From the journal Nature to your problem set: Carbon nanotubes

Prof. McEuen in our Department of Physics studies the vibrations of tiny tubes (with nanometer diameter!) made entirely of carbon. The propagation of transverse waves along such tubes is described by the modified wave equation

$$\mu \frac{\partial^2 y}{\partial t^2} -\tau \frac{\partial^2 y}{\partial x^2} +F \frac{\partial^4 y}{\partial x^4} = 0$$ (2)

where $\mu$ and $\tau$ (as in lecture) are the the linear mass density and tension respectively, and $F$ is an elastic parameter independent of the tension and characteristic of the nanotube.
Note: This modified equation also describes wires with stiffness (such as steel piano wires) rather than simple strings, something about which an astute student in the morning section asked.

(a)

Show that the standing wave $y(x,t) = A \cos(\omega t)\cos(kx)$ is a solution to the nanotube wave equation (2), and derive the dispersion relation $\omega = f(k)$.

(b)

Typical values for these tubes are $\mu=3 \times 10^{-15}$ kg/m, $\tau=0.5 \times 10^{-9}$ N, and $F=4\times10^{-26}$ N$\cdot$m. Prof. McEuen's tubes are typically 1 $\mu$m=$10^{-6}$ m in length. Assuming a wavelength of 2 $\mu$m for the fundamental (lowest frequency) mode, how important is the correction term $F$? To answer this, compute the ratio $\omega/(c k)$ where $c$ is the speed you would expect for a normal string; i.e., if the new, $F$ term were not there. Given your results, for work good to a few percent, should Prof. McEuen consider the $F$ term for the first few modes of his tubes?

(c)

Below what wavelength (in Angstroms, 1 Å=$10^{-10}$m) does the new $F$-term change the frequency significantly? Specifically, determine the wavelength $\lambda$ such that the frequency is twice as high as you would expect from the usual dispersion relation $\omega = c k$.