Grade 11

Advanced

8 min

Lesson 8 of 12

Standing Waves and Resonance

Not Started

The formation of stationary waves in strings and air columns, bridging classical waves to quantum energy levels.

Have you ever wondered why a singer can shatter a wine glass with just their voice, or why a guitar string creates a beautiful note instead of just chaotic noise?

Learning Objectives

1.Identify nodes and antinodes in various standing wave patterns
2.Calculate fundamental frequencies and harmonics for strings and air columns
3.Explain how resonance bridges classical wave mechanics to quantized energy levels

The Anatomy of a Standing Wave

A standing wave is not a wave that travels; it is a pattern of vibration that appears to stay in one place. It forms through superposition—when two waves of the same frequency and amplitude travel in opposite directions and interfere. The points that never move are called nodes, caused by total destructive interference. The points of maximum displacement are antinodes, caused by constructive interference. In a string fixed at both ends, the boundaries must always be nodes because the string is physically tethered.

Quick Check

If a standing wave on a string has 3 'loops' (antinodes), how many nodes does it have in total, including the ends?

Draw three loops and count the points where the string meets the center line.

Answer

4 nodes.

Harmonics in Strings

The simplest vibration pattern is the fundamental frequency (

f_1

), also known as the first harmonic. For a string of length

L

, the fundamental wavelength is

\\lambda = 2L

. Higher patterns, called harmonics, occur at integer multiples of the fundamental. The frequency of the

n

-th harmonic is given by:

\begin{gathered}\\ f_n = n \\cdot f_1 = \\frac{nv}{2L} \\\end{gathered}

where

v

is the wave speed on the string. Each harmonic adds exactly one node and one antinode to the pattern.

A guitar string is 0.65 m long and has a wave speed of 400 m/s. Find the fundamental frequency.

1. Identify the formula: $f_1 = \\frac{v}{2L}$ 2. Substitute the values: $f_1 = \\frac{400}{2 \\times 0.65}$ 3. Calculate: $f_1 \\approx 307.7$ Hz.

Air Columns and Boundary Conditions

Air columns in pipes create standing waves based on their ends. An open pipe (both ends open) has antinodes at both ends, behaving like a string:

f_n = \\frac{nv}{2L}

. However, a closed pipe (one end closed) has a node at the closed end and an antinode at the open end. This asymmetry means only odd harmonics (

n = 1, 3, 5...

) can exist. The fundamental frequency for a closed pipe is:

\begin{gathered}\\ f_1 = \\frac{v}{4L} \\\end{gathered}

An organ pipe is closed at one end and is 1.5 m long. If the speed of sound is 340 m/s, what is the frequency of the 3rd harmonic?

1. Note that for a closed pipe, the '3rd harmonic' is the next available mode after the fundamental ( $n=3$ ). 2. Use the formula: $f_n = \\frac{nv}{4L}$ 3. Substitute: $f_3 = \\frac{3 \\times 340}{4 \\times 1.5}$ 4. Calculate: $f_3 = \\frac{1020}{6} = 170$ Hz.

Quick Check

Why can't a pipe closed at one end produce the 2nd harmonic?

Think about the shape of the wave: can you fit exactly half a wavelength into a pipe that must end in an antinode?

Answer

Because the boundary conditions require a node at one end and an antinode at the other, which only allows for odd multiples of a quarter-wavelength.

Resonance and the Quantum Bridge

Resonance occurs when an external driving frequency matches the natural frequency of a system, leading to large amplitude vibrations. This is why a specific note can shatter glass. This classical concept is the foundation of Quantum Mechanics. Electrons in an atom are modeled as standing waves. Just as a string only allows specific harmonics, an electron only exists in specific 'allowed' orbits where its de Broglie wavelength forms a standing wave. This is why energy levels are quantized—nature literally 'tunes' the atom.

A pipe closed at one end has two successive resonance frequencies at 450 Hz and 750 Hz. Find the fundamental frequency.

1. For a closed pipe, $f_n = (2n-1)f_1$ . Successive harmonics are $f_n$ and $f_{n+2}$ . 2. The difference between successive harmonics is $(n+2)f_1 - n f_1 = 2f_1$ . 3. $750 - 450 = 300$ Hz. 4. Therefore, $2f_1 = 300$ , so $f_1 = 150$ Hz.

Key Takeaways

1Nodes are points of zero displacement; Antinodes are points of maximum displacement.
2Strings and open pipes produce all integer harmonics ( $f_n = n f_1$ ).
3Closed pipes only produce odd harmonics ( $f_1, 3f_1, 5f_1...$ ) because of their asymmetric boundaries.
4Resonance explains both musical tone production and the quantized nature of electron orbits in atoms.

Quick Check

Multiple Choice

In a standing wave, what is the distance between a node and the adjacent antinode?

Multiple Choice

If the fundamental frequency of an open pipe is 200 Hz, what is the frequency of its second harmonic?

True/False

The nodes of a standing wave are the result of constructive interference.

What's Next?

Review Tomorrow

In 24 hours, try to sketch the first three harmonics for both a string and a closed pipe from memory.

Practice Activity

Blow across the top of a half-full water bottle. Calculate the fundamental frequency by measuring the length of the air column and using $v = 340$ m/s.

Browse

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Standing Waves and Resonance

Learning Objectives

The Anatomy of a Standing Wave

Harmonics in Strings

Air Columns and Boundary Conditions

Resonance and the Quantum Bridge

Key Takeaways

Quick Check

What's Next?

Standing Waves and Resonance

Learning Objectives

The Anatomy of a Standing Wave

Harmonics in Strings

Air Columns and Boundary Conditions

Resonance and the Quantum Bridge

Key Takeaways

Quick Check

What's Next?