Propagation of Mechanical Waves

Lessons

Senior 5 · Physics

Propagation of Mechanical Waves

4 lessons

Lesson 4 of 4

Waves on a Vibrating String

25 min

The inanga's strings

The inanga, Rwanda's trough zither, has strings stretched along a carved wooden trough. When the player plucks a string, a pulse runs to each end, reflects, and comes back; within a fraction of a second the string settles into stationary waves whose frequencies decide the note you hear, while the hollow trough helps the vibration reach the air. A guitar, a violin and a piano work the same way. Why does a shorter, tighter or thinner string give a higher note? This lesson answers with two results: the speed of a wave on a string, and the condition that the stationary wave must fit between the fixed ends.

The speed of a wave on a string

Send a small pulse along a string under tension $T$ with mass per unit length $μ \mu$ . Watch it from a frame moving with the pulse at speed $v$ : the pulse stands still and the string streams through it at speed $v$ . Near the top of the pulse, a short element of length $\Delta l$ follows an arc of radius $R$ , subtending an angle $2\theta$ at the centre, so $\Delta l = 2R\theta$ .

The tension pulls on each end of the element along the string. The horizontal components cancel, and the vertical components add to a force towards the centre:

F = 2T\sin\theta \approx 2T\theta = \frac{T\,\Delta l}{R} \quad\text{(small angles)}

This force provides the centripetal acceleration $\frac{v^2}{R}$ of the element's mass $\mu\Delta l$ :

\frac{T\,\Delta l}{R} = \mu\,\Delta l\,\frac{v^2}{R} \quad\Longrightarrow\quad v = \sqrt{\frac{T}{\mu}}

Both $\Delta l$ and $R$ cancel, so every small pulse, of any shape, travels at the same speed. A tighter string gives a larger restoring force and a faster wave; a heavier string has more inertia and a slower one.

Harmonics of a string fixed at both ends

Both ends are clamped, so both must be nodes. A stationary wave can fit only if the length holds a whole number of loops, each $\frac{\lambda}{2}$ long:

L = n\frac{\lambda_n}{2} \quad\Longrightarrow\quad \lambda_n = \frac{2L}{n}, \qquad f_n = \frac{v}{\lambda_n} = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, \qquad n = 1, 2, 3, \ldots

The lowest, $f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ , is the fundamental frequency or first harmonic. The others are whole-number multiples: $f_{n} = n f_{1}$ . The frequencies above the fundamental are also called overtones (the 2nd harmonic is the 1st overtone, and so on). A plucked or bowed string vibrates in several modes at once; the fundamental sets the pitch, and the mixture of harmonics, different for an inanga, a guitar and a violin, gives each instrument its quality or timbre.

The laws of vibrating strings

From $f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ :

Law	Keeping fixed	Statement
length	$T$ , $μ \mu$	$f \propto \frac{1}{L}$ : halve the length, double the frequency (an octave up)
tension	$L$ , $μ \mu$	$f \propto \sqrt{T}$ : four times the tension doubles the frequency
mass per unit length	$L$ , $T$	$f \propto \frac{1}{\sqrt{\mu}}$ ; for a wire of radius $r$ and density $\rho$ , $\mu = \pi r^2\rho$ , so $f \propto \frac{1}{r\sqrt{\rho}}$

This is how string players tune and play: they tighten a peg to raise the tension, press a finger down to shorten the vibrating length, and use thicker strings for the low notes.

Definition

The harmonics of a string fixed at both ends are its natural frequencies $f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$ , $n = 1, 2, 3, \ldots$ . The fundamental ( $n = 1$ ) is the lowest; the higher harmonics are whole-number multiples of it. Example: a string with fundamental 110 Hz also vibrates at 220 Hz, 330 Hz, 440 Hz, and so on.

Worked example 1: tuning a guitar string

The highest string of a guitar has a vibrating length of 0.65 m and a mass per unit length of $4.0 \times 10^{-4}\,\mathrm{kg\,m^{-1}}$ . What tension gives a fundamental of 330 Hz?

Solution

$v = 2Lf_1 = 2 \times 0.65 \times 330 = 429\,\mathrm{m\,s^{-1}}$ . Then $T = \mu v^2 = 4.0 \times 10^{-4} \times 429^2 = 74\,\mathrm{N}$ , roughly the weight of a 7.5 kg mass, on one thin string.

Worked example 2: harmonics and changing tension

(a) For the string of Worked example 1, list the first three harmonics. (b) A guitarist touches the string lightly at its midpoint while plucking. Which harmonic sounds loudest, and why? (c) If the tension rises by 21%, by what percentage does the fundamental rise?

Solution

(a) 330 Hz, 660 Hz, 990 Hz.

(b) The finger forces a node at the midpoint. Only modes with a node there survive: $n = 2, 4, 6, \ldots$ . The lowest of these, 660 Hz, is loudest: the note jumps up an octave. Guitarists call this a natural harmonic.

Experiment 3.2: the sonometer

Aim: to verify that $f \propto \frac{1}{L}$ at constant tension, and to find the speed of waves on the wire and its mass per unit length.

Apparatus: a sonometer (a wire over a hollow wooden box, with two movable bridges and a pulley), a 5.0 kg load (tension $T = 49\,\mathrm{N}$ ), a set of tuning forks of known frequency, small paper riders, a metre rule.

Method:

Put a light paper rider on the wire midway between the bridges.
Strike a tuning fork on a rubber pad and press its stem on the sound box.
Starting with the bridges close together, slowly move one bridge to change $L$ until the rider jumps off: the wire's fundamental is then in resonance with the fork. Measure $L$ .
Repeat for each fork. Plot $f$ against $\frac{1}{L}$ .

Illustrative results ( $T = 49\,\mathrm{N}$ ):

$f$ / Hz	$L$ / m	$\frac{1}{L}$ / $\mathrm{m^{-1}}$
256	0.433	2.31
288	0.383	2.61
320	0.347	2.88
384	0.289	3.46
426.7	0.258	3.88
512	0.217	4.61

The points lie on a straight line through the origin, confirming $f \propto \frac{1}{L}$ . Since $f = \frac{v}{2}\cdot\frac{1}{L}$ , the gradient $111\,\mathrm{m\,s^{-1}}$ equals $\frac{v}{2}$ , so $v = 222\,\mathrm{m\,s^{-1}}$ , and $\mu = \frac{T}{v^2} = \frac{49}{222^2} = 0.99 \times 10^{-3}\,\mathrm{kg\,m^{-1}}$ , about 1 g per metre.

Warning

Safety. A sonometer wire is under large tension and can snap: wear eye protection, add load gently rather than dropping masses onto the hanger, and keep feet clear of the hanging load. Strike tuning forks on a rubber pad, never on the bench or the sonometer.

Warning

Common mistakes. (1) Using $L = \lambda$ for the fundamental: it is $L = \frac{\lambda}{2}$ . (2) Using the mass of the whole string, or its density, instead of $μ \mu$ in $\mathrm{kg\,m^{-1}}$ ; convert g per m to kg per m. (3) Thinking $f \propto T$ : it is $f \propto \sqrt{T}$ . (4) Confusing the speed of the wave on the string with the speed of the sound it makes in air: they share the frequency, not the speed or the wavelength.

Tip

For any string question, write the master formula $f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$ first. Then every "what happens if" question is a ratio: cross out what stays fixed and compare what changes.

Standing waves on strings (Khan Academy). Watch how the requirement of a node at each end picks out the allowed wavelengths 2L, L, 2L⁄3 and so on, and leads to fₙ = nv ÷ 2L.

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Practice

Show that $\sqrt{\frac{T}{\mu}}$ has the units of speed.
A string 0.80 m long has waves of speed $240\,\mathrm{m\,s^{-1}}$ . Find its first three harmonic frequencies.
A wire of mass per unit length $2.5\,\mathrm{g\,m^{-1}}$ is stretched by a tension of 90 N between supports 0.60 m apart. Find its fundamental frequency.
State and explain what happens to the fundamental frequency if (a) the length is halved, (b) the tension is quadrupled, (c) a wire of twice the diameter, same material, is used.
A string vibrates in its third harmonic at 450 Hz. Find its fundamental frequency, and draw the pattern of nodes and antinodes for the third harmonic.
In a sonometer experiment at fixed tension, a 512 Hz fork resonates with 0.25 m of wire. Predict the length for a 384 Hz fork.
A sonometer wire resonates with a 256 Hz fork when the load is 4.0 kg. What load would make the same length resonate with a 512 Hz fork? Comment on whether this is sensible.
Explain why an inanga string and a guitar string tuned to the same fundamental frequency still sound different.

Show answersHide answers

$\sqrt{\frac{\mathrm{N}}{\mathrm{kg\,m^{-1}}}} = \sqrt{\frac{\mathrm{kg\,m\,s^{-2}}}{\mathrm{kg\,m^{-1}}}} = \sqrt{\mathrm{m^2\,s^{-2}}} = \mathrm{m\,s^{-1}}$ .
$f_1 = \frac{240}{2 \times 0.80} = 150\,\mathrm{Hz}$ ; 300 Hz; 450 Hz.
$v = \sqrt{\frac{90}{2.5 \times 10^{-3}}} = \sqrt{3.6 \times 10^4} = 190\,\mathrm{m\,s^{-1}}$ ; $f_1 = \frac{190}{1.20} = 158\,\mathrm{Hz}$ .
(a) Doubles ( $f \propto \frac{1}{L}$ ). (b) Doubles ( $f \propto \sqrt{T}$ ). (c) $\mu \propto r^2$ becomes 4 times as large, so $f$ halves.
$f_1 = \frac{450}{3} = 150\,\mathrm{Hz}$ . Three loops: nodes at both ends and at $\frac{L}{3}$ and $\frac{2L}{3}$ ; antinodes at $\frac{L}{6}$ , $\frac{L}{2}$ , $\frac{5L}{6}$ .
$f L$ is constant: $L = \frac{512 \times 0.25}{384} = 0.33\,\mathrm{m}$ .
$f \propto \sqrt{T}$ , so doubling $f$ needs 4 times the tension: a 16 kg load. That may exceed the wire's breaking stress; it is safer to halve the length instead.
Each instrument excites a different mixture of harmonics with different strengths, and its sound box (trough or guitar body) strengthens some frequencies more than others. The fundamental, and so the pitch, is the same; the timbre differs.

Key points

Wave speed on a string: $v = \sqrt{\frac{T}{\mu}}$ .
Fixed ends are nodes: $L = \frac{n\lambda}{2}$ , $f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} = nf_1$ .
$f \propto \frac{1}{L}$ , $f \propto \sqrt{T}$ , $f \propto \frac{1}{\sqrt{\mu}}$ .
The fundamental sets the pitch; the mix of harmonics sets the quality.

1The inanga's strings

2The speed of a wave on a string

3Harmonics of a string fixed at both ends

The laws of vibrating strings

4Experiment 3.2: the sonometer

The inanga's strings

The speed of a wave on a string

Harmonics of a string fixed at both ends

Experiment 3.2: the sonometer