In this section we derive the wave equation in an ideal case, which is nevertheless very useful for practical applications in acoustics (e.g. marimba, xylophone, tubular bells) and in engineering (stresses on beams, seismic waves).

After carefully defining the system, we will proceed in three simple steps:

1. balance of forces
2. identification of the quantity to be used as the wave function
3. writing the equation of motion

What is a bar

• Unlike the string, the bar we consider here is a true three-dimensional elastic system. Inside it, forces act that are not only longitudinal, but can be directed in any direction.
• The fact that it is a "bar", that is, an object longer than it is wide, serves in this case only to reduce the number of dimensions in the equations, but it is not a critical property (at least until the bar becomes a waveguide)
• The bar we consider is nevertheless ideal, because we assume that its elasticity is perfectly linear, that is, that the deformation is directly proportional to the force that caused it (Hooke's law).

The elastic Force is directly proportional to the Deformation

• The constant of proportionality is not in general a simple constant, but an object called a tensor, which takes into account the fact that a force exerted in one direction could actually give rise to a deformation in a completely different direction.

• However, in the simplest cases this object can be reduced to a few constants. Each of them describes a different type of deformation. This is why in what follows we examine one by one the waves that can be produced by different deformations of the bar.

Longitudinal waves

What we know

• Longitudinal waves are generated by compression (or traction).
• The corresponding deformation is a compression (or a dilation), which propagates in the same direction in which the force that produced it was exerted.
• The constant of proportionality between the deformation and the pressure that produces it is called Young's modulus Y.
• Newton's equation of motion for translations

${\displaystyle F=ma}$

What we want to derive

• We assume that a compression (or a dilation) takes place along the axis of the bar, in the x direction.
• We want to derive the equation of motion for the deformation ${\displaystyle \xi (x,t)}$ of an infinitesimal element of the bar with cross-section A and height dx, located at position x.

Proof

• The infinitesimal element of the bar is deformed into an element of length ${\displaystyle dx+d\xi }$.
• Let us remember that what propagates in the bar is the compression, and therefore the relevant quantity is ${\displaystyle \xi }$, and not x.
• The deformation is given by ${\displaystyle \frac{\partial \xi }{\partial x}}$, and, by Hooke's law, we have

${\displaystyle F(x(t))=AY\frac{\partial \xi }{\partial x}}$.

This equation gives us the value of F, but we are interested in the net force acting on the small volume element (see figure), which is

${\displaystyle F(x+dx)-F(x)=dF=\frac{\partial F}{\partial x}dx}$.

• We must therefore differentiate the previous relation with respect to x, and we obtain

${\displaystyle \frac{\partial F}{\partial x}dx=YA\frac{{\partial }^2\xi }{\partial x^2}dx}$

• By Newton's law this force must be equated to the product ma for the small element, which is

${\displaystyle ma=\rho Adx\frac{{\partial }^2\xi }{\partial t^2}}$

• Equating the two terms we find

${\displaystyle YAdx\frac{{\partial }^2}{\partial x^2}\xi (x,t)=\rho Adx\frac{{\partial }^2}{\partial t^2}\xi (x,t)}$,

which is an equation identical to the one for transverse waves on the string, that is

${\displaystyle \frac{{\partial }^2\xi }{\partial x^2}=\frac{1}{c^2}\frac{{\partial }^2\xi }{\partial t^2}}$,

except for the fact that the propagation speed is given by

${\displaystyle c=\sqrt{\frac{Y}{\rho }}}$.

Flexural waves

What we know

• Flexural waves are generated by a bending of the bar.
• Bending is produced when a pair of opposite forces, not applied at the same point, acts on the bar in a plane containing the longitudinal axis (see figure).
• If the bar were a perfectly rigid body it would tend to rotate rather than deform, so the equation of motion F=ma is not enough: we must also use its equivalent for rotation, that is

${\displaystyle M=I\frac{d^2\theta }{d{t}^2}}$,

where M is the moment of the applied force (bending moment), I the moment of inertia of the bar about an axis perpendicular to the longitudinal axis, and ${\displaystyle \theta }$ the angle of rotation.

• In this case too, perfect elasticity is expressed by a law of proportionality between the applied bending moment and the resulting curvature. The constant of proportionality Y is the elastic constant (Young's modulus) of the bar.

What we want to derive

• We assume that the bending takes place along the y axis of the bar, perpendicular to the longitudinal axis x. In this case the moment is directed along the z axis (see figure).
• We want to derive the equation of motion for the vertical position y(x,t) of an infinitesimal element of the bar of initial length dx, located at horizontal position x.

Proof

• We assume that a bending takes place along the y direction, perpendicular to the axis of the bar. This means that each small element of the bar moves perpendicularly to the axis by an amount y(x,t) that varies along the bar.
• The deformation in this case is the curvature of the bar, which is proportional to the second derivative with respect to x of the "shape" of the bar y(x).
• Hooke's law, in this case, requires the stress to be proportional to the deformation, that is

${\displaystyle M(x)=-YJ\frac{d^{2}y}{d{x}^2}}$,

where the constant of proportionality is, for convenience, split into a factor depending only on the material (Young's modulus ${\displaystyle Y}$, which has the dimensions of a pressure) and one depending only on its cross-section, called the second moment of area ${\displaystyle J}$ (which has the dimensions of an area squared, that is, of a length to the fourth power).

• Differentiating this relation once with respect to x we obtain the force ${\displaystyle F_t}$ (called the shear force)

${\displaystyle F_t(x)=-YJ\frac{d^{3}y}{d{x}^3}}$.

The minus sign depends on the convention chosen for the bending moment: here M is positive when the bar bends with its concavity downwards, that is, when the curvature ${\displaystyle d^{2}y/d{x}^2}$ is negative.

• However, as usual, we are interested in the quantity

${\displaystyle F_t(x+dx)-F_t(x)=d{F}_t=\frac{\partial F_t}{\partial x}dx}$,

which is the net force acting on the infinitesimal element (called the load of the bar). The load on the infinitesimal element is therefore

${\displaystyle d{F}_t=-YJ\frac{d^{4}y}{d{x}^4}dx}$.

• Equating the load to the product ma, according to Newton's equation, we find

${\displaystyle -YJdx\frac{{\partial }^4}{\partial x^4}y(x,t)=\rho Adx\frac{{\partial }^2}{\partial t^2}y(x,t)}$,

that is

${\displaystyle \frac{{\partial }^{4}y}{\partial x^4}=-\frac{1}{k^2}\frac{{\partial }^{2}y}{\partial t^2}}$,

with

${\displaystyle k=\sqrt{\frac{YJ}{\rho A}}}$

• NOTE: what we have obtained in this case is a fourth-order equation, and therefore completely different from the one for longitudinal waves in the bar, and from the one for transverse waves on the string. In this case we cannot expect the same type of solutions.

• Looking for a harmonic solution ${\displaystyle y=A\sin (qx-\omega t)}$, where q is the wavenumber, we obtain ${\displaystyle q^4={\omega }^2/k^2}$, that is ${\displaystyle \omega =k{q}^2}$. The phase velocity ${\displaystyle v=\omega /q=\sqrt{k\omega }}$ grows with frequency: flexural waves are dispersive, and a pulse changes shape as it propagates. For the same reason the natural frequencies of a bar are not integer multiples of the fundamental (see the page natural frequencies of the bar).

• In the derivation we have neglected the rotational inertia of the small elements of the bar (Euler-Bernoulli approximation): the result holds when the wavelength is much larger than the thickness of the bar.

Torsional waves

What we know

• Torsional waves are generated by a twisting of the bar.
• Torsion, like bending, is produced when a pair of opposite forces, not applied at the same point, acts on the bar. However, unlike bending, the forces are directed in a plane perpendicular to the axis of the bar (see figure).
• In this case too we must use the equation of motion for rotation

${\displaystyle M=I_l\frac{d^2\theta }{d{t}^2}}$,

where M is the moment of the applied force (torque), ${\displaystyle I_l}$ the moment of inertia of the bar about its axis (for a small element of length dx it is ${\displaystyle \rho I_{p}dx}$, where ${\displaystyle I_p}$ is the polar moment of inertia of the cross-section), and ${\displaystyle \theta }$ the angle of rotation.

• Hooke's law requires the torque to be directly proportional to the rotation. The constant of proportionality G is the shear modulus of the bar.

What we want to derive

• We assume that the torsion takes place about the longitudinal axis x of the bar. In this case the moment is directed along the x axis (see figure).
• We want to derive the equation of motion for the rotation ${\displaystyle \theta (x,t)}$ of an infinitesimal element of the bar of length dx, located at position x.

Proof

• We study a small rotation by an angle ${\displaystyle d\theta }$.
• Hooke's law is written

${\displaystyle M(\theta (x))=G{I}_p\frac{\partial \theta }{\partial x}}$

• As usual, we are interested in the difference between the moments at the ends of the small volume element

${\displaystyle M(x+dx)-M(x)=dM=\frac{\partial M}{\partial x}dx}$,

and therefore, differentiating Hooke's law with respect to x, we have

${\displaystyle dM=G{I}_p\frac{{\partial }^2\theta }{\partial x^2}dx}$

• Substituting into the equation of motion for rotation

${\displaystyle dM=I_p\rho \frac{{\partial }^2\theta }{\partial t^2}dx}$

this becomes

${\displaystyle \rho I_{p}dx\frac{{\partial }^2}{\partial t^2}\theta (x,t)=G{I}_{p}dx\frac{{\partial }^2}{\partial x^2}\theta (x,t)}$,

that is

${\displaystyle \frac{{\partial }^2\theta }{\partial x^2}=\frac{1}{c^2}\frac{{\partial }^2\theta }{\partial t^2}}$,

• The wave equation is therefore entirely analogous to that of longitudinal waves (or to that of transverse waves on the string), but its variable is the angle ${\displaystyle \theta }$

• The propagation speed of torsional waves is therefore (for a bar of circular cross-section; for cross-sections of other shapes the torsional stiffness is smaller than ${\displaystyle G{I}_p}$ and the waves are slower)

${\displaystyle c=\sqrt{\frac{G}{\rho }}}$

Remarks on the solutions

The same equations have the same solutions.

On the basis of this principle we can conclude that longitudinal waves in the bar, torsional waves and transverse waves on the string, although they refer to different quantities, have exactly the same properties. Flexural waves are the exception: they obey a different equation and are dispersive.

In the section wave equation for sound waves it is shown that the wave equation for sound waves also has the same form as the previous ones.

In all these cases the unifying power of the concept of wave is evident.

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