Why a Bottle Sings Higher as You Fill It With Water

Pluck a sitar string or blow across a flute's hole and you get a clear note, not a random noise. Waves bounce back and forth inside the instrument, and only certain wavelengths fit neatly between its ends.

Those allowed waves are standing waves, and they fix every note an instrument can make. Two nearby notes played together produce a throbbing called beats.

This part covers reflection of waves, standing waves, harmonics of strings and pipes, and beats. Take the speed of sound as $340$ m/s.

**At a rigid boundary a wave reflects with a phase change of $\pi$, so a crest returns as a trough; at a free boundary it reflects with no phase change, so a crest returns as a crest.

Why.** At a rigid end the string cannot move, so the incoming and reflected displacements must cancel there. For an incident wave $y=a\sin (kx-\omega t)$ meeting a rigid end at $x=0$, the reflected wave is

$$y_r=a\sin (kx+\omega t+\pi )=-a\sin (kx+\omega t)$$

At a free end, such as a string tied to a ring sliding on a smooth rod, the end moves freely and the pulse returns upright.

For sound in pipes, a closed end acts like a rigid boundary for displacement, and an open end like a free boundary.

Worked example. A pulse moves at $30$ m/s along a $6$ m string tied to a wall. It returns to the starting end after

$$t=\frac{2\times 6}{30}=0.4\text{ s}$$

— and it comes back upside down.

An everyday example. Flick a rope tied to a pole and watch the hump run to the pole and come back on the other side of the rope.

The substance. Reflection changes the phase, not the frequency or speed of the wave.

**Two identical waves travelling in opposite directions superpose to give a standing wave $y=2a\sin kx\cos \omega t$, in which nodes ($\sin kx=0$) never move and antinodes ($\sin kx=\pm 1$) vibrate with the largest amplitude $2a$.

Positions.

- Nodes**: $x=0,\frac{\lambda }{2},\lambda ,\frac{3\lambda }{2},\dots$
- Antinodes: $x=\frac{\lambda }{4},\frac{3\lambda }{4},\frac{5\lambda }{4},\dots$
- Node to node: $\frac{\lambda }{2}$; node to nearest antinode: $\frac{\lambda }{4}$

Worked example. A standing wave has $\lambda =0.8$ m and $a=2$ mm.

- Nodes at $0$, $0.4$ m, $0.8$ m; antinodes at $0.2$ m and $0.6$ m, with amplitude $4$ mm
- **At $x=0.1$ m**, the amplitude is

$$2a\left|\sin \frac{2\pi \times 0.1}{0.8}\right|=4\sin \frac{\pi }{4}\approx 2.83\text{ mm}$$

An everyday example. A plucked sitar string shows a blurred, stationary shape — widest in the middle and still at the two fixed ends.

The substance. A standing wave does not carry energy along, and all points between two neighbouring nodes vibrate in phase.

**A string fixed at both ends and a pipe open at both ends allow $f_n=\frac{nv}{2L}$ for $n=1,2,3,\dots$, while a pipe closed at one end allows only odd harmonics, $f=\frac{(2n-1)v}{4L}$.

Why.** A string needs nodes at both ends, so $L=n\frac{\lambda }{2}$. An open pipe has antinodes at both ends — the same condition. A closed pipe has a node at the closed end and an antinode at the open end, so $L=(2n-1)\frac{\lambda }{4}$.

Worked example 1 — a string. $L=0.5$ m and $v=100$ m/s:

$$f_1=\frac{100}{2\times 0.5}=100\text{ Hz}\text{harmonics: }200,300,400\text{ Hz},\dots$$

Worked example 2 — an open pipe. $L=0.5$ m:

$$f_1=\frac{340}{2\times 0.5}=340\text{ Hz}\text{next: }680\text{ Hz},\dots$$

Worked example 3 — a closed pipe. $L=0.25$ m:

$$f_1=\frac{340}{4\times 0.25}=340\text{ Hz}\text{next allowed: }3f_1=1020\text{ Hz}$$

An everyday example. A flute is an open pipe, while blowing across the mouth of a bottle makes a closed pipe.

The substance. A closed pipe has half the fundamental frequency of an open pipe of the same length and lacks the even harmonics.

**When two sounds of slightly different frequencies $f_1$ and $f_2$ overlap, their superposition swells and fades in loudness $|f_1-f_2|$ times per second — the beat frequency.

Derivation outline.** Adding $a\cos 2\pi f_{1}t$ and $a\cos 2\pi f_{2}t$:

$$s=2a\cos \left[\pi (f_1-f_2)t\right]\cos \left[\pi (f_1+f_2)t\right]$$

The first factor is a slowly changing amplitude. Loudness peaks whenever it is $+2a$ or $-2a$, which happens $|f_1-f_2|$ times a second.

Worked example 1. Tuning forks of $256$ Hz and $260$ Hz sound together:

$$f_{beat}=260-256=4\text{ beats per second}$$

Worked example 2 — finding an unknown frequency. An unknown fork gives $4$ beats per second with a $256$ Hz fork, so it is $252$ Hz or $260$ Hz. Sticking a little wax on the unknown fork lowers its frequency, and the beats rise to $6$ per second. Lowering $260$ Hz would reduce the beats, so the unknown fork must be **$252$ Hz.

An everyday example. Musicians tuning two harmoniums listen for the slow wobble between notes and adjust until it disappears.

The substance. Beats are heard only when the two frequencies are close enough** for the ear to follow the rise and fall.