I believe there is an acoustic analogy to evanescent waves, therefore a spring and mass lattice model must approximate evanescent waves in the same way that a finite difference model of the continuous medium example below must simulate evanescent waves.
In homogeneous mediums scalar acoustic fields fulfil the Helmholtz equation:
$$\left(\nabla^2 + \frac{\omega^2}{c_a^2}\right) \psi = 0$$
where $\omega$ is the waves angular frequency, $c_a = \sqrt{\frac{K}{\rho}}$ the speed of sound in the medium, $\rho$ the medium's mass density and $K$ its bulk modulus.
Therefore, if you have a plane wave travelling in a medium of relatively low speed and it is incident at a glancing angle on a plane interface with a medium of relatively high speed and if the angle of incidence $\theta$ is greater than the critical angle $\theta_c$ where:
$$\sin\theta_c = \frac{c_{a,1}}{c_{a,2}} = \sqrt{\frac{K_1\,\rho_2}{K_2\,\rho_1}}$$
where $c_{a,1}$ is the speed of sound in the first (slower) medium and $c_{a,2}$ that of the second (faster) medium, then there will be total internal reflexion. Therefore, there will be an evanescent wave tunnelling into the second (faster) medium, with a Goos-Hänchen phase shift. Scalar wave theory (i.e. waves fulfilling the Helmholtz equation) is all that is needed to account for this phenomenon in optics, so there is a precise mathematical analogy between this acoustic example and my answer here where I derive the evanescent wave and the Goos-Hänchen phase shift.
So now, if you have a discretised model of such a system: i.e. two "mediums" made of a spring and mass lattices with a plane interface between them, you can have total internal reflexion and the continuum, Helmholtz-equation-fulfilling waves as long as $\lambda = 2\pi c_a/\omega$ is much longer than the lattice spacing in both mediums. Therefore, you must get approximately evanescent waves on the high velocity medium side.
It is likely even possible for discretised exactly exponentially decaying waves to be derived mathematically in the lattice without thinking of them as approximation to the continuum Helmholtz equation modes: the lattice will support discretised sine waves. You could probably derive the theory with $z$-Transformed discretised propagation equations instead of Laplace / Fourier transformed Helmholtz equations.
I have added the acoustics tag to your question because there is definitely an acoustic element to this question. Hopefully an acoustics expert can confirm the above and describe mechanical evanescent waves more directly.
Further answer after new comments
Great, total internal reflection is what I was now going to try. In fact the system I'm looking for needn't be discrete - I'd be happy with a continuous model. I tested on discrete one because I could make band gap there. What bothers me in total internal reflection is that it seems to be fundamentally multidimensional. Do you have any ideas on getting this effect in 1 dimension?
This is a very good point. I agree with you that you need two or more dimensions for total internal reflexion and this is a fundamental limitation because TIR is a glancing phenomenon - the result of a component of the wavevector projected onto an interface such that the wavevector component $k_x^\prime$ is higher than the wavevector magnitude $k^\prime$ in the new medium so that the new axial component $k_z^\prime$, which fulfils the Fourier transformed Helmholtz equation ${k_z^\prime}^2 + {k_x^\prime}^2 = k^2$ so that ${k_z^\prime}^2 < 0$.
If you are interested in one dimension, then I believe that truly there are no truly one dimensional evanescent plane waves.
For general background, you may wish to take a look at the following document:
S.W. Rienstra & A. Hirschberg (Eindhoven University of Technology), "An Introduction to Acoustics", 14 December 2013
They deal with evanescent waves in §3.3 and one dimensional systems in §4. An interesting piece of trivia is that the acoustic coupling between the loudspeaker and the listener's ear in a Sony Walkman (the authors are perhaps letting slip their age here - presumably this comment includes things like modern earpieces for iPods and the like) is through near-field, evanescent wave transmission. This of course has the advantage of being not bothersome to other nearby people because the radiative component of the acoustic field is minimised. Evanescent waves come up again in §7.2, where they look at cut-off modes in ducts. However, all these examples are two dimensional: a circular duct supports cylindrical Bessel modes whose pressure varies like $J_n(U_{n,m} r/r_0) \cos(n\,\phi+\delta)\,e{i\,k_z\,z}$ where $r$ is the radial co-ordinate, $\phi$ the azimuthal co-ordinate, $z$ the axial co-ordinate, $r_0$ the duct radius and $U_{n,m}$ the $m^{th}$ zero of the first derivative of $n^{th}$ order first kind Bessel function $J_n$. The transverse component of the wavevector has magnitude $U_{n,m}/r_0$ and the substitution of the solution into Helmholtz's equation yields:
$$k_z^2 + \frac{U_{n,m}^2}{r_0^2} = k^2$$
thus leading to "cut-off" modes (i.e. evanescent ones with imaginary axial wavenumber) when $U_{n,m} > k\,r_0$.
In one dimension, something that might be construed as an evanescent wave is the transitional acoustic field in a tuned acoustic grating. In §4.4.1 of the Rienstra &Hirschberg reference above, they derive the transmission $T$ and reflection $R$ coefficients across a plane interface and they are precisely analogous to these electromagnetic case for transverse electromagnetic waves on two-conductor, translationally invariant transmission-lines, namely:
$$R = \frac{Z_1-Z_2}{Z_1+Z_2};\quad T = \frac{2\,Z_2}{Z_1+Z_2}$$
where $Z_1, Z_2$ are the acoustic impedances of the two mediums in question. Thus if we assemble a stack of alternating, uniform media, we can build lossless acoustic reflectors, and the wave penetrating into the reflector dwindles exponentially with depth as the grating losslessly converts the forward running into the reflected field. If:
$$\psi_j=\left(\begin{array}{c}a_j\\b_j\end{array}\right)$$
represents the field at the $j^{th}$ interface in the stack, then the transfer matrix across the interface from medium 1 into medium 2 is:
$$K_j=\frac{1}{1+R}\left(\begin{array}{cc}-R&1\\1&-R\end{array}\right)$$
so that the transfer matrix through a layer of thickness $t_1$ of medium 1 followed by a layer of thickness $t_2$ of medium 2 is:
$$P=\frac{1}{1-R^2}\left(\begin{array}{cc}e^{i\,k_1\,t_1}&0\\0&e^{-i\,k_1\,t_1}\end{array}\right)\left(\begin{array}{cc}-R&1\\1&-R\end{array}\right)\left(\begin{array}{cc}e^{i\,k_2\,t_2}&0\\0&e^{-i\,k_2\,t_2}\end{array}\right)\left(\begin{array}{cc}R&1\\1&R\end{array}\right)$$
which has eigenvalues:
$$\lambda = a\pm\sqrt{a^2-b^2}$$
where:
$$a = \cos (k_1 t_1-k_2 t_2)-R^2\,\cos (k_1 t_1+k_2 t_2));\qquad b = 1-R^2$$
By checking stationary points of this expression we can show that the eigenvalues both lie inside the closed unit circle and that there are choices for $k_1\,t_1\,k_2 \,t_2$ for any $R$ such that both eigenvalues lie strictly within the unit circle. Therefore, because $\psi_{2\,j} = P^j \psi_0$, we see that this means $|\psi_j|^2$ decreases exponentially with grating cycle number $j$ and is of the order of $|\lambda_{max}|^{2\,j} |\psi_0|^2$, where $\lambda_{max}$ is the maximum magnitude eigenvalue, which we can choose to have either exactly unit magnitude, or strictly less than unit magnitude. So the grating steadily converts forwards running waves into backwards running waves as the wave penetrates deeper into the reflector. This, I believe, is a pretty good analogue for the penetration and gradual conversion of wave direction that happens beyond a totally internally reflecting optical interface.
We can even chirp the grating to get essentially any frequency response of the reflexion that we like. This is what is done in one-dimensional gratings in one-moded fibre optic waveguides. Leon Paladian at the University of Sydney came up with a general method to design the band structure of the photonic crystal (grating) to achieve a given frequency response in the Mid 1990s. Check his personal page (and here) for publications in this field. I should declare a friendship with Leon, but was not involved in this particular work.
Best Answer
The energy of an element of a traveling wave is not constant. Halliday-Resnick-Krane is right. For a string of density $\mu$ and tension $T$ the kinetic energy of an element $dx$ is $$dK=\frac 12\mu dx\left(\frac{\partial \xi}{\partial t}\right)^2.$$ For the potential energy we have $$dU=Tdl,$$ where $dl$ is the stretched amount of the string. A small section $dh$ of the string is the hypotenuse of a right angle triangle with basis $dx$ and height $\frac{\partial \xi}{\partial x}dx$. Hence the amount stretched is $$dl=\sqrt{dx^2+\left(\frac{\partial \xi}{\partial x}\right)^2dx^2}-dx=\frac{dx}{2}\left(\frac{\partial \xi}{\partial x}\right)^2.$$ In the last equation we neglect higher order terms in $\left(\frac{\partial \xi}{\partial x}\right)$ since we assume small displacements. Then $$dU=\frac{Tdx}{2}\left(\frac{\partial \xi}{\partial x}\right)^2.$$ Therefore $$dE=dK+dU=\frac 12\mu \left(\frac{\partial \xi}{\partial t}\right)^2dx+\frac{T}{2}\left(\frac{\partial \xi}{\partial x}\right)^2dx.$$
For a progressive harmonic wave $\xi(x,t)=A\cos(kx-\omega t)$ we get $$dE=\frac 12\mu\omega^2 A^2\sin^2(kx-\omega t)dx+\frac{1}{2}Tk^2A^2\sin^2(kx-\omega t)dx.$$ Using $v^2=T/\mu$ and $\omega=vk$ we get $$dE=\mu\omega^2A^2\sin^2(kx-\omega t)dx,$$ which is not constant.
Remember that energy is being transmitted by the wave along the rope. The source of energy being the harmonic oscillator that generates the wave at one end of the string. So it is not a problem that the energy at each point is not constant. Another important point: The reason the result is quite different to what we expect when we think about simple harmonic motion (which gives constant energy for the particle) is that the element of the rope is not only moving transversely. A wave in a rope always have a longitudinal component. This was implicit when we computed the potential energy and assumed the stretched amount was the $dl=dh-dx$. Notice that when the element $dx$ has displacement $A$, it is at rest having vanishing kinetic energy. Moreover this element is not stretched, $\frac{\partial \xi}{\partial x}=0$, giving vanishing elastic potential energy. This does agree with the equation obtained for $dE$.