\(n=1\))

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

General Harmonic Formula (\(n\)-th)

\[ f_n = n \cdot f_1 = \frac{nv}{2L} \]

where \(n = 1, 2, 3, 4, ...\)

Wavelength Relation

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

Harmonic Calculation

An open pipe has a fundamental frequency (\(f_1\)) of 262 Hz (Middle C). Calculate the frequency of its 4th harmonic.

Workspace:

Hz

Open Pipe Field Work

Instructions: Using the "Physics Classroom" style (no end lines), draw the standing wave displacement envelopes. Ensure Antinodes (A) are at both open ends. Label every Node (N) and Antinode (A).

n = 1

Fundamental

harmonic length:

L = ______ λ

Drawing Workspace

n = 2

2nd Harmonic

harmonic length:

L = ______ λ

Drawing Workspace

n = 3

3rd Harmonic

harmonic length:

L = ______ λ

Drawing Workspace

V. Closed-End Air Columns

Closed-end air columns are asymmetrical systems. One end is open, but the other is permanently sealed by a rigid barrier. This physical constraint dictates that the air molecules at the sealed end are immobile.

The Node-Antinode Rule:

Resonance only occurs if the pattern maintains a displacement node at the closed end and a displacement antinode at the open end. The simplest pattern that fits is exactly one-quarter of a wavelength (\(\frac{1}{4}\lambda\)).

The Odd Harmonic Series

Because of the asymmetrical boundary, closed pipes only support odd harmonics. Even-numbered multiples (n=2, 4...) would require two nodes or two antinodes, which is physically impossible for a closed-open system.

The Octave Rule:

For the same tube length \(L\), a closed pipe plays a frequency exactly one octave lower (half the frequency) of an open pipe.

Technical Reference: Closed

Fundamental Formula (\(n=1\))

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

General Harmonic Formula (\(n\)-th)

\[ f_n = n \cdot f_1 = \frac{nv}{4L} \ ]

n = 1, 3, 5, 7, ... (ODD ONLY)

Wavelength Relation

\[ \lambda_n = \frac{4L}{n} \ ]

Instrument Timbre

Instruments like the clarinet resonate as closed pipes. Because they skip even harmonics, their "harmonic series" is less dense. This is why a clarinet sounds "hollow" or "woody" compared to the brighter, fuller sound of an open flute.

Closed Pipe Field Work

Instructions: Draw the displacement envelopes. The sealed end (left) must be a Node (N). The open end (right) must be an Antinode (A). Label every Node (N) and Antinode (A).

n = 1

Fundamental

harmonic length:

L = ______ λ

Drawing Workspace

n = 3

3rd Harmonic

harmonic length:

L = ______ λ

Drawing Workspace

n = 5

5th Harmonic

harmonic length:

L = ______ λ

Drawing Workspace

VII. Displacement vs. Pressure

Up to this point, we have visualized harmonics as displacement waves (particle motion). However, sound is also a pressure wave (density change). The physics of resonance remains the same, but the representation flips.

The 90° Phase Difference

"A point where particles don't move (Displacement Node) is a point where they are smashed together (Pressure Antinode)."

Displacement Wave Rules

Closed Wall = NODE

Open Air = ANTINODE

Pressure Wave Rules

Closed Wall = ANTINODE

Open Air = NODE

Reference Summary Chart

VIII. Acoustic Mastery Assessment

1. Synthesis: A clarinet (closed pipe) and a flute (open pipe) have the exact same physical length. Which instrument will play a higher fundamental pitch? Use your wavelength formulas to prove your answer.

Draft your explanation here...

2. Calculation: An organ pipe is 2.5 meters long. Speed of sound is 340 m/s. Find the frequency of the 3rd Harmonic for both types of tubes.

Open Pipe

f_3 = ____________ Hz

Closed Pipe

f_3 = ____________ Hz

3. Boundary Violation: A student claims to have found the 2nd harmonic for a closed pipe. Using the node-antinode rules, explain why the 2nd harmonic is mathematically forbidden in a closed-open system.

Draft your explanation here...

2

Interference

The reflected wave travels back toward the mouth of the pipe. As it travels, it "meets" new incoming waves. When these waves align perfectly, they combine to form a larger pattern.

3

Resonance

If the wavelength is a "perfect fit" for the tube length, the wave is reinforced with each bounce. The air column locks into a stable vibration pattern called a Standing Wave.

The "Fit" Requirement

For resonance to occur, the Standing Wave pattern must have nodes and antinodes in the correct places. If the wavelength is slightly off, the waves will interfere destructively and cancel out. The specific frequencies that sustain themselves are called Harmonics.

IV. Open-End Air Columns

Open-end pipes have openings at both ends. To satisfy the boundary conditions, the wave pattern must show a displacement antinode (A) at both ends of the tube.

The Fundamental Pattern (n=1):

The simplest way to put an antinode at each end is to have one node in the exact center. This pattern represents exactly one-half of a wavelength (\(\frac{1}{2}\lambda\)).

Technical Mastery: Open Pipes

Harmonic Relation

\[ f_n = n \cdot f_1 \]

where \(n = 1, 2, 3, 4, ...\)

Length Relation

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

Each harmonic adds \(\frac{1}{2}\lambda\)

Quick Calculation Check:

An open pipe has a fundamental (\(f_1\)) of 300 Hz. What is its 3rd harmonic frequency?

Answer: \(f_3\) = ______________ Hz

Open Pipe Workspace

Draw the displacement envelopes for each harmonic. Open pipes show the Physics Classroom style (no end vertical lines). Ensure Antinodes (A) are at the boundaries.

01

Fundamental (n=1)

02

2nd Harmonic (n=2)

03

3rd Harmonic (n=3)

VI. Closed-End Air Columns

Closed-end air columns are asymmetrical. One end is open, while the other is permanently sealed by a rigid barrier. This physical barrier dictates that the air molecules at the sealed end cannot move.

The Odd Harmonic Series:

Resonance only occurs if the pattern maintains a displacement node (N) at the closed end and a displacement antinode (A) at the open end. Because of this N-A requirement, closed pipes only support odd harmonics.

The fundamental pattern fits exactly one-quarter of a wavelength (\(\frac{1}{4}\lambda\)).

Technical Mastery: Closed Pipes

Harmonic Relation

\[ f_n = n \cdot f_1 \]

n = 1, 3, 5, 7, ... (ODD ONLY)

Length Relation

\[ L = \frac{n \cdot \lambda}{4} \]

Skips even harmonic quarters

Did You Know?

Because they skip even harmonics, closed-end instruments (like the clarinet) have a uniquely "hollow" or "woody" timbre compared to the fuller, brighter sound of a flute or saxophone.

Closed Pipe Workspace

Draw the displacement envelopes. The sealed end (left) must be a Displacement Node (N). The open end (right) is a Displacement Antinode (A). Label N and A.

01

Fundamental (n=1)

03

3rd Harmonic (n=3)

05

5th Harmonic (n=5)

VII. Displacement vs. Pressure

Up to this point, we have visualized harmonics as displacement waves (tracking how far air particles move). However, sound is also a pressure wave (tracking how dense air molecules become).

The 90-Degree Phase Rule

"A point of zero motion (Node) is a point of maximum pressure build-up (Antinode)."

Displacement Boundaries

Closed: Node (N)

Open: Antinode (A)

Pressure Boundaries

Closed: Antinode (A)

Open: Node (N)

Final Master Question

You have an open pipe and a closed pipe of the same length. Prove mathematically using the fundamental formulas which pipe plays the lower note.

2

Interference

The reflected wave travels back toward the mouth of the pipe. As it travels, it "meets" new incoming waves. When these waves align perfectly, they combine to form a larger pattern.

3

Resonance

If the wavelength is a "perfect fit" for the tube length, the wave is reinforced with each bounce. The air column locks into a stable vibration pattern called a Standing Wave.

The "Fit" Requirement

For resonance to occur, the Standing Wave pattern must have nodes and antinodes in the correct places according to boundary conditions. If the wavelength is slightly off, the waves will interfere destructively and cancel out. The specific frequencies that sustain themselves are called Harmonics.

IV. Open-End Air Columns

Open-end pipes have openings at both ends. To satisfy the boundary conditions, the wave pattern must show a displacement antinode (A) at both ends of the tube.

The Fundamental Pattern (n=1):

The simplest way to put an antinode at each end is to have one node in the exact center. This pattern represents exactly one-half of a wavelength (\(\frac{1}{2}\lambda\)).

Technical Mastery: Open Pipes

Harmonic Frequency Relation

\(f_n = n \cdot f_1\)

\(n = 1, 2, 3, 4, ...\)

Fundamental Base (\(f_1\))

\(f_1 = \frac{v}{2L}\)

Speed over double-length

Quick Calculation Check:

An open pipe has a fundamental (\(f_1\)) of 300 Hz. What is its 3rd harmonic frequency?

Calculation: \(f_3 = \) ______________ Hz

Open Pipe Workspace

Draw the displacement envelopes for each harmonic. Open pipes show the Physics Classroom style (no end vertical lines). Label every Node (N) and Antinode (A).

n=1

Fundamental

A - N - A

n=2

2nd Harmonic

A-N-A-N-A

n=3

3rd Harmonic

A-N-A-N-A-N-A

V. Closed-End Air Columns

Closed-end air columns are asymmetrical. One end is open, while the other is permanently sealed by a rigid barrier. This physical barrier dictates that the air molecules at the sealed end cannot move.

The Node-Antinode Rule:

Resonance only occurs if the pattern maintains a displacement node (N) at the closed end and a displacement antinode (A) at the open end. Because of this requirement, closed pipes only support odd harmonics.

The fundamental pattern represents exactly one-quarter of a wavelength (\(\frac{1}{4}\lambda\)).

Technical Mastery: Closed Pipes

Harmonic Frequency Relation

\(f_n = n \cdot f_1\)

n = 1, 3, 5, 7... (ODD ONLY)

Fundamental Base (\(f_1\))

\(f_1 = \frac{v}{4L}\)

Speed over quad-length

Instrument Timbre:

Because they skip even harmonics, closed-end instruments (like the clarinet) have a uniquely "hollow" or "woody" sound. The first overtone is the 3rd harmonic, which is much higher in pitch than the 2nd harmonic overtone of a flute.

Closed Pipe Workspace

Draw the displacement envelopes. The sealed end (left) must be a Displacement Node (N). The open end (right) must be an Antinode (A). Label N and A.

n=1

Fundamental

N - A

n=3

3rd Harmonic

N - A - N - A

n=5

5th Harmonic

N-A-N-A-N-A

VII. Displacement vs. Pressure

Up to this point, we have visualized harmonics as displacement waves (tracking how far air particles move). However, sound is also a pressure wave (tracking how dense air molecules become).

The Phase Shift

"A point of zero motion (Displacement Node) is a point of maximum pressure build-up (Pressure Antinode)."

Displacement Rules

Closed: NODE

Open: ANTINODE

Pressure Rules

Closed: ANTINODE

Open: NODE

Final Master Question

You have an open pipe and a closed pipe of the same length (L). Using your formulas, explain which pipe plays a lower note and by how many octaves.

2

Interference

The reflected wave travels back toward the mouth of the pipe. As it travels, it "meets" new incoming waves. When these waves align perfectly, they combine to form a larger pattern.

3

Resonance

If the wavelength is a "perfect fit" for the tube length, the wave is reinforced with each bounce. The air column locks into a stable vibration pattern called a Standing Wave.

The "Fit" Requirement

For resonance to occur, the Standing Wave pattern must have nodes and antinodes in the correct places according to boundary conditions. If the wavelength is slightly off, the waves will interfere destructively and cancel out. The specific frequencies that sustain themselves are called Harmonics.

IV. Open-End Air Columns

Open-end pipes have openings at both ends. To satisfy the boundary conditions, the wave pattern must show a displacement antinode (A) at both ends of the tube.

The Fundamental Pattern (n=1):

The simplest way to put an antinode at each end is to have one node in the exact center. This pattern represents exactly one-half of a wavelength (\(\frac{1}{2}\lambda\)).

Technical Mastery: Open Pipes

Harmonic Frequency Relation

\(f_n = n \cdot f_1\)

\(n = 1, 2, 3, 4, ...\)

Fundamental Base (\(f_1\))

\(f_1 = \frac{v}{2L}\)

Speed over double-length

Quick Calculation Check:

An open pipe has a fundamental (\(f_1\)) of 300 Hz. What is its 3rd harmonic frequency?

Calculation: \(f_3 = \) ______________ Hz

Open Pipe Workspace

Draw the displacement envelopes for each harmonic. Open pipes show the Physics Classroom style (no end vertical lines). Label every Node (N) and Antinode (A).

n=1

Fundamental

A - N - A

n=2

2nd Harmonic

A-N-A-N-A

n=3

3rd Harmonic

A-N-A-N-A-N-A

V. Closed-End Air Columns

Closed-end air columns are asymmetrical. One end is open, while the other is permanently sealed by a rigid barrier. This physical barrier dictates that the air molecules at the sealed end cannot move.

The Node-Antinode Rule:

Resonance only occurs if the pattern maintains a displacement node (N) at the closed end and a displacement antinode (A) at the open end. Because of this requirement, closed pipes only support odd harmonics.

The fundamental pattern represents exactly one-quarter of a wavelength (\(\frac{1}{4}\lambda\)).

Technical Mastery: Closed Pipes

Harmonic Frequency Relation

\(f_n = n \cdot f_1\)

n = 1, 3, 5, 7... (ODD ONLY)

Fundamental Base (\(f_1\))

\(f_1 = \frac{v}{4L}\)

Speed over quad-length

Instrument Timbre:

Because they skip even harmonics, closed-end instruments (like the clarinet) have a uniquely "hollow" or "woody" sound. The first overtone is the 3rd harmonic, which is much higher in pitch than the 2nd harmonic overtone of a flute.

Closed Pipe Workspace

Draw the displacement envelopes. The sealed end (left) must be a Displacement Node (N). The open end (right) must be an Antinode (A). Label N and A.

n=1

Fundamental

N - A

n=3

3rd Harmonic

N - A - N - A

n=5

5th Harmonic

N-A-N-A-N-A

VII. Displacement vs. Pressure

Up to this point, we have visualized harmonics as displacement waves (tracking particle motion). However, sound is also a pressure wave.

The Phase Shift

"A point of zero motion (Displacement Node) is a point of maximum pressure build-up (Pressure Antinode)."

Displacement Rules

Closed: NODE

Open: ANTINODE

Pressure Rules

Closed: ANTINODE

Open: NODE

Final Master Question

Using your fundamental formulas, explain which pipe type (open vs. closed) plays a lower note for a given length \(L\).

Acoustic Reference Summary

Open Air Columns

Boundaries Antinode - Antinode

Harmonic Series (\(n\)) 1, 2, 3, 4, 5... (All)

Frequency Base

\(f_1 = \frac{v}{2L}\)

Harmonic Relation

\(f_n = n \cdot f_1\)

Instruments: Flute, Recorder, Organ (Open)

Closed Air Columns

Boundaries Node - Antinode

Harmonic Series (\(n\)) 1, 3, 5, 7... (Odd)

Frequency Base

\(f_1 = \frac{v}{4L}\)

Harmonic Relation

\(f_n = n \cdot f_1\)

Instruments: Clarinet, Bottle, Organ (Closed)

Remember: Displacement and Pressure are inverse. A boundary that is a Node for particle movement is an Antinode for air pressure. Use the rules above for displacement-based drawings.

2

Interference

The reflected wave travels back toward the mouth of the pipe. As it travels, it "meets" new incoming waves. When these waves align perfectly, they combine to form a larger pattern.

3

Resonance

If the wavelength is a "perfect fit" for the tube length, the wave is reinforced with each bounce. The air column locks into a stable vibration pattern called a Standing Wave.

The "Fit" Requirement

For resonance to occur, the Standing Wave pattern must have nodes and antinodes in the correct places according to boundary conditions. If the wavelength is slightly off, the waves will interfere destructively and cancel out. The specific frequencies that sustain themselves are called Harmonics.

IV. Open-End Air Columns

Open-end pipes have openings at both ends. To satisfy the boundary conditions, the wave pattern must show a displacement antinode (A) at both ends of the tube.

The Fundamental Pattern (n=1):

The simplest way to put an antinode at each end is to have one node in the exact center. This pattern represents exactly one-half of a wavelength (\(\frac{1}{2}\lambda\)).

Technical Mastery: Open Pipes

Harmonic Frequency Relation

\(f_n = n \cdot f_1\)

\(n = 1, 2, 3, 4, ...\)

Fundamental Base (\(f_1\))

\(f_1 = \frac{v}{2L}\)

Speed over double-length

Quick Calculation Check:

An open pipe has a fundamental (\(f_1\)) of 300 Hz. What is its 3rd harmonic frequency?

Calculation: \(f_3 = \) ______________ Hz

Open Pipe Workspace

Draw the displacement envelopes for each harmonic. Open pipes show the Physics Classroom style (no end vertical lines). Label every Node (N) and Antinode (A).

n=1

Fundamental

A - N - A

n=2

2nd Harmonic

A-N-A-N-A

n=3

3rd Harmonic

A-N-A-N-A-N-A

V. Closed-End Air Columns

Closed-end air columns are asymmetrical. One end is open, while the other is permanently sealed by a rigid barrier. This physical barrier dictates that the air molecules at the sealed end cannot move.

The Node-Antinode Rule:

Resonance only occurs if the pattern maintains a displacement node (N) at the closed end and a displacement antinode (A) at the open end. Because of this requirement, closed pipes only support odd harmonics.

The fundamental pattern represents exactly one-quarter of a wavelength (\(\frac{1}{4}\lambda\)).

Technical Mastery: Closed Pipes

Harmonic Frequency Relation

\(f_n = n \cdot f_1\)

n = 1, 3, 5, 7... (ODD ONLY)

Fundamental Base (\(f_1\))

\(f_1 = \frac{v}{4L}\)

Speed over quad-length

Instrument Timbre:

Because they skip even harmonics, closed-end instruments (like the clarinet) have a uniquely "hollow" or "woody" sound. The first overtone is the 3rd harmonic, which is much higher in pitch than the 2nd harmonic overtone of a flute.

Closed Pipe Workspace

Draw the displacement envelopes. The sealed end (left) must be a Displacement Node (N). The open end (right) must be an Antinode (A). Label N and A.

n=1

Fundamental

N - A

n=3

3rd Harmonic

N - A - N - A

n=5

5th Harmonic

N-A-N-A-N-A

VII. Displacement vs. Pressure

Up to this point, we have visualized harmonics as displacement waves (tracking particle motion). However, sound is also a pressure wave.

The Phase Shift

"A point of zero motion (Displacement Node) is a point of maximum pressure build-up (Pressure Antinode)."

Displacement Rules

Closed: NODE

Open: ANTINODE

Pressure Rules

Closed: ANTINODE

Open: NODE

Final Master Question

Using your fundamental formulas, explain which pipe type (open vs. closed) plays a lower note for a given length \(L\).

Acoustic Reference Summary

Open Air Columns

Boundaries Antinode - Antinode

Harmonic Series (\(n\)) 1, 2, 3, 4, 5... (All)

Fundamental Frequency

\(f_1 = \frac{v}{2L}\)

Harmonic Relation

\(f_n = n \cdot f_1\)

EXAMPLES: Flute, Recorder, Open Organ

Closed Air Columns

Boundaries Node - Antinode

Harmonic Series (\(n\)) 1, 3, 5, 7... (Odd)

Fundamental Frequency

\(f_1 = \frac{v}{4L}\)

Harmonic Relation

\(f_n = n \cdot f_1\)

EXAMPLES: Clarinet, Bottle, Stopped Organ

REMINDER: Displacement and Pressure are inverse. A boundary that is a NODE for particle movement is an ANTINODE for air pressure. All drawings in this workbook use Displacement.

Analysis Questions

1. THE OCTAVE RULE: Compare the Fundamental Frequency (\(f_1\)) of the Open Pipe (Table 1) to the Closed Pipe (Table 2). Which frequency is higher? By what factor?

2. THE MISSING DATA: In Table 2, were you able to find a "2nd harmonic" where \(f_2 = 2 \cdot f_1\)? Why does the physics of the boundary prevent this resonance from happening?

3. PREDICTION: If you were to open the sealed end of the closed pipe while it was playing its fundamental frequency, would the pitch go up or down? Explain using wavelength patterns.

Percent Error Calculation

Theoretical Velocity at 20°C is 343 m/s.

Open Pipe Error

___________ %

Closed Pipe Error

___________ %

V. Discussion & Conclusion

Analytical Reflection

1. Compare the calculated velocities from Dataset A and Dataset B. Were they consistent? Based on the physics of sound, explain why the speed of sound should be the same regardless of whether the tube is open or closed.

Error Source Identification

2. Discuss the sources of experimental error in your simulation data collection. Why might the calculated velocities deviate slightly from the theoretical 343 m/s? (Think about the precision of finding the exact resonant frequency).

Synthesis & Application

3. Final Conclusion: State clearly whether your results support or refute your hypothesis. How did the comparison of the two different air column types validate the universal properties of longitudinal waves?

Report Submission Checklist

Complete 10 trials total // Show velocity work // Calculated percent error // Formal discussion

Error Analysis (Percent Error)

Open Tube Accuracy

Show math here...

___________ %

Closed Tube Accuracy

Show math here...

___________ %

V. Discussion & Conclusion

Analytical Comparison

1. Compare the calculated velocities from Dataset A (Open) and Dataset B (Closed). Why is it scientifically significant that the velocity remained approximately constant despite the change in boundary conditions and resonant patterns?

Wave Properties Identification

2. Describe the physical relationship between the calculated wavelength and the tube length. How did the "Wavelength Formula" column in your data table change when you switched from an open tube to a closed tube?

Conclusion

3. Final Claim: Based on your experimental data, how does the boundary of a medium affect the velocity of waves within that medium? Discuss any potential sources of human error in identifying the exact frequency of resonance.

Formal Submission Requirements

5 trials per tube type // Correct wavelength formulas used // Percent error calculations shown // All discussion questions answered with physical evidence.

V. Discussion & Conclusion

Harmonic Pattern Analysis

1. Examine the calculated "Ratio" columns for both tube types. How did the sequence of resonant frequencies (\(f_n\)) differ between the open and closed tubes? Use your data to justify why the closed tube is said to "skip over" the even harmonics.

Wavelength-to-Length Relationship

2. Based on your "Calculated \(\lambda\)" and "Constant Length" values, determine the fractional relationship between length (\(L\)) and wavelength (\(\lambda\)) for the fundamental (\(n=1\)) of both tubes. (e.g., in an open tube, \(L = \_\_\_\_ \lambda\)).

Synthesis of Boundary Rules

3. Final Conclusion: Summarize how the physical boundary condition (Node vs. Antinode) limits which wavelengths are allowed to resonate in a pipe. How did your data support the node-antinode requirements for standing waves?

Formal Report Checklist

10 total frequency trials // Wavelength formulas derived // Ratios calculated for all harmonics // Synthesis discussion complete.

Qualitative Observations

Observe the standing wave visual pattern for each resonant state. Record the visual differences between open and closed tubes as frequency increases.

V. Discussion & Conclusion

Harmonic Pattern Analysis

1. Examine the calculated "Ratio" columns for both tube types. How did the sequence of resonant frequencies (\(f_n\)) differ? Why did the closed tube "skip" the even-numbered harmonics?

Velocity Consistency

2. Look at the "Velocity" column in your datasets. How consistent was the calculated speed of sound across the different trials and harmonics? Explain why the medium dictates the velocity rather than the boundary condition.

Synthesis of Boundary Rules

3. Final Conclusion: Summarize how the physical boundary condition (Node vs. Antinode) limits which wavelengths are allowed to resonate. Use your experimental ratios as evidence.

Formal Report Checklist

10 total frequency trials // Velocity calculated for each trial // Ratios calculated for all harmonics // Synthesis discussion complete.

3. Final Synthesis (The Node-Antinode Rule)

Summary: The physical boundaries dictate the "allowed" wave patterns. Open ends (Antinodes) allow air to oscillate; closed ends (Nodes) restrict it. This means Open Pipes fit \(\frac{1}{2}\lambda\) segments, while Closed Pipes fit \(\frac{1}{4}\lambda\) segments (and only odd multiples). The experimental ratios of 1:2:3 vs 1:3:5 provide direct mathematical proof of these geometric constraints.

Common Student Errors

• • Precision Error: Students not finding the maximum stable amplitude in the simulation.
• • Harmonic Mismatch: Misidentifying the 2nd resonant state of a closed pipe as \(n=2\) (it's actually \(n=3\)).
• • Calculation: Using \(L\) instead of \(\lambda\) for the velocity formula.