Theory of the String Telegraph


  1. Introduction
  2. Longitudinal Vibrations
  3. Application of the Theory
  4. References


The string telegraph, invented by Robert Hooke (see String Telegraph), is really a telephone, and the best string is a steel wire. Sound is transmitted over the wire as longitudinal vibrations. The theory is simple, and has many applications to similar systems, so it is worth-while studying.

Longitudinal Vibrations

Consider a wire of uniform cross-sectional area S cm2 made of a material with density ρ gm/cm2, and of arbitrary length. The maximum transverse dimension of the wire is to be much less than a wavelength of the vibrations in the wire, a condition that is usually very well satisfied. The tension in the wire is a matter of indifference, so long as the wire is reasonably taut, and not in contact with a surface, so that friction can be neglected.

As in the case of sound waves, it is convenient to describe the vibrations by the displacement ξ(z,t) of a point on the wire ("particle") that is at position z when the wire is at rest. From this function, it is easy to find the strain ε = ΔL/L in the wire, which is ε = ∂ξ/∂z. The change in tension in the wire due to this strain will be F = SYε, where Y is the adiabatic Young's modulus of the wire material. For solids, such as steel, this is close enough to the usual isothermal Young's modulus that we can neglect the difference. The net force on a length dz of the wire is the difference of the tensions at its two ends, which is SY(∂2ξ/∂z2)dz. This is equal to the mass of the length dz, ρSdz, times its acceleration. The result is the wave equation, ∂2ξ/∂t2 = c22ξ/∂z2, with a phase velocity c = √(Y/ρ). Note that the area S has cancelled out of the result, so the area of the wire does not affect the propagation characteristics.

Now we can use all the tools we have for dealing with the wave equation. Let us assume we have solutions of the type ξ = Aej(ωt - kz), where A is a phasor amplitude, ω is the angular frequency 2πf, and k is the propagation constant, 2π/λ, where λ is the wavelength. The magnitude of k is given by k = ω/c, which we discover by substituting the solution in the wave equation. Positive and negative values correspond to waves propagating in the +z and -z directions, respectively.

For a wave propagating in the +z direction, the tension F = -jkSYA, and for a wave in the other direction, F = jkSYA. The combination kY = ω(Y/c) = ωρc. The particle velocity for either wave is v = jωA. Therefore, the ratio F/v = ±ρcS, which is a mechanical impedance. The energy in the wave will be I = avg[|v|2ρcS] erg/s = ω2ρcS|A|2/2, assuming that A is the peak amplitude. This allows us to calculate the particle displacement when we know the energy in the wave.

Now suppose that at z = 0 two wires of different characteristics are joined. The displacement ξ must be continuous, and the tension in one wire must equal the tension in the other. Let a wave of amplitude A be incident from the left in wire 1, a wave of amplitude B be reflected and travel to the left in wire 1, and a wave of amplitude C propagated to the right in wire 2. The continuity of displacement gives the condition A + B = C, while the equality of forces gives (A - B)ρcS = Cρ'c'S', where primes refer to wire 2, and no primes to wire 1. These equations are easily solved for the unknowns B and C.

The results are B = [(ρcS - ρ'c'S')/(ρcS + ρ'c'S')] A and C = [2ρcS/(ρcS + ρc'S')] A. Let's look at the two limiting cases when the impedances are very different. First, if ρcS is much less than ρ'c'S', C is about zero, and B = -A. This is the case of a rigid wall, or rare-to-dense reflection. The incident wave is reversed in phase, and all the energy is reflected from the boundary. If ρcS is much greater than ρ'c'S', then C = 2A and B = A. The incident wave is reflected without phase change in this dense-to-rare reflection, so all the energy is again reflected. Do not be troubled by the fact that C = 2A. The intensity of this wave will be ρ'c'S'|C|2/2 = 4(ρ'c'S'/ρcS)(ρcS|A|2/2), or approximately zero.

If ρ'c'S' = ρcS, or the impedances are equal, then B = 0, and all the energy is transmitted. C has just the correct value to make this so, as you can easily check.

Application of the Theory

The Young's modulus of steel is 30 x 106 psi, or 2.07 x 1012 dyne/cm2, and its density is ρ = 7.86 g/cm3. The phase velocity of longitudinal waves in a steel wire is then c = 5130 m/s. This should not be confused with the phase velocity of bulk longitudinal waves (which is somewhat greater), or with the phase velocity of transverse waves in the wire (which is much less). The impedance ρc = 4.04 x 106 g/s-cm2.

Aluminium has Y = 6.9 x 1011 dyne/cm2 and ρ = 2.71 g/cm3, both smaller than for steel, but c = 5040 m/s is not much different. ρc = 1.37 x 106 g/s-cm2 for Al. Copper's Y = 1.172 x 1012 and its ρ = 8.9 g/cm3, so c = 3630 m/s and ρc = 3.23 x 106 g/s-cm2. For rubber, Y = 2.76 x 107 and ρ = 1.1 g/cm3 are typical values. Therefore, c = 50 m/s and ρc = 5500 g/s-cm2.

Air has ρc = 43 g/s-cm2 and c = 332 m/s at STP. The impedance is very much less than for steel, so that at an air-steel interface, the transmitted energy is only about 10-5. It is very hard to couple acoustic longitudinal waves to longitudinal waves in solids. This forms one of the major problems in designing a string telegraph.

Let us suppose we use #80 steel wire, 0.0343 mm or 0.0135" in diameter. The area of this wire is 4.24 x 10-4 cm2, or 1.4314 x 10-4 in2. 100 m of this wire weighs 72 g. If we assume a working stress of 50,000 psi, the tension should be 7.16 lb. Coupling with the air at both ends could be by a circular diaphragm of 10 cm diameter or 78.54 cm2 area. The mechanical impedance of the wire will be 3696 g/s, while that of the diaphragm may be estimated at (43)(78.54) = 3377 g/s. The match is, therefore, pretty good. The diaphragm will not have an impedance this high if it simply radiates in free air. It would be better for it to work into a volume connected to an exponential horn.

Since the threshold sensitivity of the ear is about 10-16 W/cm2 at 1000 Hz, a power of 10-8 W transmitted through the wire should give sufficient audibility. The peak amplitude of the wave will be 5.2 x 10-3 cm, or 52 μm, under these conditions. This seems quite practical, so sufficient power can be transmitted on such a small wire.

The effects of different kinds of wire supports should be investigated if an actual string telegraph is constructed. Knife edges should be satisfactory if they can move longitudinally enough that particle displacements can be considered as free. The use of a fine wire makes these supports difficult to arrange. On the other hand, a resilient support may be preferable. Fine wire hangers could also be tried. It should not be difficult to get a range of 100 m with a string telegraph using fine steel wire if care is taken with the receivers. Even the tin can might form a reasonable substitute for a proper diaphragm and horn.


J. W. S. Rayleigh, Theory of Sound (New York: Dover, 1945). Vol I, pp. 234-235 and 242-253.

Return to Wave Index

Composed by J. B. Calvert
Created 22 September 2003
Last revised