Standing Waves — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Formation of standing waves via superposition, boundary conditions for fixed/free ends, node/antinode relationships, harmonic frequency/wavelength formulas, string wave speed, and problem-solving strategies for fixed-fixed, open-open, and closed-end media.

You should already know: The wave superposition principle and interference of coherent waves. The fundamental wave relation $v=f\lambda$. Boundary behavior for waves reflecting off fixed and free ends.

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 Standing Waves?

Standing waves (also called stationary waves) are a special case of wave interference that occurs when two identical, coherent waves traveling in opposite directions superpose on the same linear medium. Unlike traveling waves that transfer net energy continuously through a medium, standing waves have fixed locations of zero displacement (nodes) and maximum displacement (antinodes) that do not move over time. The net energy transfer along the medium is zero for a stable standing wave.

According to the AP Physics 1 Course and Exam Description (CED), standing waves contribute approximately 3-4% of the total exam score, and they appear in both multiple choice (MCQ) and free response (FRQ) sections. They are often combined with concepts of tension, wave speed, proportional reasoning, and resonance, making them a common topic for multi-concept questions. The convention used throughout this guide is: $L$ = length of the medium, $n$ = harmonic number (integer), $v$ = speed of the wave in the medium, $f_1$ = fundamental (first harmonic) frequency.

2. Nodes, Antinodes, and Boundary Conditions

Boundary conditions are the constraints imposed by the ends of the medium that determine which standing wave patterns can form. For a fixed end (a string tied to a wall, the closed end of an air tube), displacement at the end is always zero, so fixed ends are always nodes. For a free end (the open end of an air tube, a string tied to a freely sliding ring), displacement can be maximum, so free ends are always antinodes.

Nodes are points of permanent destructive interference (total displacement always equals zero), while antinodes are points of permanent constructive interference (maximum displacement amplitude). The distance between two adjacent nodes is $\lambda /2$, and the distance between a node and its nearest adjacent antinode is $\lambda /4$. This relationship comes directly from the standing wave displacement equation $y(x,t)=2A\sin (kx)\cos (\omega t)$, where zeros of the sine function are nodes and maxima are antinodes.

Worked Example

A student measures the distance between the first node and the 5th node in a standing wave on a string as 78 cm. What is the wavelength of the traveling waves that formed this standing wave?

1. Recall that adjacent nodes are separated by $\lambda /2$, so the number of gaps between nodes equals the difference between the node numbers.
2. Between node 1 and node 5, there are $5-1=4$ intervals, each of length $\lambda /2$.
3. Set up the equation for total distance: $4\times (\lambda /2)=2\lambda =78$ cm.
4. Solve for $\lambda$: $\lambda =78/2=39$ cm = 0.39 m.

Exam tip: Always count the number of intervals between nodes/antinodes, not the number of nodes/antinodes themselves. Students often incorrectly use the number of nodes instead of the number of gaps, leading to a factor of 2 error that is easily avoided.

3. Harmonics for Fixed-Fixed and Open-Open Media

The most common standing wave scenario in AP Physics 1 is a medium with two identical boundary conditions: both ends fixed (a vibrating guitar string) or both ends open (an open-ended flute). For these cases, both ends are matching boundaries: both nodes for fixed-fixed, both antinodes for open-open. This means the total length $L$ of the medium must equal an integer multiple of $\lambda /2$, because we start and end at the same boundary type.

The wavelength and frequency formulas for fixed-fixed and open-open media are: $${\lambda }_n=\frac{2L}{n},f_n=n\frac{v}{2L}=n{f}_1,n=1,2,3,...$$ where $n=1$ is the first harmonic (fundamental frequency), $f_1=v/(2L)$ is the fundamental frequency. For strings, wave speed is given by $v=\sqrt{T/\mu }$, where $T$ is tension and $\mu$ is mass per unit length, so we can substitute this into the frequency formula to relate $f$ to $T$, $L$, and $\mu$, a common AP exam question.

Worked Example

A 0.75 m long guitar string has a total mass of 4.5 g, and is tightened to a tension of 100 N. What is the frequency of the 2nd harmonic?

1. Calculate mass per unit length: $\mu =m/L=0.0045$ kg / 0.75 m = 0.0060 kg/m.
2. Calculate wave speed on the string: $v=\sqrt{T/\mu }=\sqrt{100/0.0060}\approx 129$ m/s.
3. For 2nd harmonic, $n=2$, so use the fixed-fixed frequency formula: $f_2=2v/(2L)=v/L=129/0.75=172$ Hz.

Exam tip: Proportional reasoning questions about frequency vs. tension, length, or mass per unit length are extremely common in MCQ. Remember: $f\propto \sqrt{T}$, $f\propto 1/L$, and $f\propto 1/\sqrt{\mu }$ for a string.

4. Harmonics for Closed-End (Fixed-Open) Media

The other common AP Physics 1 scenario is a tube closed at one end and open at the other (a stopped pipe, like an empty bottle you blow across). Here we have mixed boundary conditions: the closed end is a node, the open end is an antinode. This means the length of the tube must equal an odd multiple of $\lambda /4$: a half-wavelength between a node and antinode would end at a node, which violates the open end antinode requirement.

The wavelength and frequency formulas for closed-end (fixed-open) media are: $${\lambda }_n=\frac{4L}{n},f_n=n\frac{v}{4L}=n{f}_1,n=1,3,5,...$$ Only odd harmonics exist for closed-end tubes; even $n$ values do not satisfy the boundary condition, so no stable standing wave forms. The fundamental frequency $f_1=v/(4L)$ is lower than the fundamental of an open-open tube of the same length.

Worked Example

A 22 cm tall empty water bottle acts as a closed-end tube, with the closed end at the bottom and open end at the top. The speed of sound in air is 340 m/s. What is the fundamental frequency, and what is the next highest harmonic that can form?

1. Convert length to meters: $L=0.22$ m, confirm boundary conditions: one closed, one open.
2. Fundamental frequency uses $n=1$ (the first odd integer): $f_1=v/(4L)=340/(4\times 0.22)\approx 386$ Hz.
3. The next highest harmonic for a closed-end tube is the next odd integer after 1, which is $n=3$.
4. Calculate $f_3=3f_1=3\times 386\approx 1160$ Hz.

Exam tip: Never use even harmonic numbers for a closed-end tube. If an FRQ asks for all possible harmonics, explicitly state that only odd $n$ are allowed to earn full points.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Counting the number of nodes instead of the number of intervals between nodes when solving for wavelength. For example, between node 1 and node 3, using 3 intervals instead of 2. Why: Students confuse counting discrete points with counting the gaps between them, which is how wavelength is measured. Correct move: Always subtract the lower node number from the higher node number to get the number of $\lambda /2$ intervals.
• Wrong move: Using the open-open frequency formula for a closed-end tube, or including even harmonics in a list for a closed-end tube. Why: Students memorize the more common fixed-fixed/open-open formula first, and forget the exception for mixed boundary conditions. Correct move: Before writing any frequency formula, explicitly write down the boundary condition for each end of the medium, and confirm which formula applies.
• Wrong move: Confusing wave speed on a string with the speed of sound. For example, using $v=340$ m/s for a wave on a vibrating string. Why: Students encounter sound waves in tubes right after string standing waves, so they mix up the wave speed values. Correct move: Always note what medium the wave is traveling through: use $v=\sqrt{T/\mu }$ for strings, and 340 m/s for sound in air unless stated otherwise.
• Wrong move: Assuming the fundamental frequency always has one full wavelength along the length $L$. Why: Students generalize from fixed-fixed where the fundamental has $\lambda =2L$, so they incorrectly apply that to all cases. Correct move: Derive the wavelength from boundary conditions every time: for fixed-fixed/open-open, ${\lambda }_1=2L$; for closed-end, ${\lambda }_1=4L$.
• Wrong move: Using wavelength in centimeters when calculating frequency without converting to meters to match wave speed units (usually m/s). Why: Students get lazy with unit conversions, and keep length in centimeters when plugging into formulas. Correct move: Convert all lengths to meters before calculating frequency when wave speed is given in m/s.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A guitar string of length 0.75 m vibrates at a fundamental frequency of 200 Hz. The guitarist presses the string against a fret, shortening the vibrating length to 0.50 m, keeping tension and mass per unit length unchanged. What is the new fundamental frequency? A) 133 Hz B) 200 Hz C) 300 Hz D) 400 Hz

Worked Solution: The fundamental frequency for a fixed-fixed string is $f_1=v/(2L)$. Wave speed $v$ depends only on tension and mass per unit length, which are unchanged here, so $f$ is inversely proportional to $L$. This gives the ratio $\frac{f_{new}}{f_{old}}=\frac{L_{old}}{L_{new}}=\frac{0.75}{0.50}=1.5$. Multiplying by the original frequency gives $f_{new}=1.5\times 200=300$ Hz. The most common error is inverting the ratio to get 133 Hz (option A). Correct answer: C.

Question 2 (Free Response)

A student studies standing waves in a tube that is open at both ends. The speed of sound in the room is 340 m/s, and the tube has a total length of 1.0 m. (a) Calculate the fundamental frequency of the tube. (b) Draw the standing wave pattern for the 3rd harmonic, label the nodes and antinodes, and calculate its frequency. (c) Explain why no stable standing wave can form at a frequency of 200 Hz in this tube.

Worked Solution: (a) For an open-open tube, all boundaries are antinodes, so fundamental frequency is $f_1=v/(2L)=340/(2\times 1.0)=170$ Hz. (b) For 3rd harmonic, $n=3$, so $f_3=3f_1=3\times 170=510$ Hz. The standing wave pattern has an antinode at each end, with 3 antinodes total and 2 nodes along the length of the tube. (c) For a stable standing wave in an open-open tube, the frequency must be an integer multiple of the fundamental frequency. 200 Hz is not an integer multiple of 170 Hz, so it cannot satisfy the boundary condition of antinodes at both ends, so no stable standing wave forms.

Question 3 (Application / Real-World Style)

A pipe organ designer needs a closed-end pipe that produces a fundamental frequency of 32 Hz (a low bass note) for a concert hall, where the speed of sound is 343 m/s. What length of closed-end pipe is required? If the designer accidentally builds an open-open pipe of the same length, what fundamental frequency will it produce, and how will this change the sound a listener hears?

Worked Solution:

1. For a closed-end pipe, $f_1=v/(4L)$, so rearrange to solve for $L$: $L=v/(4f_1)$.
2. Plug in values: $L=343/(4\times 32)\approx 2.7$ m, which is the required length for the closed-end pipe.
3. For an open-open pipe of the same length, fundamental frequency is $f_{1,open}=v/(2L)=2\times (v/(4L))=2f_{1,closed}=64$ Hz.
4. In context: 64 Hz is one octave higher than the intended 32 Hz, so the pipe will produce a note that is perceived as twice as high in pitch as the designer intended.

7. Quick Reference Cheatsheet

8. What's Next

Standing waves are the foundation for understanding all resonant phenomena, which are a core recurring concept in AP Physics 1. Next, you will apply standing wave interference principles to beat frequency, which occurs when two waves of slightly different frequencies superpose, producing an amplitude-modulated beat pattern. Without mastering the boundary conditions and harmonic relationships covered here, you will not be able to correctly solve resonance or beat frequency problems that regularly appear on the AP exam. Standing waves also connect to mechanical resonance of simple harmonic oscillators, a related topic in Unit 7 that often pairs with standing wave questions in FRQs. The key follow-on topics you will study next are: Beat Frequency Wave Interference and Superposition Sound Waves Mechanical Resonance