1
$\begingroup$

A bending wave in a metal bar or string is simply called a transverse wave because the macroscopic oscillation is transverse. However, paradoxically, in a frame co-moving with the atoms, the atoms are oscillating longitudinally. The outer boundary of a curved section is being stretched, while the inner boundary is being compressed, somewhat like a bimetal.

Is the longitudinal motion in the co-moving frame really irrelevant to the term 'transverse wave'? Or is there a less paradoxical terminology?

$\endgroup$
  • $\begingroup$ There’s also transverse motion that leads to the stretching. For all reasonably-strong materials, isn’t that larger? $\endgroup$ – Bob Jacobsen Jan 8 at 15:26
0
$\begingroup$

We normally distinguish between a shear wave and a compression wave.

A bending wave of the type you describe is a pure shear wave in the limit of zero amplitude, but as you say it also involves longitudinal motion when the amplitude of wave is large. However in many (most?) cases the amplitudes aren't large enough for this to be an issue and the interpretation of the wave is straightforward.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Wouldn't a wave in a metal string be shear free, at least a standing wave? For example in a node of the standing wave, in the co-moving frame, the atom lattice rotates in accordance with the local direction of the string. My intuition says that rotation prevents shear, although I admit it isn't proper mathematics. $\endgroup$ – jkien Sep 26 '17 at 16:59
0
$\begingroup$

I don't think that it will be a transverse wave. A wave is considered transverse if the displacement vector is orthogonal to the wave vector. If we consider the vertical ($z$) direction to be the direction for deflection and the horizontal ($x$) direction the direction for extension, we have the following for Euler-Bernoulli Beams.

\begin{align} &u_x = u_0 - x\frac{d w}{d x}\, ,\\ &u_z = w(x)\, , \end{align}

where $u_0$ is the extension of the mid-axis.

So, we can see that the displacement in the horizontal direction depends on the deflection, and the displacement won't be completely transverse.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.