Transverse and Longitudinal Waves — Question 5

Question 5

Density follows local stretch, not displacement amplitude alone

A line with initial mass per unit length $\mu_0$ has uniform longitudinal displacement gradient $\partial u/\partial X=-.020$ over a short region. Mass is conserved.

1. Find the current mass per unit length exactly within this one-dimensional deformation model.

2. Compare with the first-order density approximation and explain the sign.

Question 5 - Solution

Density follows local stretch, not displacement amplitude alone

Method and mathematical working

Follow a material segment so its mass is fixed. Divide this mass by the deformed, rather than undeformed, length.$$\begin{aligned}dm&=\mu_0dX=\mu\,dx,\\ \frac{\mu}{\mu_0}&=\frac{1}{1+u_X}=1-u_X+u_X^2-\cdots.\\[4pt]\frac{\mu}{\mu_0}&=\frac1{1-0.020}=1.020408\ldots,\qquad \left(\frac{\mu}{\mu_0}\right)_{\mathrm{linear}}=1.020.\end{aligned}$$

(a) Calculation and conclusion

A material segment of initial length $dX$ has unchanged mass $\mu_0dX$ and current length $.980dX$. Thus $\mu=\mu_0/.980=1.020408\mu_0$.

(b) Interpretation and limits

To first order, $(1+u_X)^{-1}\approx1-u_X$, giving $\mu\approx1.020\mu_0$. A negative displacement gradient shortens the segment, so fixed mass occupies less length and density increases. A uniform displacement with zero gradient would not change density.

Physical check

The first neglected term is of order $u_X^2$. For a two-percent strain its small size explains why the linear approximation is close, but not identical, to the reciprocal result.