Honors + AP Physics: HS-PS4 Waves and Their Applications

Honors + AP: Acoustics - Sound, Resonance, and Wave Interactions

This unit extends the waves and sound portion of the Waves and Radiation page. Students use sound as a physical wave system to investigate reflection, Doppler shift, shock waves, interference, beats, standing waves, and resonance in closed pipes.

Unit Question: How can sound reveal the behavior of waves that cannot easily be seen?

NGSS Standards Alignment

This acoustics unit is built around the high school NGSS wave standards. The strongest match is HS-PS4-1, because students use mathematical relationships among wave speed, wavelength, and frequency for sound waves in air. The inquiry lab also supports HS-PS4-5 when students communicate how wave behavior is used in devices such as sonar, ultrasound, microphones, and musical instruments.

HS-PS4-1

Use mathematical representations to support a claim about relationships among frequency, wavelength, and speed of waves traveling in various media.

HS-PS4-5

Communicate technical information about how devices use wave behavior and wave interactions with matter to transmit and capture information and energy.

HS-PS4-2

Evaluate questions about information transmission. This is a secondary connection through microphones, ultrasound, sonar, and audio recording technology.

HS-PS4 Topic Arrangement

Wave properties, information technologies, instrumentation, and applications of wave interactions.

Science and Engineering Practices

  • SEP Using Mathematics and Computational Thinking: Students use \(v=f\lambda\), echo timing, Doppler equations, and percent error.
  • SEP Planning and Carrying Out Investigations: Students collect resonance data from a closed-pipe air column.
  • SEP Analyzing and Interpreting Data: Students identify resonance lengths and use them to calculate wavelength and wave speed.
  • SEP Obtaining, Evaluating, and Communicating Information: Students explain how acoustic technologies use wave behavior.

Disciplinary Core Ideas

  • DCI PS4.A Wave Properties: Wavelength, frequency, speed, reflection, interference, resonance, and wave superposition.
  • DCI PS4.C Information Technologies and Instrumentation: Sonar, ultrasound, microphones, speakers, and musical instruments.

Crosscutting Concepts

  • CCC Patterns: Harmonic frequencies, resonance lengths, and beat frequency patterns.
  • CCC Cause and Effect: Changes in motion, path length, and boundary conditions cause observable sound effects.
  • CCC Systems and System Models: Pipes, strings, air columns, and listening environments are modeled as interacting wave systems.

Foundations of Sound and Acoustics

HS-PS4-1 HS-PS4-5 Using Math PS4.A Patterns

Sound is a mechanical longitudinal wave. It requires a medium such as air, water, metal, or another material. In air, sound travels as repeating regions of compression and rarefaction. The air molecules do not travel all the way from the source to your ear; instead, energy is transferred through local vibrations of the medium.

\[ v=f\lambda \]

Frequency

The number of vibrations per second. Frequency is measured in hertz, \(\text{Hz}\), and is related to pitch.

Wavelength

The distance between repeating parts of a wave, such as compression to compression.

Amplitude

The size of the pressure variation. Larger amplitude usually means louder sound.

Speed of Sound

The speed depends on the medium and temperature. At room temperature in air, a common estimate is \(343\ \text{m/s}\).

Conceptual Question: Why can sound travel through a wall but not through empty space?
Show Solution
Sound needs particles in a medium to vibrate and pass energy along. A wall has matter that can vibrate, but a vacuum has no medium to carry a mechanical sound wave.
Calculation Question: A tuning fork has a frequency of \(512\ \text{Hz}\). If sound travels at \(343\ \text{m/s}\), what is the wavelength?
Show Solution
\[\lambda=\frac{v}{f}=\frac{343}{512}=0.670\ \text{m}\]

Echoes and Reflection of Sound

HS-PS4-1 HS-PS4-5 Using Math PS4.A / PS4.C Cause & Effect

An echo occurs when sound reflects from a surface and returns to the listener after a noticeable delay. Echoes connect wave reflection to real technologies such as sonar, ultrasound imaging, and echolocation.

\[ d=\frac{vt}{2} \]

The factor of \(2\) appears because the measured time usually includes the trip from the source to the reflecting surface and back to the listener.

Conceptual Question: Why are echoes more noticeable in a canyon or empty gym than in a small carpeted room?
Show Solution
Hard, distant surfaces reflect sound well and create a time delay large enough to distinguish the reflected sound from the original. Soft materials absorb sound and nearby walls return the sound too quickly to be heard as a separate echo.
Calculation Question: A climber shouts and hears an echo \(1.80\ \text{s}\) later. If \(v=343\ \text{m/s}\), how far away is the cliff?
Show Solution
\[d=\frac{vt}{2}=\frac{(343)(1.80)}{2}=309\ \text{m}\]

The Doppler Effect

HS-PS4-1 HS-PS4-5 Developing Models PS4.A / PS4.C Cause & Effect

The Doppler effect is the apparent change in frequency caused by relative motion between a wave source and an observer. When a source moves toward an observer, wavefronts are compressed and the observed frequency is higher. When the source moves away, wavefronts are stretched and the observed frequency is lower.

\[ f' = f\left(\frac{v \pm v_o}{v \mp v_s}\right) \]

For a stationary observer and a moving source, a useful classroom form is:

\[ f' = f\left(\frac{v}{v-v_s}\right)\quad \text{approaching} \qquad f' = f\left(\frac{v}{v+v_s}\right)\quad \text{receding} \]
Conceptual Question: Why does an ambulance siren seem to drop in pitch as it passes you?
Show Solution
Before passing, the ambulance moves toward you, so wavefronts arrive more frequently. After passing, it moves away, so wavefronts arrive less frequently. The source frequency has not changed; the observed frequency has changed.
Calculation Question: A car horn emits \(500\ \text{Hz}\) while the car approaches a stationary observer at \(30\ \text{m/s}\). Use \(v=343\ \text{m/s}\). What frequency does the observer hear?
Show Solution
\[f'=500\left(\frac{343}{343-30}\right)=548\ \text{Hz}\]

Sonic Booms and Shock Waves

HS-PS4-1 Developing Models PS4.A Patterns

A sonic boom forms when an object travels faster than sound through a medium. The object outruns its own pressure waves, causing wavefronts to pile up into a shock wave. A common misconception is that the boom happens only at the instant the object breaks the sound barrier. In reality, a supersonic object continuously creates a shock wave as it travels.

\[ \text{Mach Number}=\frac{v_{\text{object}}}{v_{\text{sound}}} \]
Conceptual Question: Why is a sonic boom a shock wave instead of just a louder version of normal sound?
Show Solution
Normal sound waves spread separately. A supersonic source catches up to and compresses many wavefronts into a sharp pressure change, creating a shock wave.
Calculation Question: A jet travels at \(686\ \text{m/s}\) through air where the speed of sound is \(343\ \text{m/s}\). What is its Mach number?
Show Solution
\[M=\frac{686}{343}=2.00\] The jet is traveling at Mach 2.

Interference and Superposition

HS-PS4-1 HS-PS4-5 Analyzing Data PS4.A / PS4.C Systems & Models

Interference occurs when waves overlap. The displacement or pressure variation at a point is the sum of the individual wave effects. This is called superposition.

Constructive Interference

Waves arrive in phase and reinforce each other.

\[ \Delta x=m\lambda \]

Destructive Interference

Waves arrive out of phase and reduce or cancel each other.

\[ \Delta x=\left(m+\frac{1}{2}\right)\lambda \]
Conceptual Question: Why can two speakers playing the same tone create quiet spots in a room?
Show Solution
At some points, the sound from one speaker travels a path that is half a wavelength different from the other speaker. The waves arrive out of phase and partially cancel.
Calculation Question: Two speakers emit \(686\ \text{Hz}\) sound. If \(v=343\ \text{m/s}\), what path difference gives the first destructive interference?
Show Solution
\[\lambda=\frac{343}{686}=0.500\ \text{m}\] First destructive interference occurs at \(\Delta x=\frac{\lambda}{2}=0.250\ \text{m}\).

Beat Frequencies

HS-PS4-1 Using Math PS4.A Patterns

Beat frequencies occur when two waves with nearly equal frequencies interfere. The listener hears a single tone that gets louder and softer at a steady rate. Musicians use beats to tune instruments: as the two frequencies become closer, the beat frequency becomes slower.

\[ f_{\text{beat}}=|f_1-f_2| \]
Classroom demonstration: Use two tone generators or two tuning forks with nearby frequencies, such as \(440\ \text{Hz}\) and \(442\ \text{Hz}\). Students should hear two loudness pulses per second.
Conceptual Question: Why do beats disappear when two instruments are perfectly in tune?
Show Solution
If the frequencies are equal, their phase relationship does not drift over time. The alternating reinforcement and cancellation stops, so the beat frequency becomes zero.
Calculation Question: Two tuning forks produce frequencies of \(256\ \text{Hz}\) and \(260\ \text{Hz}\). What beat frequency is heard?
Show Solution
\[f_{\text{beat}}=|260-256|=4\ \text{Hz}\]

Standing Waves and Resonance

HS-PS4-1 HS-PS4-5 Developing Models PS4.A / PS4.C Systems & Models

A standing wave forms when waves reflect and interfere in a stable pattern. In sound systems, resonance occurs when a driving frequency matches one of the natural standing-wave patterns of the system. At resonance, sound becomes much louder because energy is transferred efficiently into the oscillation.

Open End

An open end of an air column behaves like a displacement antinode and pressure node.

Closed End

A closed end behaves like a displacement node and pressure antinode.

Standing wave and resonance diagrams showing closed-open pipe, open-open pipe, and fixed string harmonics with nodes and antinodes.
Standing waves and resonance shown with node and antinode patterns for closed-open pipes, open-open pipes, and fixed strings.
\[ L=\frac{\lambda}{4},\ \frac{3\lambda}{4},\ \frac{5\lambda}{4},\dots \]

Successive resonance lengths in a closed pipe are separated by half a wavelength:

\[ L_{n+1}-L_n=\frac{\lambda}{2} \]
Conceptual Question: Why does a closed pipe support only odd quarter-wavelength resonances?
Show Solution
The closed end must be a displacement node, and the open end must be a displacement antinode. The patterns that satisfy those two boundary conditions are \(\lambda/4\), \(3\lambda/4\), \(5\lambda/4\), and so on.

Why Harmonics and Resonance Exist

HS-PS4-1 HS-PS4-5 Developing Models PS4.A Patterns Cause & Effect

Every physical system that supports waves has boundary conditions—constraints at the edges that dictate how the wave must behave at those points. Boundary conditions are the fundamental reason that only certain wavelengths and frequencies can sustain themselves inside a system. A wave that does not satisfy its boundary conditions will destructively interfere with itself and die out. A wave that does satisfy them will constructively reinforce with each reflection, growing into a large, stable oscillation called a standing wave. The set of frequencies that produce standing waves are the system's harmonics, and the amplification that occurs when a driving force matches one of those frequencies is resonance.

Boundary Conditions: The Origin of Harmonics

Consider a wave bouncing back and forth inside a confined system. Each time it reflects, it overlaps with incoming waves. For most frequencies the reflected and incoming waves are out of step and cancel. But at specific frequencies, the round-trip path length is an exact number of half-wavelengths, so the reflections line up perfectly and produce standing waves. The pattern at the boundaries determines which set of harmonics is allowed:

Pipes (Air Columns)

A closed end forces a displacement node; an open end forces a displacement antinode. A pipe closed at one end supports only odd-numbered harmonics because only odd multiples of a quarter wavelength fit those mixed boundary conditions. A pipe open at both ends supports all harmonics because identical boundary conditions at each end allow both even and odd multiples of a half wavelength.

Strings

Both ends of a string are fixed, creating nodes at each end. The string must fit an integer number of half-wavelengths between its endpoints. This means all harmonics are possible: the fundamental is one half-wavelength, the second harmonic is two half-wavelengths, and so on. The frequency of each harmonic is an integer multiple of the fundamental.

Architecture

A room or concert hall is a three-dimensional resonant system. Sound reflects between walls, floor, and ceiling, creating room modes—standing wave patterns at specific frequencies determined by the room's dimensions. Architects design concert halls to spread these modes evenly and add absorptive materials so no single frequency dominates, avoiding the muddy reverberation that comes from strong resonance at a few frequencies.

Why Resonance Amplifies Sound

Resonance is not a new source of energy—it is efficient energy transfer. When a driving force vibrates at the same frequency as a natural mode of the system, each push arrives exactly in phase with the existing oscillation. The amplitude grows with every cycle, like pushing a swing at just the right moment. At any other frequency, the pushes are sometimes in phase and sometimes out of phase, so the oscillation stays small. This is why a tuning fork held over a pipe at the right length produces a dramatic increase in loudness, and why a singer sustaining the right pitch can set a wine glass vibrating until it shatters.

Resonance in Architecture

The physics of resonance extends directly into the built environment. Every enclosed space has a set of resonant frequencies called room modes, determined by the distances between opposing surfaces. For a rectangular room the lowest mode along a single dimension has a half-wavelength equal to that dimension's length:

\[ f_{\text{mode}} = \frac{v}{2L} \]

A small practice room with one wall 3.4 m away from the opposite wall has a first mode near 50 Hz—a low bass frequency that booms and muddies the sound. Concert hall designers combat this by using non-parallel walls, diffusive surfaces, and absorptive panels at calculated positions to break up strong standing-wave patterns. The Elbphilharmonie in Hamburg, for example, uses algorithmically carved wall panels to scatter sound in thousands of directions, preventing any single room mode from dominating.

Problematic Resonance

The Tacoma Narrows Bridge (1940) collapsed when wind excited a torsional resonance mode. Soldiers break step crossing footbridges to avoid driving a resonance at the bridge's natural frequency. Tall buildings sway in wind gusts that happen to match a structural mode.

Designed Resonance

Organ pipes, guitar bodies, and wind instruments are deliberately built to resonate at useful frequencies. A violin body amplifies the strings' vibrations by resonating over a broad range. Whispering galleries in domed buildings focus sound along curved surfaces using constructive interference of reflections.

Conceptual Question: A new auditorium has a persistent low hum at around 85 Hz even when no sound system is running. A physicist measures the distance between the front and rear walls to be 2.0 m. Explain what is likely happening and suggest a fix.
Show Solution
The distance between the walls creates a room mode at \(f = v/(2L) = 343/(2 \times 2.0) = 85.8\ \text{Hz}\), very close to 85 Hz. Environmental noise near that frequency is being amplified by resonance. Possible fixes include adding absorptive panels on the rear wall to reduce reflection, or installing a bass trap tuned to that frequency range to dissipate the energy.
Conceptual Question: Explain why a tuning fork sounds much louder when its stem is pressed against a tabletop.
Show Solution
The tuning fork alone has small tines that push very little air. When pressed to a table, the fork drives the large surface area of the table at the fork's frequency. The table has many closely spaced resonance modes, and if any of them are near the fork frequency, the table vibrates efficiently and pushes much more air, producing a louder sound. Even without a perfect frequency match, the large surface radiates sound more effectively than the small tines.

Harmonics in Pipes

HS-PS4-1 HS-PS4-5 Using Math Developing Models PS4.A / PS4.C Patterns

A pipe supports standing waves at specific frequencies determined by its length and whether each end is open or closed. The lowest allowed frequency is the fundamental (first harmonic). Higher harmonics are integer multiples of the fundamental that also satisfy the boundary conditions. The full set of harmonics gives an instrument its timbre—a clarinet (effectively a closed pipe) sounds different from a flute (effectively an open pipe) because they emphasize different harmonics.

Open Pipe (Open at Both Ends)

Both ends are displacement antinodes. The fundamental fits one half-wavelength into the pipe length. All integer harmonics are present.

\[ f_n = n\,\frac{v}{2L} \qquad n = 1,\,2,\,3,\,4,\dots \]
Pipe harmonic diagrams comparing closed-open and open-open pipes with first, third, and fifth modes for closed pipes and first three modes for open pipes.
Pipe harmonics shown side by side: closed-open pipes support odd harmonics, while open-open pipes support each integer harmonic.

Closed Pipe (Closed at One End)

The closed end is a displacement node and the open end is a displacement antinode. The fundamental fits one quarter-wavelength into the pipe. Only odd harmonics are present because only odd multiples of a quarter wavelength satisfy these mixed boundary conditions.

\[ f_n = n\,\frac{v}{4L} \qquad n = 1,\,3,\,5,\,7,\dots \quad\text{(odd only)} \]
Closed-open pipe resonance diagrams showing the first, third, and fifth harmonics with labeled nodes and antinodes.
Closed-open pipe resonance supports the fundamental plus odd harmonics because one end is a node and the other is an antinode.

Comparing Open and Closed Pipes

PropertyOpen PipeClosed Pipe
Boundary conditionsAntinode at both endsNode at closed end, antinode at open end
Fundamental wavelength\(\lambda_1 = 2L\)\(\lambda_1 = 4L\)
Fundamental frequency\(f_1 = v/(2L)\)\(f_1 = v/(4L)\)
Harmonics presentAll: \(f_1, 2f_1, 3f_1, \dots\)Odd only: \(f_1, 3f_1, 5f_1, \dots\)
Instrument examplesFlute, organ pipe (open)Clarinet, bottle, closed organ pipe
Conceptual Question: A flute and a clarinet have air columns of the same length. Why does the clarinet sound about an octave lower?
Show Solution
A flute behaves approximately like an open pipe with \(f_1 = v/(2L)\). A clarinet behaves approximately like a closed pipe with \(f_1 = v/(4L)\). For the same length, the clarinet's fundamental is half the flute's fundamental, which is one octave lower.
Calculation Question: An open organ pipe is \(0.65\ \text{m}\) long. Using \(v = 343\ \text{m/s}\), find the frequencies of the first three harmonics.
Show Solution
\[f_1 = \frac{v}{2L} = \frac{343}{2(0.65)} = 264\ \text{Hz}\] \[f_2 = 2f_1 = 528\ \text{Hz}\] \[f_3 = 3f_1 = 792\ \text{Hz}\]
Calculation Question: A closed pipe produces a fundamental frequency of \(130\ \text{Hz}\). What are the next two resonant frequencies?
Show Solution
Only odd harmonics exist in a closed pipe: \[f_3 = 3f_1 = 3(130) = 390\ \text{Hz}\] \[f_5 = 5f_1 = 5(130) = 650\ \text{Hz}\]
Calculation Question: A student blows across the top of a bottle and hears a tone at \(220\ \text{Hz}\). The air column inside is \(0.39\ \text{m}\). Is this consistent with a closed-pipe model? Use \(v = 343\ \text{m/s}\).
Show Solution
For a closed pipe: \[f_1 = \frac{v}{4L} = \frac{343}{4(0.39)} = 220\ \text{Hz}\] Yes, the measured frequency matches the closed-pipe fundamental exactly. The bottom of the bottle acts as the closed end and the opening acts as the open end.

Harmonics on Strings

HS-PS4-1 HS-PS4-5 Using Math Developing Models PS4.A / PS4.C Patterns Cause & Effect

A vibrating string fixed at both ends is one of the clearest examples of harmonic behavior. Because both endpoints are fixed, they must be displacement nodes. The string must therefore fit an integer number of half-wavelengths between its endpoints. The lowest frequency that satisfies this condition is the fundamental (first harmonic), and every higher harmonic adds one more half-wavelength loop to the pattern.

The Harmonic Series on a String

\[ f_n = n\,\frac{v}{2L} = \frac{n}{2L}\sqrt{\frac{T}{\mu}} \qquad n = 1,\,2,\,3,\,4,\dots \]

Here \(L\) is the string length, \(v\) is the wave speed on the string, \(T\) is the tension, and \(\mu\) is the linear mass density (mass per unit length, in kg/m). The wave speed on a string is \(v = \sqrt{T/\mu}\).

Wavelength Pattern

The wavelength of the \(n\)-th harmonic is: \[\lambda_n = \frac{2L}{n}\] As the harmonic number increases, the wavelength gets shorter and the frequency gets higher.

What Controls Pitch?

A string's fundamental frequency depends on three things: length (shorter = higher pitch), tension (tighter = higher pitch), and mass density (thinner/lighter = higher pitch). Musicians use all three: pressing a fret shortens the string, tuning pegs change tension, and different strings have different thicknesses.

Illustrated Harmonic Modes

The diagrams below show the displacement pattern for each harmonic on a string of length \(L\). Nodes (points of zero displacement) are shown in green and antinodes (points of maximum displacement) are shown in gold. The dashed line shows the equilibrium position of the string.

Standing waves on a fixed-fixed string showing the first, second, and third harmonics with nodes and antinodes.
String harmonics shown for fixed-fixed ends: each higher harmonic adds another half-wavelength loop between the endpoints.

Harmonic Summary Table

Harmonic \(n\)LoopsNodesAntinodesWavelength \(\lambda_n\)Frequency \(f_n\)
1 (fundamental)121\(2L\)\(f_1\)
2232\(L\)\(2f_1\)
3343\(2L/3\)\(3f_1\)
4454\(L/2\)\(4f_1\)
\(n\)\(n\)\(n+1\)\(n\)\(2L/n\)\(nf_1\)

Factors That Affect String Frequency

Musical Connections

When a guitar string vibrates, it does not produce just one frequency. It vibrates simultaneously in its fundamental and many overtones. The relative strength of each harmonic is what gives the instrument its characteristic timbre. Lightly touching the string at its midpoint while plucking damps the fundamental and odd harmonics, forcing the string to vibrate at the second harmonic—the familiar "harmonic" technique that guitarists and violinists use. Touching at the one-third point isolates the third harmonic, and so on. The note's pitch jumps up by an octave (2nd harmonic), an octave plus a fifth (3rd), two octaves (4th), etc.

Sample Questions

Conceptual Question 1: A guitar string is plucked and produces its fundamental frequency. Without changing the tension or string, what could a guitarist do to double the fundamental frequency?
Show Solution
Since \(f_1 = v/(2L)\), doubling the frequency requires halving the vibrating length. The guitarist can press the string at its midpoint (the 12th fret on a guitar), cutting the effective length in half.
Conceptual Question 2: Two identical strings are tuned to the same tension. One string is touched lightly at its midpoint and plucked. The other is not touched and is plucked normally. Compare the pitches heard.
Show Solution
The untouched string vibrates primarily at its fundamental \(f_1\) along with overtones. The string touched at the midpoint is forced to have a node at the center, which suppresses the fundamental and all odd harmonics. It vibrates at the second harmonic \(2f_1\), sounding one octave higher than the other string's fundamental.
Conceptual Question 3: Why do bass guitar strings appear thicker than treble strings?
Show Solution
Bass strings need a lower fundamental frequency. Since all strings on the instrument have roughly the same length and are tuned to similar tension ranges, the only remaining way to lower the frequency is to increase the linear mass density \(\mu\). A thicker (or wound) string has more mass per unit length, which reduces the wave speed and lowers the frequency: \(f_1 = \frac{1}{2L}\sqrt{T/\mu}\).
Calculation Question 1: A violin string is \(0.33\ \text{m}\) long. The wave speed on the string is \(280\ \text{m/s}\). Find the fundamental frequency and the frequencies of the 2nd and 3rd harmonics.
Show Solution
\[f_1 = \frac{v}{2L} = \frac{280}{2(0.33)} = 424\ \text{Hz}\] \[f_2 = 2f_1 = 848\ \text{Hz}\] \[f_3 = 3f_1 = 1{,}272\ \text{Hz}\]
Calculation Question 2: A string has a length of \(0.80\ \text{m}\), a tension of \(200\ \text{N}\), and a linear mass density of \(0.0050\ \text{kg/m}\). Find the wave speed on the string and the fundamental frequency.
Show Solution
\[v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200}{0.0050}} = \sqrt{40{,}000} = 200\ \text{m/s}\] \[f_1 = \frac{v}{2L} = \frac{200}{2(0.80)} = 125\ \text{Hz}\]
Calculation Question 3: A standing wave on a \(1.20\ \text{m}\) string has 4 loops (antinodes). What harmonic is this? What is the wavelength? If the wave speed is \(150\ \text{m/s}\), what is the frequency?
Show Solution
Four loops means the 4th harmonic (\(n = 4\)). \[\lambda_4 = \frac{2L}{n} = \frac{2(1.20)}{4} = 0.60\ \text{m}\] \[f_4 = \frac{v}{\lambda_4} = \frac{150}{0.60} = 250\ \text{Hz}\] The fundamental would be \(f_1 = 250/4 = 62.5\ \text{Hz}\).
Calculation Question 4: A piano wire produces a fundamental frequency of \(262\ \text{Hz}\) (middle C). The wire is \(0.65\ \text{m}\) long with a linear mass density of \(0.0090\ \text{kg/m}\). What tension must the wire be under?
Show Solution
Start from the fundamental frequency formula and solve for tension: \[f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\] \[2Lf_1 = \sqrt{\frac{T}{\mu}}\] \[T = \mu(2Lf_1)^2 = (0.0090)\bigl[2(0.65)(262)\bigr]^2\] \[T = (0.0090)(340.6)^2 = (0.0090)(116{,}008) = 1{,}044\ \text{N}\]
Calculation Question 5: A \(0.50\ \text{m}\) string vibrates at its third harmonic with a frequency of \(900\ \text{Hz}\). What is the fundamental frequency? What is the wave speed on the string? What is the wavelength of the third harmonic?
Show Solution
\[f_1 = \frac{f_3}{3} = \frac{900}{3} = 300\ \text{Hz}\] \[v = 2Lf_1 = 2(0.50)(300) = 300\ \text{m/s}\] \[\lambda_3 = \frac{2L}{3} = \frac{2(0.50)}{3} = 0.333\ \text{m}\] Check: \(v = f_3\lambda_3 = (900)(0.333) = 300\ \text{m/s}\) ✓
Challenge Question: A string of length \(L\) and fundamental frequency \(f_1\) is replaced by a string of the same material and tension but twice the diameter. By what factor does the fundamental frequency change?
Show Solution
The linear mass density \(\mu\) depends on the cross-sectional area of the string. Doubling the diameter quadruples the cross-sectional area, so \(\mu\) increases by a factor of 4. \[f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \propto \frac{1}{\sqrt{\mu}}\] If \(\mu\) increases by 4, \(f_1\) decreases by \(\sqrt{4} = 2\). The new fundamental frequency is half the original: \(f_1' = f_1/2\), which is one octave lower.

Inquiry Lab: Measuring the Speed of Sound with a Closed Pipe

HS-PS4-1 HS-PS4-5 Planning Investigations Analyzing Data PS4.A / PS4.C Patterns

Essential Question: How can resonance in an air column be used to determine wavelength and measure the speed of sound?

In this lab, students use a known tuning fork frequency and a closed-pipe resonance tube to determine the wavelength of the sound wave and calculate the experimental speed of sound.

Materials

  • Resonance tube or clear graduated tube
  • Water reservoir or tall cylinder partially filled with water
  • Known tuning fork, such as \(256\ \text{Hz}\), \(512\ \text{Hz}\), or another assigned frequency
  • Rubber striker or soft tuning fork mallet
  • Meter stick or measuring tape
  • Thermometer, if available
  • Data table and calculator

Safety and Setup

  • Do not strike tuning forks on lab tables or hard metal surfaces.
  • Keep water away from electrical devices and outlets.
  • Keep the tuning fork above the tube opening; do not hit the glass or plastic tube.
  • Work quietly enough to hear resonance clearly.

Investigation Procedure

  1. Predict: Use \(v=f\lambda\) and \(v\approx343\ \text{m/s}\) to estimate the wavelength for your tuning fork.
  2. Predict the first resonance: Estimate \(L_1\approx\lambda/4\).
  3. Create the closed pipe: Raise or lower the water level so the air column above the water changes length. The water surface acts like the closed end.
  4. Strike the tuning fork: Hold it near the tube opening without touching the tube.
  5. Find resonance: Slowly change the air column length until the sound becomes noticeably louder.
  6. Record \(L_1\): Measure the air column length from the water surface to the open end of the tube.
  7. Find \(L_2\): Continue changing the air column length until the next loud resonance occurs.
  8. Calculate wavelength: Use \(L_2-L_1=\lambda/2\), so \(\lambda=2(L_2-L_1)\).
  9. Calculate speed: Use \(v=f\lambda\).
  10. Evaluate: Compare your result to the accepted value and identify likely sources of error.

Data Table

TrialTuning Fork Frequency \(f\)First Resonance \(L_1\)Second Resonance \(L_2\)\(\lambda=2(L_2-L_1)\)\(v=f\lambda\)
1
2
3

Analysis Questions

  1. Why does the sound become louder at resonance?
  2. Why is the water surface treated as the closed end of the pipe?
  3. Why is using \(L_2-L_1\) often better than using only \(L_1\)?
  4. What would happen to the resonance lengths if you used a higher-frequency tuning fork?
  5. How would a warmer classroom affect the speed of sound?
  6. What sources of error could make your measured speed too high or too low?

Sample Calculation

Sample: A \(512\ \text{Hz}\) tuning fork produces resonances at \(0.168\ \text{m}\) and \(0.503\ \text{m}\). Find the wavelength and speed of sound.
Show Solution
\[\lambda=2(L_2-L_1)=2(0.503-0.168)=0.670\ \text{m}\] \[v=f\lambda=(512)(0.670)=343\ \text{m/s}\]

Conclusion Claim

Write a paragraph using the claim-evidence-reasoning format:

  • Claim: State your measured speed of sound.
  • Evidence: Use your resonance lengths, calculated wavelength, and frequency.
  • Reasoning: Explain why resonance in a closed pipe allowed you to determine wavelength and speed.

Engineering Extension: Design a Simple Wind Instrument

HS-PS4-1 HS-PS4-5 Constructing Explanations Communicating Information PS4.A / PS4.C Systems & Models

Design Challenge: Use wave speed, frequency, wavelength, and boundary conditions to design a simple wind instrument that produces a target frequency.

Design Constraints

  • Your instrument must produce at least one measurable target frequency.
  • You must explain whether it behaves more like an open pipe or a closed pipe.
  • You must calculate the expected length before testing.
  • You must test the sound using a tuner app, frequency analyzer, or classroom sensor.
  • You must revise your design based on evidence.

Deliverables

Design Sketch

Include dimensions, open/closed ends, and where the sound is produced.

Physics Calculation

Show how frequency, wavelength, and pipe length are related.

Test Data

Record the target frequency, measured frequency, and percent error.

Revision Explanation

Explain how your design changed and why.

Rubric

ScoreDescription
1Design is attempted but wave behavior is not clearly connected to the instrument.
2Design produces sound, but calculations or explanations are incomplete.
3Design uses relevant wave equations and includes basic testing.
4Design, calculations, measurements, and explanations are accurate and clearly connected.
5Design includes thoughtful revision, strong evidence, and a clear explanation of boundary conditions and resonance.
Note: Solutions are hidden in expandable sections so students can check their work after attempting each problem. This unit addresses NGSS HS-PS4: Waves and Their Applications. Standards information from nextgenscience.org.