Longitudinal Waves on Thin Elastic Rod

Consider a thin uniform elastic rod of length $l$ and cross-sectional area $A$. Let us examine the longitudinal oscillations of such a rod. These oscillations are often, somewhat loosely, referred to as sound waves. It is again convenient to let $x$ denote position along the rod. Thus, in equilibrium, the two ends of the rod lie at $x=0$ and $x=l$. Suppose that a longitudinal wave causes an $x$-directed displacement $\psi(x,t)$ of the various elements of the rod from their equilibrium positions. Consider a thin section of the rod, of length $\delta x$, lying between $x-\delta x/2$ and $x+\delta x/2$. The displacements of the left and right boundaries of the section are $\psi(x-\delta x/2,t)$ and $\psi(x+\delta x/2,t)$, respectively. Thus, the change in length of the section, due to the action of the wave, is $\psi(x+\delta x/2,t)-\psi(x-\delta x/2,t)$. Now, strain in an elastic rod is defined as change in length over unperturbed length (Love 1944). Thus, the strain in the section of the rod under consideration is

$$\epsilon(x,t) = \frac{ \psi(x+\delta x/2,t)-\psi(x-\delta x/2,t)}{\delta x}.$$ (5.10)

In the limit $\delta x\rightarrow 0$, this becomes

$$\epsilon(x,t) = \frac{\partial \psi(x,t)}{\partial x}.$$ (5.11)

It is assumed that the strain is small compared to unity; that is, $\vert\epsilon\vert\ll 1$. Stress, $\sigma(x,t)$, in an elastic rod is defined as the elastic force per unit cross-sectional area (ibid.). In a conventional elastic material, the relationship between stress and strain (for relatively small strains) takes the simple form

$$\sigma = Y\,\epsilon.$$ (5.12)

Here, $Y$ is a constant, with the dimensions of pressure, that is known as the Young's modulus (ibid.). If the strain in a given element is positive then the stress acts to lengthen the element, and vice versa. (Similarly, in the spring-coupled mass system investigated in the preceding section, the external forces exerted on an individual spring act to lengthen it when its extension is positive, and vice versa.)

Consider the motion of a thin section of the rod lying between $x-\delta x/2$ and $x+\delta x/2$. If $\rho$ is the mass density of the rod then the section's mass is $\rho\,A\,\delta x$. The stress acting on the left boundary of the section is $\sigma(x-\delta x/2) = Y\,\epsilon(x-\delta x/2)$. Because stress is force per unit area, the force acting on the left boundary is $A\,Y\,\epsilon(x-\delta x/2)$. This force is directed in the minus $x$-direction, assuming that the strain is positive (i.e., the force acts to lengthen the section). Likewise, the force acting on the right boundary of the section is $A\,Y\,\epsilon(x+\delta x/2)$, and is directed in the positive $x$-direction, assuming that the strain is positive (i.e., the force again acts to lengthen the section). Finally, the mean longitudinal (i.e., $x$-directed) acceleration of the section is $\partial^{\,2}\psi(x,t)/\partial t^{\,2}$. Hence, the section's longitudinal equation of motion becomes

$$\rho\,A\,\delta x\,\frac{\partial^{\,2}\psi(x,t)}{\partial t^{\,2}} = A\,Y\left[\epsilon(x+\delta x/2,t)-\epsilon(x-\delta x/2,t)\right].$$ (5.13)

In the limit $\delta x\rightarrow 0$, this expression reduces to

$$\rho\,\frac{\partial^{\,2}\psi(x,t)}{\partial t^{\,2}} = Y\,\frac{\partial\epsilon(x,t)}{\partial x},$$ (5.14)

or

$$\frac{\partial^{\,2}\psi}{\partial t^{\,2}} = v^{\,2}\,\frac{\partial^{\,2}\psi}{\partial x^{\,2}},$$ (5.15)

where

$$v=\sqrt{\frac{Y}{\rho}}$$ (5.16)

is a constant having the dimensions of velocity, which turns out to be the propagation speed of longitudinal waves along the rod (see Section 6.2), and use has been made of Equation (5.11). Equation (5.15) has the same mathematical form as Equation (4.30), which governs the motion of transverse waves on a uniform string. This implies that longitudinal and transverse waves in continuous dynamical systems (i.e., systems with an infinite number of degrees of freedom) can be described using the same mathematical equation.

In order to solve Equation (5.15), we need to specify boundary conditions at the two ends of the rod. Suppose that the left end of the rod is fixed; that is, it is clamped in place such that it cannot move. This implies that $\psi(0,t)=0$. Suppose, on the other hand, that the left end of the rod is free; that is, it is not attached to anything. This implies that $\sigma(0,t)=0$, because there is nothing that the end can exert a force (or a stress) on, and vice versa. It follows from Equations (5.11) and (5.12) that $\partial\psi(0,t)/\partial x=0$. Likewise, if the right end of the rod is fixed then $\psi(l,t)=0$, and if the right end is free then $\partial\psi(l,t)/\partial x=0$.

Suppose, for the sake of argument, that the left end of the rod is free, and the right end is fixed. It follows that $\partial\psi(0,t)/\partial x=0$, and $\psi(l,t)=0$. Let us search for normal modes of the form

$$\psi(x,t) = A\,\cos(k\,x)\,\cos(\omega\,t-\phi),$$ (5.17)

where $A>0$, $k>0$, $\omega>0$, and $\phi$ are constants. The preceding expression automatically satisfies the boundary condition $\partial\psi(0,t)/\partial x=0$. The other boundary condition is satisfied provided

$$\cos(k\,l) = 0,$$ (5.18)

which yields

$$k\,l = (n-1/2)\,\pi,$$ (5.19)

where $n$ is an integer. As usual, the imposition of the boundary conditions leads to a quantization of the possible mode wavenumbers. Substitution of Equation (5.17) into the equation of motion, (5.15), yields the normal mode dispersion relation

$$\omega = k\,v.$$ (5.20)

The preceding dispersion relation is consistent with the previously derived dispersion relation (5.9), given that $m=\rho\,A\,a$ and $K=A\,Y/a$. Here, $a$ is the interplane spacing, $m$ the mass of a section of the rod containing a single plane of atoms, and $K$ the effective force constant between neighboring atomic planes.

It follows, from the previous analysis, that the $n$th longitudinal normal mode of an elastic rod, of length $l$, whose left end is free, and whose right end is fixed, is associated with the characteristic displacement pattern

$$\psi_n(x,t) = A_n\,\cos\left[(n-1/2)\,\pi\,\frac{x}{l}\right]\cos(\omega_n\,t-\phi_n),$$ (5.21)

where

$$\omega_n = (n-1/2)\,\frac{\pi\,v}{l}.$$ (5.22)

Here, $A_n$ and $\phi_n$ are constants that are determined by the initial conditions. It can be demonstrated that only those normal modes whose mode numbers are positive integers yield unique displacement patterns. Equation (5.21) describes a standing wave whose nodes (i.e., points where $\psi=0$ for all $t$) are evenly spaced a distance $l/(n-1/2)$ apart. The boundary condition $\psi(l,t)=0$ ensures that the right end of the rod is always coincident with a node. On the other hand, the boundary condition $\partial\psi(0,t)/\partial x=0$ ensures that the left hand of the rod is always coincident with a point of maximum amplitude oscillation [i.e., a point where $\cos(k\,x)=\pm 1$]. Such a point is known as an anti-node. The anti-nodes associated with a given normal mode lie halfway between the corresponding nodes. According to Equation (5.22), the normal mode oscillation frequencies depend linearly on mode number. Finally, it can be shown that, in the long wavelength limit $k\,a\ll 1$, the normal modes and normal frequencies of a uniform elastic rod specified in Equations (5.21) and (5.22) are analogous to the normal modes and normal frequencies of a linear array of identical spring-coupled masses specified in Equations (5.7) and (5.8), and pictured in Figures 5.2 and 5.3.

Figure 5.4: Time evolution of the normalized longitudinal displacement of a thin elastic rod whose left end is free (and is struck with a hammer at $t=0$), and whose right end is fixed. The top-left, top-right, middle-left, middle-right, bottom-left, and bottom-right panels correspond to $t/\tau =0.01$, $0.02$, $0.04$, $0.08$, $0.16$, $0.32$, $0.64$, and $1.28$, respectively.

Because Equation (5.15) is linear, its most general solution is a linear combination of all of the normal modes; that is,

$$\psi(x,t) = \sum_{n'=1,\infty} A_{n'}\,\cos\left[(n'-1/2)\,\pi\,\frac{x}{l}\right]\,\cos\left[(n'-1/2)\,\pi\,\frac{v\,t}{l}-\phi_{n'}\right].$$ (5.23)

The constants $A_n$ and $\phi_n$ are determined from the initial displacement,

$$\psi(x,0) = \sum_{n'=1,\infty} A_{n'}\,\cos \phi_{n'}\,\cos\left[(n'-1/2)\,\pi\,\frac{x}{l}\right],$$ (5.24)

and the initial velocity,

$$\dot{\psi}(x,0) = \frac{\pi\,v}{l}\sum_{n'=1,\infty} (n'-1/2)\,A_{n'}\,\sin \phi_{n'}\,\cos\left[(n'-1/2)\,\pi\,\frac{x}{l}\right].$$ (5.25)

It can be demonstrated that [cf., Equation (4.53)]

$$\frac{2}{l}\int_0^l \cos\left[(n-1/2)\,\pi\,\frac{x}{l}\right]\,\cos\left[(n'-1/2)\,\pi\,\frac{x}{l}\right]dx = \delta_{n,n'}.$$ (5.26)

Thus, multiplying Equation (5.24) by $(2/l)\,\cos[(n-1/2)\,\pi\,x/l]$, and integrating in $x$ from 0 to $l$, we obtain

$$C_n = \frac{2}{l} \int_0^l\psi(x,0)\,\cos\left[(n-1/2)\,\pi\,\frac{x}{l}\right]dx = A_n\,\cos\phi_n,$$ (5.27)

where use has been made of Equations (5.26) and (4.54). Likewise, Equation (5.25) gives

$$S_n =\frac{2}{v\,(n-1/2)\,\pi} \int_0^l\dot{\psi}(x,0)\,\cos\left[(n-1/2)\,\pi\,\frac{x}{l}\right]dx = A_n\,\sin\phi_n.$$ (5.28)

Finally, $A_n = (C_n^{\,2}+S_n^{\,2})^{1/2}$ and $\phi_n=\tan^{-1}(S_n/C_n)$.

Suppose, for the sake of example, that the rod is initially at rest, and that its left end is hit with a hammer at $t=0$ in such a manner that a section of the rod lying between $x=0$ and $x=a$ (where $a<l$) acquires an instantaneous velocity $V_0$. It follows that $\psi(x,0)=0$. Furthermore, $\dot{\psi}(x,0)=V_0$ if $0\leq x\leq a$, and $\dot{\psi}(x,0)=0$ otherwise. It can be demonstrated that these initial conditions yield $C_n=0$, $\phi_n=\pi/2$,

$$A_n =S_n= \frac{V_0\,a}{v}\,\frac{2}{\pi}\,\frac{\sin[(n-1/2)\,\pi\,a/l]}{(n-1/2)^{\,2}\,\pi\,a/l},$$ (5.29)

and

$$\psi(x,t) = \sum_{n=1,\infty} A_n\,\cos\left[(n-1/2)\,\pi\,\frac{x}{l}\right]\sin\left[(n-1/2)\,\pi\,\frac{t}{\tau}\right],$$ (5.30)

where $\tau=l/v$. Figure 5.4 shows the time evolution of the normalized rod displacement, $\hat{\psi}(x,t)= (v/V_0\,a)\,\psi(x,t)$, calculated from the preceding equations using the first 100 normal modes (i.e., $n=1,100$), and choosing $a/l=0.1$. It can be seen that the hammer blow generates a displacement wave that initially develops at the free end of the rod ($x/l=0$), which is the end that is struck, propagates along the rod at the velocity $v$, and reflects off the fixed end ($x/l=1$) at time $t/\tau=1$ with no phase shift. (The wave front is traveling from the left to the right in all panels except the final one, where it is traveling from right to left.)