Transverse and Longitudinal Waves — Question 4

Question 4

Compression depends on the spatial derivative of displacement

A uniform line has small longitudinal displacement $u(X)=A\sin kX$ at an instant, with $Ak\ll1$. Neighboring material labels are separated by $dX$.

1. Derive their current separation to first order and locate maximum compression.

2. At a positive displacement maximum, is the compression also maximal? Explain.

Question 4 - Solution

Compression depends on the spatial derivative of displacement

Method and mathematical working

Compare a pair of neighboring material labels, expand their displacement difference to first order, and read compression from the sign of the resulting strain.$$\begin{aligned}u(X+dX)-u(X)&\simeq u_XdX,\\ dx-dX&=u_XdX.\\[4pt]\frac{\Delta x-\Delta X}{\Delta X}&\simeq u_X=Ak\cos kX,\\\text{maximum compression: }\cos kX&=-1\Rightarrow kX=(2m+1)\pi.\end{aligned}$$

(a) Calculation and conclusion

The separation is $dx=[1+\partial u/\partial X]dX=[1+Ak\cos kX]dX$. Compression is greatest where $\cos kX=-1$, or $kX=(2n+1)\pi$. There $u=0$ and the displacement slope is most negative.

(b) Interpretation and limits

At $kX=\pi/2+2n\pi$, displacement is maximal but its slope is zero. Neighboring elements have nearly equal displacements and their first-order separation is unchanged. Absolute displacement and local strain measure different things.

Physical check

At a displacement extremum the slope is zero. This means adjacent labels move almost together, so the largest displacement is not the point of strongest first-order compression.