Unit 6 Waves and Sound: Mechanical Waves, Sound, and Standing-Wave Resonance

Properties of Waves

A wave is a traveling disturbance that transfers energy and momentum from one place to another without permanently transporting matter along with it. That “energy without net matter flow” idea is the reason waves show up everywhere in physics: they let one part of a system influence another at a distance through a medium (sound, water waves, waves on a string) or through fields (electromagnetic waves, which you’ll meet in physical optics).

What a wave is (and what it is not)

When you flick a rope, you see a pulse travel down the rope. Individual pieces of the rope mostly move up and down, but the disturbance pattern moves along the rope. This separation between motion of the medium and motion of the pattern is central.

Two common wave types (based on how the medium oscillates relative to the direction the wave travels):

  • Transverse wave: the medium oscillates perpendicular to the direction of wave propagation (wave on a string, many water surface waves).
  • Longitudinal wave: the medium oscillates parallel to the direction of propagation (sound in air).

A common misconception is that “longitudinal waves don’t have amplitude” because you don’t see the motion as easily. They do: the oscillation is in pressure/density (compressions and rarefactions), and the size of those pressure variations is the wave’s amplitude.

Key wave quantities and how they connect

To describe a repeating wave, you use a small set of quantities that are tightly linked:

  • Amplitude: the maximum displacement (or pressure variation, etc.) from equilibrium.
  • Wavelength: the spatial period—the distance between repeating points (crest-to-crest, compression-to-compression).
  • Period: time for one cycle of oscillation.
  • Frequency: cycles per second.
  • Wave speed: how fast the pattern moves through space.

Their most important relationships are:

f = \frac{1}{T}

v = f\lambda

Here, f is frequency (Hz), T is period (s), v is wave speed (m/s), and \lambda is wavelength (m).

Why these matter: AP questions often give you two of these quantities and expect you to infer the others, or to reason about what changes when the medium changes versus when the source changes.

What sets wave speed (source vs medium)

A very testable idea is that wave speed is determined by the properties of the medium, not by how hard you shake the source.

  • If you shake a string faster (increase f), the wavelength changes so that v = f\lambda still holds.
  • If you change the string’s tension or mass density, the speed changes—and then the wavelength adjusts for a given driving frequency.

Students often mix this up by assuming “bigger amplitude means faster wave.” In ideal linear media, amplitude affects energy carried, not speed.

Energy transport and intensity

Waves carry energy. For many mechanical waves, increasing amplitude increases energy transport.

For sound (developed more in the next section), a key quantity is intensity, the power delivered per area:

I = \frac{P}{A}

If a source radiates sound roughly equally in all directions (spherical spreading), the area grows like a sphere:

A = 4\pi r^2

so intensity falls with distance:

I = \frac{P}{4\pi r^2}

This inverse-square relationship is a frequent reasoning target: doubling distance makes intensity one-fourth.

Superposition: the rule that explains interference

The principle of superposition says that when two (or more) waves overlap in space, the resulting displacement is the algebraic sum of the individual displacements.

Superposition matters because it is the engine behind:

  • Interference (constructive and destructive)
  • Standing waves (a special interference pattern)
  • Many resonance phenomena

A subtle but important point: destructive interference does not “destroy energy.” It redistributes energy to other locations (and sometimes into other forms if damping is present). On an exam, “the wave disappears” is usually a red flag.

Interference patterns (conceptual level)

When two sinusoidal waves of the same frequency overlap, the phase relationship determines the net amplitude.

  • Constructive interference: waves arrive “in step,” producing larger amplitude.
  • Destructive interference: waves arrive “out of step,” producing smaller amplitude or cancellation.

For path-based reasoning (very common), the condition for constructive and destructive interference for waves of the same wavelength is:

\Delta r = m\lambda

\Delta r = \left(m + \frac{1}{2}\right)\lambda

Here \Delta r is path difference and m is an integer (0, 1, 2, …). These conditions show up more in physical optics, but the same logic applies to sound (for example, two speakers producing quiet and loud spots in a room).

Reflection at boundaries (including phase inversion)

When a wave reaches a boundary (like a pulse on a string reaching an end), some energy can reflect and some can transmit.

For transverse waves on a string, whether the reflected pulse flips depends on the boundary condition:

  • Fixed end: the end cannot move, so the reflected wave undergoes a phase inversion (a crest reflects as a trough).
  • Free end: the end can move, so the reflected wave does not invert.

This matters because standing wave formation depends on the reflection behavior at boundaries.

Worked example 1: Using v = f\lambda

A wave travels along a rope at speed v = 12\text{ m/s}. The source vibrates with frequency f = 3.0\text{ Hz}. Find the wavelength.

Reasoning: The medium sets speed, the source sets frequency, and wavelength adjusts accordingly.

\lambda = \frac{v}{f}

\lambda = \frac{12}{3.0} = 4.0\text{ m}

Worked example 2: Intensity change with distance

A small siren radiates sound approximately uniformly. At r = 2.0\text{ m} the intensity is I_1. What is the intensity at r = 6.0\text{ m}?

Reasoning: Use inverse-square spreading.

I \propto \frac{1}{r^2}

\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} = \frac{2.0^2}{6.0^2} = \frac{4}{36} = \frac{1}{9}

So I_2 = \frac{1}{9}I_1.

Exam Focus
  • Typical question patterns:
    • Given two of v, f, \lambda, solve for the third and explain what changes when the source frequency changes versus when the medium changes.
    • Conceptual questions about superposition (what happens when pulses overlap) and whether energy is “lost.”
    • Boundary reflection questions: identify whether the reflection is inverted and connect it to standing wave formation.
  • Common mistakes:
    • Treating wave speed as controlled by amplitude or by frequency of the source (in most ideal AP contexts, speed is set by the medium).
    • Saying destructive interference eliminates energy; instead, emphasize redistribution.
    • Mixing up fixed-end versus free-end reflection (fixed end inverts, free end does not).

Sound Waves and Resonance

Sound is a mechanical longitudinal wave: it requires a medium, and it propagates through oscillations in pressure and density. You hear sound because your eardrum responds to pressure variations at audio frequencies.

How sound travels: compressions and rarefactions

In air, sound consists of alternating regions of higher-than-average pressure (compressions) and lower-than-average pressure (rarefactions). Air molecules oscillate back and forth; they do not travel with the wave over long distances.

Why this matters: Many students picture sound as “particles moving from speaker to ear.” That picture leads to wrong reasoning about speed and frequency. The correct picture is an oscillation pattern that propagates through interactions between neighboring molecules.

Speed of sound: what affects it

The speed of sound depends on the medium’s elastic and inertial properties. In fluids (liquids and gases), a standard relationship is:

v = \sqrt{\frac{B}{\rho}}

Here B is the bulk modulus (how hard it is to compress the medium) and \rho is density. Larger B tends to increase speed; larger \rho tends to decrease speed.

For AP-style reasoning, you’re often expected to know qualitative trends:

  • Sound travels faster in solids than in liquids than in gases (solids are generally much “stiffer”).
  • In a given medium, changing the source frequency does not significantly change the wave speed (in typical non-dispersive conditions).

Frequency, pitch, and the “source vs medium” idea (again)

  • Pitch is your perception closely related to frequency.
  • When you increase the driving frequency of a speaker, the sound in the same air has the same speed, so the wavelength decreases.

Use the same core relationship:

v = f\lambda

This also helps with Doppler and resonance later.

Sound intensity and the decibel scale

The physical quantity tied to loudness is intensity (power per area). Because human hearing spans a huge range, intensity is often expressed in decibels (dB).

The sound level is defined by:

\beta = 10\log_{10}\left(\frac{I}{I_0}\right)

Here \beta is sound level in dB, I is intensity, and I_0 is a reference intensity typically taken as:

I_0 = 1.0\times 10^{-12}\text{ W/m}^2

Key ideas that are often tested:

  • A change of +10 dB corresponds to a factor of 10 increase in intensity.
  • Because it’s logarithmic, adding sources is easiest by adding intensities, not dB values (unless the problem explicitly sets it up for a known rule).

Resonance: why it creates big amplitudes

Resonance happens when a system that can oscillate (like an air column, a string, or a mass-spring) is driven at (or near) one of its natural frequencies. Energy transfer from the driver to the oscillator becomes especially efficient, leading to large amplitude oscillations (limited by damping and nonlinearity).

Resonance matters because it explains:

  • Why certain notes sound much louder on instruments
  • Why a room can “boom” at specific frequencies
  • Why standing waves appear in strings and air columns

A helpful way to think about it:

  • The driver supplies energy each cycle.
  • If the system oscillates in sync with the driver, the driver repeatedly “pushes at the right time,” building amplitude.
  • If the driver is off-frequency, pushes sometimes oppose the motion and energy transfer is less efficient.

A common misconception is that resonance requires matching the driving frequency to “the frequency of the wave in the medium.” The medium supports many wavelengths; resonance is about the system’s boundary conditions selecting specific modes.

Beats: interference in time

If you listen to two tones with close frequencies, you often hear a pulsing loudness called beats. Beats come from superposition: the combined wave alternates between more-constructive and more-destructive interference.

The beat frequency is:

f_{\text{beat}} = \left|f_1 - f_2\right|

This is used for tuning instruments: when beats slow to near zero, frequencies match.

Doppler effect (sound)

The Doppler effect is the change in perceived frequency due to relative motion between the source and the observer.

Conceptually:

  • If the source and observer move closer, wavefronts arrive more frequently, so measured frequency increases.
  • If they move apart, measured frequency decreases.

A commonly used AP sound Doppler formula is:

f' = f\left(\frac{v + v_o}{v - v_s}\right)

Here f' is the observed frequency, f is the source frequency, v is the speed of sound in the medium, v_o is the observer’s speed (positive when moving toward the source), and v_s is the source’s speed (positive when moving toward the observer).

Students frequently flip signs. A safer approach is to reason physically first (should f' be bigger or smaller?), then choose signs to match that.

Worked example 1: Decibels and intensity ratios

A sound’s intensity increases by a factor of 100. By how many dB does the sound level increase?

Use the definition:

\Delta \beta = 10\log_{10}\left(\frac{I_2}{I_1}\right)

\Delta \beta = 10\log_{10}(100) = 10\cdot 2 = 20\text{ dB}

So a 100x intensity increase corresponds to +20 dB.

Worked example 2: Beats

Two tuning forks have frequencies f_1 = 256\text{ Hz} and f_2 = 260\text{ Hz}. What beat frequency do you hear?

f_{\text{beat}} = \left|260 - 256\right| = 4\text{ Hz}

You would hear loudness pulses 4 times per second.

Worked example 3: Doppler effect sign reasoning

A stationary observer hears a siren of frequency f = 900\text{ Hz} on an ambulance approaching at v_s = 30\text{ m/s}. Take the speed of sound as v = 340\text{ m/s}. Find the observed frequency.

Observer is stationary, so v_o = 0. Source approaches, so v_s is positive in the denominator with a minus sign.

f' = f\left(\frac{v}{v - v_s}\right)

f' = 900\left(\frac{340}{340 - 30}\right) = 900\left(\frac{340}{310}\right)

f' \approx 900(1.097) \approx 987\text{ Hz}

Check: approaching should sound higher than 900 Hz, and it does.

Exam Focus
  • Typical question patterns:
    • Convert between intensity ratios and dB changes using the logarithmic definition.
    • Use inverse-square spreading plus dB to compare loudness at different distances.
    • Resonance reasoning: identify when amplitude is maximized and connect to natural frequencies and damping.
  • Common mistakes:
    • Adding dB values directly when combining sources; the physically additive quantity is intensity.
    • Confusing pitch with intensity (frequency with amplitude).
    • Doppler sign errors; always predict higher/lower first, then apply signs to match.

Standing Waves

A standing wave is a stable wave pattern formed by the interference of two waves of the same frequency and amplitude traveling in opposite directions (often an incident wave and its reflection). Instead of the pattern moving along the medium, you get fixed points that always have zero displacement and points that oscillate with maximum displacement.

Standing waves are a centerpiece of AP waves and sound because they connect superposition, boundary conditions, and resonance in one idea: only certain wavelengths “fit” in a system, and those correspond to resonant frequencies.

Nodes and antinodes

In a standing wave:

  • A node is a point that never moves (displacement always zero).
  • An antinode is a point that oscillates with maximum amplitude.

Why this matters: Boundary conditions at the ends of a string or air column force either nodes or antinodes at specific locations. From that, you can derive allowed wavelengths and thus resonant frequencies.

A common misconception is that nodes mean “no energy anywhere.” Energy is still stored and exchanged between kinetic and potential forms locally in the medium; nodes are points of zero displacement, not necessarily “no physics.”

Boundary conditions: strings vs air columns

Standing-wave problems are mostly about translating a physical end condition into node/antinode requirements.

Strings (transverse displacement):

  • Fixed end: displacement must be zero, so a fixed end is a node.

Air columns (sound, longitudinal pressure/displacement):

At an open end of a pipe, the air is free to move, so the displacement is large (displacement antinode). At a closed end, air cannot move, so displacement is zero (displacement node).

Many students memorize “open equals antinode, closed equals node” without noting it refers to displacement. Pressure standing waves are shifted: pressure node corresponds to displacement antinode, and pressure antinode corresponds to displacement node. If a question mentions pressure explicitly, you must switch.

Allowed wavelengths and harmonics on a string (both ends fixed)

Consider a string of length L fixed at both ends. Both ends are nodes. The simplest standing wave (fundamental) fits half a wavelength into the length.

Allowed wavelengths:

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

Allowed frequencies (using f = v/\lambda):

f_n = \frac{nv}{2L}

Here n = 1, 2, 3, \dots is the harmonic number, v is wave speed on the string.

Interpretation: higher n means more “loops” (antinodes) along the string and higher frequency.

Pipes with both ends open (open-open)

An open-open pipe has displacement antinodes at both ends. The pattern of allowed wavelengths matches the “both ends the same” case like a fixed-fixed string (just with displacement antinodes instead of nodes).

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

f_n = \frac{nv}{2L}

Again n = 1, 2, 3, \dots.

Pipes with one end closed (open-closed)

For an open-closed pipe, the closed end is a displacement node and the open end is a displacement antinode. The fundamental is a quarter-wavelength in the length.

Allowed wavelengths:

\lambda_n = \frac{4L}{2n - 1}

Allowed frequencies:

f_n = \frac{(2n - 1)v}{4L}

Here n = 1, 2, 3, \dots. This structure means only odd harmonics appear: 1st, 3rd, 5th, …

A very common error is to plug n = 2 into the open-closed formula and call it the “second harmonic.” It’s the second allowed mode, but it corresponds to the 3rd harmonic relative to the fundamental frequency.

Resonance as “what frequencies fit”

When you drive a string or an air column, resonance happens when the driving frequency matches one of these allowed frequencies. Physically, the reflected wave lines up in phase with the incoming wave in just the right way to reinforce a stable standing pattern.

This is why instruments have discrete notes: their geometry and boundary conditions select specific resonant frequencies.

Worked example 1: Fixed-fixed string resonant frequencies

A string of length L = 0.80\text{ m} has wave speed v = 120\text{ m/s}. Find the first three resonant frequencies.

Use:

f_n = \frac{nv}{2L}

Fundamental:

f_1 = \frac{1\cdot 120}{2(0.80)} = \frac{120}{1.6} = 75\text{ Hz}

Second harmonic:

f_2 = 2f_1 = 150\text{ Hz}

Third harmonic:

f_3 = 3f_1 = 225\text{ Hz}

Notice the linear scaling with n for this boundary condition.

Worked example 2: Open-closed pipe fundamental and next mode

An open-closed tube has length L = 0.25\text{ m} and contains air where v = 340\text{ m/s}. Find the fundamental frequency and the next resonant frequency.

For the fundamental, use:

f_1 = \frac{v}{4L}

f_1 = \frac{340}{4(0.25)} = \frac{340}{1.0} = 340\text{ Hz}

The next allowed mode corresponds to n = 2 in the open-closed formula:

f_2 = \frac{3v}{4L} = 3f_1 = 1020\text{ Hz}

So the resonances are at 340 Hz, 1020 Hz, 1700 Hz, … (odd multiples).

Worked example 3: Finding length for resonance

A student wants an open-open tube to resonate at f = 512\text{ Hz} at the fundamental. Take v = 340\text{ m/s}. What tube length is needed?

For open-open fundamental:

f_1 = \frac{v}{2L}

Solve for L:

L = \frac{v}{2f_1} = \frac{340}{2(512)} = \frac{340}{1024} \approx 0.332\text{ m}

So the tube should be about 0.33 m long.

Connecting standing waves to interference and reflection

It’s worth explicitly linking ideas across the unit:

  • Reflection at boundaries creates a backward-traveling wave.
  • Superposition of forward and backward waves produces interference.
  • When the geometry allows reinforcement at fixed points, you get a standing wave.
  • The corresponding frequencies are exactly the resonant frequencies.

This “chain” is a powerful way to explain answers on free-response: you’re not just quoting a formula; you’re describing a physical mechanism.

Exam Focus
  • Typical question patterns:
    • Determine whether ends are nodes or antinodes (string vs open/closed pipe), then write the correct relationship between L and \lambda.
    • Find resonant frequencies given L and v, or find L given a resonant frequency.
    • Conceptual questions tying resonance to superposition and boundary reflections.
  • Common mistakes:
    • Using the wrong boundary condition model (treating an open-closed pipe like an open-open pipe).
    • Confusing displacement and pressure nodes/antinodes in sound waves.
    • Labeling modes incorrectly in open-closed pipes (only odd harmonics occur; the “second mode” is the 3rd harmonic).