Supersonic Speeds, Shock Waves — Question 8

Question 8

A finite-strength front need not make the Mach angle

A straight front is attached to a source moving at $v=800\,\mathrm{m/s}$ through a still medium with small-disturbance speed $c=320\,\mathrm{m/s}$. Its front angle to the backward flight axis is $\beta$. The front propagates normally into undisturbed material at speed $D=400\,\mathrm{m/s}$; geometry gives $v\sin\beta=D$.

1. Find the small-disturbance Mach angle and the actual front angle under the supplied model.

2. Explain why substituting $1/M$ for $\sin\beta$ would be wrong here. What condition recovers the Mach-angle result?

Question 8 - Solution

A finite-strength front need not make the Mach angle

(a) Method and calculation

The source Mach number is $M=v/c=2.5$, giving$$\mu=\sin^{-1}(c/v)=\sin^{-1}(0.4)=23.578^\circ.$$The actual normal front speed instead gives$$\beta=\sin^{-1}(D/v)=\sin^{-1}(0.5)=30^\circ.$$

(b) Interpretation and further calculation

The angle relation involves the speed at which the front advances normal to itself. Here $D>c$, so $\beta>\mu$. Replacing $D$ by $c$ would assume the front is a weak acoustic disturbance. The limit $D\to c$ recovers $\beta\to\mu$; determining $D$ from front strength would require additional material and conservation relations.

Physical check

The given model requires $D\le v$ for a real attached-front angle. It does not equate small-signal sound speed with finite-strength front speed.