Sound Waves — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Longitudinal nature of sound, speed of sound relations, beat frequency, standing sound waves in open/closed tubes, and non-relativistic Doppler effect for sound, including core AP problem-solving techniques.

You should already know: Basic wave properties (wavelength, frequency, superposition), longitudinal vs transverse wave definitions, algebraic proportional reasoning.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Sound Waves?

Sound is a mechanical longitudinal pressure wave that requires a material medium to propagate — it cannot travel through a vacuum, a fact frequently tested on AP multiple choice. In AP Physics 1 CED, Sound Waves makes up 4-6% of the total exam weight, appearing regularly in both MCQ and FRQ sections, often combined with other Unit 7 topics like superposition and standing waves.

Unlike transverse waves (e.g., waves on a string), sound wave particles oscillate parallel to the direction of wave travel, creating alternating regions of higher pressure (compressions) and lower pressure (rarefactions). Common notation conventions used throughout this guide: $v$ for wave speed, $f$ for frequency, $\lambda$ for wavelength, $f_b$ for beat frequency, $L$ for tube length, and $v_s$ for speed of the sound source. Standard room temperature speed of sound in dry air is approximately $343\text{m/s}$, a value you can assume in problems unless stated otherwise. AP Physics 1 focuses on applying fundamental wave principles to sound, rather than advanced acoustic properties, so all test questions align with the core concepts outlined below.

2. Speed of Sound and Fundamental Wave Relation

The fundamental wave relationship $v=f\lambda$ holds for all waves, including sound, but the speed of sound depends only on the medium, not the frequency or wavelength of the source. This is a key distinction: when sound travels from one medium to another, frequency remains constant (it is determined by how often the source oscillates, which does not change when the wave enters a new material), while speed and wavelength change.

Stiffer, less compressible media have faster sound speeds: sound travels roughly 4 times faster in water than in air, and 15 times faster in steel than in air. Intuition for the formula: speed is how fast a compression wave moves through the medium, so stiffer media transmit compression faster, leading to higher $v$. Since $f$ is fixed by the source, $\lambda$ scales directly with $v$ across media.

Worked Example

A 220 Hz tuning fork produces sound in room temperature air ($v=343\text{m/s}$), then the same tuning fork is held underwater, where the speed of sound is $1480\text{m/s}$. Find the wavelength of sound in air and in water.

1. Frequency is determined by the source, so $f=220\text{Hz}$ in both media, it does not change.
2. Rearrange $v=f\lambda$ to solve for wavelength: $\lambda =\frac{v}{f}$.
3. For air: ${\lambda }_{air}=\frac{343}{220}\approx 1.56\text{m}$.
4. For water: ${\lambda }_{water}=\frac{1480}{220}\approx 6.73\text{m}$. The result makes sense: higher speed in water gives longer wavelength at the same frequency.

Exam tip: Always remember frequency is determined by the source, and speed is determined by the medium — never flip these two properties when solving for wavelength across different media.

3. Beat Frequency

Beats are periodic variations in loudness caused by the superposition of two sound waves with slightly different frequencies. When the waves are in phase, they interfere constructively and produce a loud pulse; when they are out of phase, they interfere destructively and produce quiet. The rate of these loudness pulses is the beat frequency.

The formula for beat frequency comes directly from the difference in frequencies of the two sources: $$f_b=|f_1-f_2|$$ The absolute value ensures beat frequency is positive, as it describes a rate of pulses. A key implication: the unknown frequency of a source can only be narrowed down to two possible values ($f_{known}\pm f_b$), so you must use additional information (e.g., how beat frequency changes when you adjust the unknown frequency) to pick the correct value. This is the most common problem format for beats on the AP exam.

Worked Example

A piano tuner tunes an A string against a standard 440 Hz tuning fork. They hear 4 beats per second, and when they slightly tighten the string (which increases the string's frequency), the beat frequency increases to 6 beats per second. What was the original frequency of the piano string before tightening?

1. Original beat frequency is 4 Hz, so the original string frequency is either $440+4=444\text{Hz}$ or $440-4=436\text{Hz}$.
2. When the string is tightened, its frequency increases, and the beat frequency (the absolute difference between string frequency and 440 Hz) also increases.
3. If the original frequency was 444 Hz, increasing it further moves it farther from 440 Hz, increasing the difference (and beat frequency), which matches the observation. If the original frequency was 436 Hz, increasing it would move it closer to 440 Hz, decreasing the beat frequency, which contradicts the observation.
4. The original frequency of the string is 444 Hz.

Exam tip: When solving beat frequency tuning problems, always test both possible values of the unknown frequency against the change in beat frequency after the source frequency is adjusted — AP problems almost always require you to eliminate the wrong candidate.

4. Standing Sound Waves in Open and Closed Tubes

When sound travels down a tube and reflects off the ends, incident and reflected waves interfere to form standing waves, the basis of most woodwind and brass instruments. The pattern of standing waves depends on boundary conditions at the ends of the tube:

• An open end (open to air): pressure is fixed at atmospheric pressure, so this is a pressure node = displacement antinode (air particles move freely here).
• A closed end (blocked by a solid surface): air cannot move, so this is a displacement node = pressure antinode.

For a tube of length $L$:

1. Open-open tube (both ends open): Both ends are displacement antinodes, so the allowed wavelengths are ${\lambda }_n=\frac{2L}{n}$ for $n=1,2,3,...$ (all positive integers, all harmonics exist). Frequency is: $$f_n=n\frac{v}{2L}$$
2. Closed-open tube (one end closed, one open): Closed end is node, open end is antinode, so allowed wavelengths are ${\lambda }_n=\frac{4L}{n}$ for $n=1,3,5,...$ (only odd positive integers, only odd harmonics exist). Frequency is: $$f_n=n\frac{v}{4L}$$

Worked Example

A 0.75 m long tube is open at both ends, speed of sound is 343 m/s. (a) Find the frequency of the 3rd harmonic. (b) If the tube is closed at one end, find the frequency of the 3rd harmonic.

1. For part (a), open tube, 3rd harmonic corresponds to $n=3$. Use $f_n=n\frac{v}{2L}$: $f_3=3\times \frac{343}{2\times 0.75}=\frac{1029}{1.5}=686\text{Hz}$.
2. For part (b), closed tube, harmonics are numbered by their multiple of the fundamental, so the 3rd harmonic still corresponds to $n=3$ (only odd harmonics exist, so 1st = 1, 2nd does not exist, 3rd = 3).
3. Use the closed tube formula $f_n=n\frac{v}{4L}$: $f_3=3\times \frac{343}{4\times 0.75}=\frac{1029}{3}=343\text{Hz}$.

Exam tip: Always confirm boundary conditions first: remember that closed ends are displacement nodes, open ends are displacement antinodes, and only odd harmonics exist for tubes closed at one end — this is the most commonly tested fact about standing sound waves on the AP exam.

5. Doppler Effect for Sound

The Doppler effect is the shift in observed frequency caused by relative motion between a sound source and an observer. When source and observer move towards each other, observed frequency is higher than the source frequency; when they move away from each other, observed frequency is lower. For AP Physics 1, we use the non-relativistic Doppler formula for sound in a stationary medium: $$f_o=f_s\frac{v\pm v_o}{v\pm v_s}$$ Where $f_o$ = observed frequency, $f_s$ = source frequency, $v$ = speed of sound, $v_o$ = speed of observer, $v_s$ = speed of source. The sign rule is simple: any motion towards the other increases $f_o$, so: add $v_o$ if the observer moves towards the source, subtract $v_s$ if the source moves towards the observer. Reverse the signs for motion away: subtract $v_o$ for observer moving away, add $v_s$ for source moving away.

Intuition: An observer moving towards the source encounters more wavefronts per second, increasing frequency. A source moving towards the observer compresses wavefronts in front of it, shortening wavelength and increasing frequency.

Worked Example

A cyclist rides away from a stationary 1000 Hz factory siren at 8 m/s. Speed of sound is 343 m/s. What frequency does the cyclist observe?

1. The source (siren) is stationary, so $v_s=0$. The observer (cyclist) is moving away from the source, so we subtract $v_o$ from the numerator.
2. Substitute into the formula: $f_o=1000\times \frac{343-8}{343+0}=1000\times \frac{335}{343}\approx 977\text{Hz}$.
3. Check the result: observer moving away should give a lower frequency than 1000 Hz, which matches our result.

Exam tip: To get the sign right every time, after calculating, confirm that motion towards gives higher frequency and motion away gives lower frequency. If your result contradicts this, flip the signs.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Claims that frequency of sound changes when it moves from air to water, and solves for wavelength using a new frequency instead of the original source frequency. Why: Students confuse which property is determined by the source vs the medium, incorrectly assuming both speed and frequency change with medium. Correct move: Always note that frequency is fixed by the source, so frequency is constant when crossing between media; only speed and wavelength change.
• Wrong move: For beats, when given the beat frequency and known frequency, picks the wrong unknown frequency without testing the change in beat frequency. Why: Students forget there are two possible solutions, and default to picking the frequency lower than the known value. Correct move: Always write both possible values ($f_{known}\pm f_b$) then eliminate the one that contradicts the given change in beat frequency.
• Wrong move: For a tube closed at one end, uses the open tube formula $f=nv/(2L)$ or counts even harmonics. Why: Students mix up standing waves on strings (fixed at both ends, same as open tube harmonics) with standing waves in closed tubes, forgetting the boundary condition difference. Correct move: Before calculating harmonic frequency, confirm if the tube is open-open or closed-open, and write down that closed-open only has odd harmonics with $f_n=nv/(4L)$.
• Wrong move: Flips the signs in the Doppler effect formula, leading to a frequency that changes in the wrong direction (e.g., lower frequency when source moves towards observer). Why: Students memorize sign rules backwards instead of using physical intuition. Correct move: After calculating, check that motion towards increases frequency and motion away decreases it; flip signs if your result contradicts this.
• Wrong move: Claims that sound can travel through a vacuum, because all waves can travel through vacuum. Why: Students confuse mechanical waves (sound) with electromagnetic waves (light), which do not need a medium. Correct move: Always remember sound is a mechanical pressure wave, so it requires a medium to propagate and cannot travel through vacuum.

7. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A 100 Hz tuning fork produces sound in room temperature air. Which of the following changes will definitely increase the wavelength of the sound wave produced, assuming the speed of sound in any medium is constant for that medium? (A) Increasing the frequency of oscillation of the tuning fork (B) Decreasing the frequency of oscillation of the tuning fork (C) Moving the tuning fork from air into a medium with a lower speed of sound (D) Increasing the amplitude of the sound wave produced by the tuning fork

Worked Solution: We start from the fundamental relation $v=f\lambda$, which rearranges to $\lambda =v/f$. Wavelength depends on speed of sound (set by medium) and frequency (set by source). For (A), increasing $f$ decreases $\lambda$, so A is incorrect. For (B), decreasing $f$ with constant $v$ (same medium) increases $\lambda$, so B is correct. For (C), moving to a slower medium keeps $f$ constant but decreases $v$, so $\lambda$ decreases, C is incorrect. For (D), amplitude affects intensity, not wavelength, so D is incorrect. Correct answer: B.

Question 2 (Free Response)

A student builds a simple wind instrument from a hollow tube that is open at both ends. The tube has a length of 0.80 m, and the speed of sound in the tube is 340 m/s. (a) Calculate the frequency of the fundamental (1st harmonic) of this instrument. (b) The student closes one end of the tube with a stopper. How many harmonics are present in the audible range (20 Hz to 20,000 Hz) for this closed tube? (c) Compare the wavelength of the 2nd harmonic of the original open tube to the wavelength of the fundamental of the closed tube. Justify your answer.

Worked Solution: (a) For an open-open tube, the fundamental frequency is $n=1$: $$f_1=\frac{nv}{2L}=\frac{1(340)}{2(0.80)}=212.5\text{Hz}\approx 210\text{Hz}$$ (b) For a closed-open tube, $f_n=\frac{nv}{4L}$ where $n=1,3,5,...$ We need $f_n\le 20000\text{Hz}$: $$n\le \frac{20000(4)(0.80)}{340}\approx 188.2$$ The largest odd $n$ is 187, so the number of harmonics is $(187+1)/2=94$. (c) 2nd harmonic of open tube: $\lambda =2L/n=2(0.80)/2=0.80\text{m}$. Fundamental of closed tube: $\lambda =4L/n=4(0.80)/1=3.2\text{m}$. The wavelengths are different: the fundamental of the closed tube is 4 times longer than the 2nd harmonic of the open tube.

Question 3 (Application / Real-World Style)

A bat flying towards a stationary cave wall emits a 40 kHz ultrasonic sound pulse, flying at 15 m/s. Speed of sound is 343 m/s. What frequency does the bat detect in the reflected pulse coming from the stationary wall?

Worked Solution: This is a two-step Doppler shift: first the wall acts as a stationary observer receiving sound from the moving bat, then the wall acts as a stationary source reflecting the sound, and the bat is a moving observer receiving the reflected sound. Step 1: First shift, source (bat) moving towards stationary wall: $f_{wall}=40000\times 343/(343-15)\approx 41817\text{Hz}$. Step 2: Second shift, bat (observer) moving towards stationary wall: f_{bat} = 41817 \times (343+15)/343 \approx 43700\ \text{Hz} = 43.7\ \{kHz}. In context, the bat detects a higher frequency than it emitted, which it uses to calculate its speed relative to the wall for echolocation.

8. Quick Reference Cheatsheet

9. What's Next

Sound waves are the primary real-world application of wave superposition and standing wave principles from Unit 7, and mastery of this topic is required to solve many multi-concept FRQ questions on the AP Physics 1 exam. Next in the AP Physics 1 syllabus, you will apply the wave principles you learned here to standing waves on strings, and deepen your understanding of two-source interference patterns. Without mastering sound wave boundary conditions, beat frequency, and Doppler effect reasoning, you will struggle to compare standing wave systems (tubes vs strings), a common AP exam question type. This topic also builds a foundation for mechanical wave behavior that transfers to electromagnetic wave topics in AP Physics 2.

• Standing Waves on Strings
• Superposition and Interference
• Wave Properties and Basics
• Doppler Effect