Reflection and standing waves

Reflection at a boundary

Narrated video. Transcript:

Tie one end of a rope firmly to a wall and send a crest along it. When the pulse reaches the wall, it comes back upside down, as a trough.

Why? At a rigid wall, the rope cannot move, so the displacement there must always be zero. The reflected wave must exactly cancel the incident wave at the wall.

Think of it with Newton’s third law. The pulse pulls the wall upwards. The wall pulls the rope downwards, and that downward tug runs back as a trough.

In the language of phase, reflection at a rigid boundary adds a phase change of π, or 180 degrees. If the incident wave is a sin of k x minus ω t, the reflected wave is minus a sin of k x plus ω t.

Now tie the end to a light ring that slides freely on a smooth vertical rod. This is an open boundary. The end is free to move.

Send the same crest. The ring flies up, and the pulse returns as a crest, the same way up. At an open boundary there is no phase change.

Air in a pipe behaves the same way. At a closed end, the air cannot move: it acts like a rigid boundary. At an open end, the air moves freely: it acts like an open boundary.

When a wave meets the boundary between two different media, part of it is reflected and part is transmitted into the second medium.

Standing waves on a string

Narrated video. Transcript:

Now keep sending a steady wave towards the fixed end. The incident and reflected waves travel in opposite directions on the same string, and they superpose.

Add them: a sin of k x minus ω t, plus a sin of k x plus ω t. The sum is y equals 2a sin k x, times cos ω t.

This is no longer a function of x minus v t. The pattern does not travel at all. It is called a standing wave, or a stationary wave. Every element oscillates with the same frequency, but with its own amplitude, 2a sin k x, and all the points in one loop move up and down together.

Where sin k x is zero, the amplitude is zero. These points, called nodes, never move. They sit at x equals n λ by 2: half a wavelength apart.

Halfway between nodes, the amplitude is the largest, 2a. These points are antinodes. A node and its neighbouring antinode are a quarter wavelength apart.

On a string of length L fixed at both ends, both ends must be nodes. So the length must hold a whole number of half wavelengths: L equals n λ by 2, so λ equals 2L by n.

Since ν equals v over λ, only certain frequencies are allowed: ν equals n v by 2L, for n equals 1, 2, 3 and so on. These are the normal modes of the string.

The lowest, n equals 1, is the fundamental mode, or first harmonic: one loop. n equals 2 is the second harmonic, with a node in the middle, and so on.

Take the steel wire from before: 0.72 metres long, with waves at 93 metres per second. Its fundamental is 93 divided by 1.44, about 65 hertz, and its second harmonic is about 129 hertz.

When a sitar player presses a string against a fret, L gets shorter and the note rises. Tightening the peg raises v, and the note rises too.

Try it: modes of a string

Interactive simulation. Things to try:

• Step through the harmonics (↑ ↓ keys). Count the nodes N and antinodes A each time.
• Tick “Show the two travelling waves” to see the standing wave as their sum.
• Every allowed frequency is a whole number times ν₁. Make the tension 4× bigger: all of them double.
• Open the Reflection tab to compare a rigid end with a free end.

Standing waves in air columns

Narrated video. Transcript:

Blow across the top of an empty bottle and it hums. The air inside is vibrating in a standing wave. Here the wave is longitudinal: the air moves to and fro along the pipe.

At a closed end, the air cannot move, so there is a displacement node. At an open end, the air moves the most, so there is a displacement antinode. We ignore the small end correction.

Take a pipe closed at one end. With a node at one end and an antinode at the other, the length can be λ by 4, 3λ by 4, 5λ by 4 and so on.

So L equals an odd number of quarter wavelengths, and ν equals 2n plus 1, times v by 4L. The fundamental is v by 4L, and only the odd harmonics, 3 ν one, 5 ν one, and so on, appear.

Now open both ends, like a bansuri. There is an antinode at each end, so L equals n λ by 2, just like a string, and ν equals n v by 2L. All harmonics are present.

For the same length, the open pipe’s fundamental is twice that of the closed pipe.

Example: a pipe 30 centimetres long is open at both ends. Take the speed of sound as 330 metres per second. Its fundamental is 330 divided by 0.6, which is 550 hertz.

A 1.1 kilohertz source is exactly twice that, so it excites the second harmonic, and the pipe resonates loudly.

Now close one end. The fundamental becomes 330 divided by 1.2, which is 275 hertz. The allowed frequencies are 275, 825, 1375 hertz. 1100 hertz is 4 times 275, an even harmonic, so there is no resonance.

Try it: resonance in a pipe

Interactive simulation. Things to try:

• The dots are layers of air moving along the pipe. The blue graph shows how far they move.
• Use “Higher mode” to jump between resonances. In the closed pipe, count which multiples of ν₁ appear.
• At a resonance, the closed end is always a node (N) and the open end an antinode (A).
• Switch to “Open at both ends” at the same length: the fundamental doubles.

Inside a standing wave

A string vibrates in a standing wave. Which statement is true?

1. Every point on the string vibrates with the same amplitude
2. Neighbouring nodes are one full wavelength apart
3. All the points between two neighbouring nodes move in step
4. The pattern moves along the string at the wave speed

Which harmonic?

A pipe 30.0 cm long is open at both ends, and the speed of sound is 330 m/s. A 1.1 kHz source makes it resonate. Which mode is excited, and what happens if one end is closed?

1. The 2nd harmonic; with one end closed there is no resonance
2. The fundamental; with one end closed it becomes the 2nd harmonic
3. The 2nd harmonic; with one end closed it becomes the 4th harmonic
4. The 4th harmonic; with one end closed it becomes the 2nd harmonic

Summary

• Standing wave: y = 2a sin kx cos ωt
• Nodes / antinodes: x = nλ/2 / x = (n + ½)λ/2
• String fixed at both ends; pipe open at both ends: ν = nv/2L, n = 1, 2, 3, …
• Pipe closed at one end: ν = (2n + 1)v/4L, n = 0, 1, 2, …

• At a rigid boundary a wave reflects with a phase change of π (upside down); at an open boundary, with no phase change.
• Two equal waves travelling in opposite directions make a standing wave: nodes never move, antinodes swing with amplitude 2a. Nodes are λ/2 apart.
• A string fixed at both ends, or a pipe open at both ends, has all harmonics: ν₁, 2ν₁, 3ν₁, …
• A pipe closed at one end has a node at the closed end and only odd harmonics: ν₁, 3ν₁, 5ν₁, … with ν₁ = v/4L.