We demonstrate the inadequacy of relying on computation without insight. We also find the vibrations of a beam.
Although elasticity is encountered in school physics, where the bulk modulus, shear modulus and Young's modulus are explained, mechanical waves in a solid medium are not, except perhaps for asserting that the speed of longitudinal waves in a bar or wire is the square root of the ratio of Young's modulus to the density. Even if we bypass the complex problem of the elasticity of real substances, and assume an isotropic, homogeneous medium, the problem of determining all the phenomena of wave propagation in solids is daunting. Surfaces have a strong effect, and there are wave modes guided by the surfaces (Rayleigh and Love waves), as well as longitudinal and transverse modes in the bulk, all of which interact with each other at a surface. Sources are another cause of anxiety. All practical waves have sources, are limited in extent, and interact with surfaces. We know all the fundamentals, and linearization is an excellent approximation, but working out the details in any particular case, or giving general rules, is difficult. Even the availability of great computing power only buries the investigator in piles of figures, and gives little insight . A slight misconception in boundary conditions, or in material behaviour, or in control of computational errors, may vitiate all the results.
The most interesting area of application of mechanical waves is geophysical. Natural waves, whose sources are crustal movements, tides or deep disturbances, give the only information on the interior of the earth, and are interesting on their own. Artificial waves, from explosions intentional and unintentional, or mechanical pounding, are used to explore the upper layers of the crust, scientifically, or for structural traps for petroleum. In this case the medium is not isotropic or homogeneous, but horizontally stratified, which introduces further complication.
It is easy to find the wave equations for the main types of bulk plane waves in an isotropic medium, since we can use principal axes. Let's begin with shear waves or S waves, where the displacement is perpendicular to the direction of motion, which we take as the x-axis. If y is the lateral displacement, the shear strain is simply dy/dx. Consider an element of thickness dx and unit area. When we equate the mass times acceleration of this element to the difference in the shear forces on its two faces, we immediately find the wave equation for y, and the wave velocity shown in the Figure.
If the displacement y is in the same direction as x, we have a longitudinal wave. If the medium is a wire or bar of limited cross-sectional area, a longitudinal strain will be accompanied by a transverse strain, given by Poisson's ratio, because the transverse stresses must be zero. If the wavelength is much longer than the transverse dimensions, the inertia of this movement will be negligible, and we can neglect it. In that case, the stress is given by Young's modulus, and we can again write down the wave equation for y at once.
However, if the longitudinal wave propagates in an extended medium, and the displacement is purely longitudinal, the transverse strains e2 and e3 will vanish. Since the dilatation Δ = e1 in this case, the tensile stress p1 = (λ + 2μ)e1. The modulus can also be written in terms of the bulk and shear moduli as κ + 4μ/3. Let's call these longitudinal waves P waves. They can also be called compressional waves. Note that finite transverse stresses p2 = p3 = (λ/λ + 2μ)p1 exist.
In isotropic, homogeneous bulk matter, then, we have a plane P wave and two polarizations of plane S waves in any direction that propagate independently, and can be superimposed. When such waves strike an ideal plane boundary between different media, they are reflected and refracted according to the usual laws, but now there is the complication of different phase velocities and change of wave type. The boundary conditions are the continuity of displacement and stress at the interface. These six conditions determine the amplitudes of the six reflected and refracted waves. The problem is straightforward and soluble, but complicated.
Transverse waves can exist on a wire or beam. When the restoring force is provided by the tension on a wire, and the stiffness of the wire is neglected, we have the familiar case of transverse waves on a string. On the other hand, consider a beam with no longitudinal force, the stiffness providing the restoring force for sideways movement. Let z be the displacement, and consider a thin slice of thickness dx, cross-sectional area A, and moment of inertia of area about the neutral axis of bending of I. The shear forces F and bending moment M act on this slice as shown in the Figure. It is straightforward to write down the equations of linear and rotational motion for the slice. The moment of inertia is the mass of the slice times I, and the angle of rotation is dz/dx. We assume that the beam never departs far from its equilibrium position, so that we can assume that angles are small. The shear force can be eliminated between the two equations of motion, and the bending moment can be expressed in terms of the second derivative of the displacement as shown. The final equation for z(x,t) is not the simple wave equation, but it is linear.
We can try a travelling wave solution of the form z = exp[j(ωt - kx)], which satisfies the equation provided c2 = ω2/k2 = (Y/ρ)(Ik2/A)/(1+Ik2/A). The term Ik2/A is the square of 2π(radius of gyration/wavelength), and is in all practical cases very small. The neglect of this factor in the denominator corresponds to the neglect of the second term on the left of the equation for z(x,t), which comes from rotation. In deriving the equation, we could have simply used F = dM/dx, and neglected the rotation from the beginning. However, including it gives us better insight. Therefore, we have c = 2πsqrt(IY/Aρ)/λ, and the propagation is strongly dispersive, short wavelengths travelling faster than longer.
If k4 = (ρA/YI)ω2, the corresponding harmonic solution of the equation is z = (A cosh kx + B sinh kx + C cos kx + D sin kx)exp(jωt). If we take the origin x = 0 at the centre of the bar, we can consider the even modes A cosh kx + B cos kx and the odd modes B sinh kx + D sin kx separately. The lowest mode will turn out to be even. At the ends of the bar, x = +L/2, -L/2, the bending moment and shear force must vanish, so d2z/dx2 = d3z/dx3 = 0. This leads to the condition that tanh (kL/2) + tan (kL/2) = 0. The solutions of this transcendental equation are very closely kL/2 = (n - 1/4)π, where n = 1, 2, ..... If you want more accurate figures, an iteration or two by, say, Newton's Method, will give them. The fundamental mode has λ = 4L/3 and frequency f = (9π/8L2)sqrt(YI/Aρ). Once k is known, the ratio of the constants can be found. The solution is z = C(cos kL/2 cosh kx + cosh kL/2 cos kx) exp(jωt). The nodes are at a distance 0.244 of the length from the ends. The antisymmetric modes have solutions kL/2 = (n + 1/4)π, and sin and sinh replace cos and cosh.
To see if our result is anywhere near reasonable, let us find the frequency of a steel bar 2 cm broad, 1 cm deep and 16 cm long. Taking ρ = 7.849 g/cm3, I = bh3/12 = 1/6 cm4 and Y = 2.139 x 1012 dy/cm2, we find f = 2080 Hz, a reasonable answer. A bar of half the area, 1 cm by 0.5 cm, would have a frequency of 1/4 this, or 520 Hz. The first overtone of the bar, which is an antisymmetric mode, has a wavelength 3/5 that of the fundamental, and so a frequency higher by a factor 25/9, or about 2.78. This will not be a harmonic (it is about an octave and a third). If it is suppressed by the mounting of the bar, or the way in which the bar is struck, the bar will sound principally the fundamental. That the frequencies of a bar were proportional to 32, 52, 72, etc., was found empirically.
If the ends of the bar are pinned, the problem resembles that of a beam bridge. Now z = d2z/dx2 = 0 at x = +L/2 and -L/2. The symmetrical vibrations are C cos kx, with cos mL/2 = 0 or k = (2n-1)π/L, and C sin kx, with k = 2nπ/L, n = 1, 2, .... The frequencies of vibration are proportional to the squares of the natural numbers, i.e., 1, 4, 9, etc.
A tuning fork is like two bars clamped at one end. The boundary conditions for a bar clamped at x = -L/2 and free at x = +L/2 are z = dz/dx = 0 at the clamped end, and d2z/dx2 = d3z/dx3 = 0 at the free end. One type of vibration has z = A cosh kx + D sin mx, and the condition coth kL/2 = tan kL/2. Approximately, kL/2 = (n + 1/4)π, with n = 0, 1, 2, .... The other type has z = B sinh kx + C cos kx, and the condition coth kL/2 = - tan kL/2. Here, kL/2 = (n - 1/4)π, n = 1, 2, 3, etc., approximately. The approximation is not as close in this case as in the preceding ones. Using an accurate value for the lowest mode, λ = 3.35L and f = (0.56/L)sqrt(YI/Aρ). One can compare the prediction of this formula with the frequency of an actual tuning fork.
H. Lamb, The Dynamical Theory of Sound, 2nd ed. (London: Edward Arnold, 1925), Chapter IV.
Composed by J. B. Calvert
Created 25 June 2000