Standing Waves and Resonance

A wave that seems to stand still

A wave goes along a string. It reaches the end, and it comes back. Now the wave and its reflection are on the string together. By superposition, they add up. The pattern that we get does not seem to travel. It is a standing wave.

• Nodes are the points that never move.
• Antinodes are the points that vibrate the most. They are halfway between two nodes.
• Two neighbouring nodes are $\tfrac{\lambda}{2}$ apart. A node and the next antinode are $\tfrac{\lambda}{4}$ apart.

A string fixed at both ends

Both ends are held, so both ends are nodes. The string can vibrate only in patterns that fit a whole number of half wavelengths into its length $L$.

The lowest frequency, $f_1 = \dfrac{v}{2L}$, is the fundamental. The other frequencies are whole number multiples of it. We call them harmonics. The $n$th harmonic has $n$ loops.

You should find that each extra loop adds one fundamental to the frequency, and that a longer string gives lower notes.

An example. A guitar string is $0.65\ \text{m}$ long, and waves go along it at $260\ \text{m/s}$. The fundamental is $f_1 = \dfrac{260\ \text{m/s}}{2 \times 0.65\ \text{m}} = 200\ \text{Hz}$. The second harmonic is $400\ \text{Hz}$, and the third is $600\ \text{Hz}$.

Pipes

The air in a pipe vibrates too. At a closed end, the air cannot move, so there is a node. At an open end, the air moves freely, so there is an antinode.

• Open at both ends. There is an antinode at each end. So $L = n\dfrac{\lambda}{2}$ and $f_n = \dfrac{nv}{2L}$. All the harmonics are possible.
• Closed at one end. There is a node at one end and an antinode at the other. So $L$ must be an odd number of quarter wavelengths, $L = (2n - 1)\dfrac{\lambda}{4}$. The frequencies are $\dfrac{v}{4L}, \dfrac{3v}{4L}, \dfrac{5v}{4L}, \ldots$ Only the odd harmonics are possible.

Resonance

Push a swing at the right moments, and it goes higher and higher. In the same way, if we push a string or an air column again and again at one of its natural frequencies, it vibrates strongly. This is resonance.

The resonance tube experiment uses this. We hold a tuning fork over a glass tube. The water level in the tube can be changed, so the air column gets longer or shorter. At certain lengths, the sound becomes loud.

The water surface is a closed end, so it is a node. At the top of the tube, the antinode is a little outside the tube. So we add an end correction $e$ to the length. For the first two resonances,

If we subtract the first equation from the second, $e$ goes away. We get $l_2 - l_1 = \dfrac{\lambda}{2}$. This is why the experiment always uses two resonance lengths. We do not need to know $e$.

Simulation: PhET Interactive Simulations, University of Colorado Boulder (CC BY 4.0).

Practise with new numbers

Try it in Python

This program finds the first five notes of a string, of a pipe that is open at both ends, and of a pipe that is closed at one end.

Remember these

• A string fixed at both ends: $f_n = \dfrac{nv}{2L}$, with all the harmonics.
• A pipe open at both ends: the same formula, with all the harmonics.
• A pipe closed at one end: $f = \dfrac{v}{4L}, \dfrac{3v}{4L}, \dfrac{5v}{4L}, \ldots$ Only the odd harmonics.
• In the resonance tube, $\lambda = 2(l_2 - l_1)$, and the end correction is $e = \dfrac{\lambda}{4} - l_1$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A string $0.6\ \text{m}$ long carries waves at $180\ \text{m/s}$. What is its fundamental, and its fourth harmonic?

2. A pipe closed at one end is $0.34\ \text{m}$ long, and sound travels at $340\ \text{m/s}$. What are its two lowest notes?

3. Why is there an antinode at the open end of a pipe?

4. In a resonance tube, why do we take the difference of two resonance lengths, and not just the first one?