# Vibrating string

Vibration, standing waves in a string. The fundamental and the first 6 overtones form a harmonic series.

A vibration in a string is a wave. Usually a vibrating string produces a sound whose frequency in most cases is constant. Therefore, since frequency characterizes the pitch, the sound produced is a constant note. Vibrating strings are the basis of any string instrument like guitar, cello, or piano.

## Wave

The speed of propagation of a wave in a string ($v$) is proportional to the square root of the tension of the string ($T$) and inversely proportional to the square root of the linear density ($\mu$) of the string:

$v = \sqrt{T \over \mu}.$

### Derivation

Let $\Delta x$ be the length of a piece of string, $m$ its mass, and $\mu$ its linear density. If the horizontal component of tension in the string is a constant, $T$, then the tension acting on each side of the string segment is given by

$T_{1x}=T_1 \cos(\alpha) \approx T.$
$T_{2x}=T_2 \cos(\beta)\approx T.$

If both angles are small, then the tensions on either side are equal and the net horizontal force is zero. From Newton's second law for the vertical component, the mass of this piece times its acceleration, $a$, will be equal to the net force on the piece:

$\Sigma F_y=-T_{2y}-T_{1y}=-T_2 \sin(\beta)-T_1 \sin(\alpha)=\Delta m a\approx\mu\Delta x \frac{\partial^2 y}{\partial t^2}.$

Dividing this expression by $T$ and substituting the first and second equations obtains

$-\frac{\mu\Delta x}{T}\frac{\partial^2 y}{\partial t^2}=\frac{T_2 \sin(\beta)}{T_2 \cos(\beta)}+\frac{T_1 \sin(\alpha)}{T_1 \cos(\alpha)}=\tan(\beta)+\tan(\alpha)$

The tangents of the angles at the ends of the string piece are equal to the slopes at the ends, with an additional minus sign due to the definition of beta. Using this fact and rearranging provides

$\frac{1}{\Delta x}\left(\left.\frac{\partial y}{\partial x}\right|^{x+\Delta x}-\left.\frac{\partial y}{\partial x}\right|^x\right)=\frac{\mu}{T}\frac{\partial^2 y}{\partial t^2}$

In the limit that $\Delta x$ approaches zero, the left hand side is the definition of the second derivative of $y$:

$\frac{\partial^2 y}{\partial x^2}=\frac{\mu}{T}\frac{\partial^2 y}{\partial t^2}.$

This is the wave equation for $y(x,t)$, and the coefficient of the second time derivative term is equal to $v^{-2}$; thus

$v=\sqrt{T\over\mu},$

where $v$ is the speed of propagation of the wave in the string. (See the article on the wave equation for more about this). However, this derivation is only valid for vibrations of small amplitude; for those of large amplitude, $\Delta x$ is not a good approximation for the length of the string piece, the horizontal component of tension is not necessarily constant, and the horizontal tensions are not well approximated by $T$.

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## Frequency of the wave

Once the speed of propagation is known, the frequency of the sound produced by the string can be calculated. The speed of propagation of a wave is equal to the wavelength $\lambda$ divided by the period $\tau$, or multiplied by the frequency $f$ :

$v = \frac{\lambda}{\tau} = \lambda f.$

If the length of the string is $L$, the fundamental harmonic is the one produced by the vibration whose nodes are the two ends of the string, so $L$ is half of the wavelength of the fundamental harmonic. Hence one obtains Mersenne's laws:

$f = \frac{v}{2L} = { 1 \over 2L } \sqrt{T \over \mu}$

where $T$ is the tension, $\mu$ is the linear density (that is, the mass per unit length), and $L$ is the length of the vibrating part of the string. Therefore:

• the shorter the string, the higher the frequency of the fundamental
• the higher the tension, the higher the frequency of the fundamental
• the lighter the string, the higher the frequency of the fundamental

Moreover, if we take the nth harmonic as having a wavelength given by $\lambda_n = 2L/n$, then we easily get an expression for the frequency of the nth harmonic:

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

And for a string under a tension T with density $\mu$, then

$f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$
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## Observing string vibrations

One can see the waveforms on a vibrating string if the frequency is low enough and the vibrating string is held in front of a CRT screen such as one of a television or a computer (not of an oscilloscope). This effect is called the stroboscopic effect, and the rate at which the string seems to vibrate is the difference between the frequency of the string and the refresh rate of the screen. The same can happen with a fluorescent lamp, at a rate that is the difference between the frequency of the string and the frequency of the alternating current. (If the refresh rate of the screen equals the frequency of the string or an integer multiple thereof, the string will appear still but deformed.) In daylight and other non-oscillating light sources, this effect does not occur and the string appears still but thicker, and lighter or blurred, due to persistence of vision.

A similar but more controllable effect can be obtained using a stroboscope. This device allows matching the frequency of the xenon flash lamp to the frequency of vibration of the string. In a dark room, this clearly shows the waveform. Otherwise, one can use bending or, perhaps more easily, by adjusting the machine heads, to obtain the same, or a multiple, of the AC frequency to achieve the same effect. For example, in the case of a guitar, the 6th (lowest pitched) string pressed to the third fret gives a G at 97.999 Hz. A slight adjustment can alter it to 100 Hz, exactly one octave above the alternating current frequency in Europe and most countries in Africa and Asia, 50 Hz. In most countries of the Americas—where the AC frequency is 60 Hz—altering A# on the fifth string, first fret from 116.54 Hz to 120 Hz produces a similar effect.

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