Traveling Sound Waves — Question 6

Question 6

A pulse accelerates material according to its pressure slope

At $t=0$ a rightward pressure pulse is $p(x,0)=P\exp(-x^2/\ell^2)$, with $P=2.0\,\mathrm{Pa}$, $\ell=0.50\,\mathrm m$, $\rho_0=1.25\,\mathrm{kg/m^3}$ and $c=320\,\mathrm{m/s}$. Use $u=p/(\rho_0c)$ and $u_t=-p_x/\rho_0$.

1. Find particle velocity and acceleration at the pulse center.

2. Find where the acceleration magnitude is greatest and evaluate it. State the acceleration signs on both sides of the center.

Question 6 - Solution

A pulse accelerates material according to its pressure slope

(a) Method and calculation

At the pressure maximum,$$u(0,0)=\frac{2}{1.25(320)}=0.0050\,\mathrm{m/s},\quad p_x(0,0)=0,\quad u_t(0,0)=0.$$The material is moving there but its instantaneous acceleration vanishes.

(b) Interpretation and further calculation

Differentiation yields$$u_t(x,0)=\frac{2Px}{\rho_0\ell^2}e^{-x^2/\ell^2}.$$For $x>0$, maximize $x e^{-x^2/\ell^2}$; its derivative vanishes when $1-2x^2/\ell^2=0$. Thus$$|x|=\ell/\sqrt2=0.3536\,\mathrm m,\quad |u_t|_{\max}=\frac{\sqrt2P}{\rho_0\ell}e^{-1/2}=2.745\,\mathrm{m/s^2}.$$Acceleration is positive ahead ($x>0$) and negative behind ($x<0$).

Physical check

The center has maximum velocity but zero acceleration, just as a smooth velocity pulse requires. Units follow from pressure gradient divided by density.