Supersonic Speeds, Shock Waves — Question 5

Section 17-9 · Physics pages 2800–2801

Question 5 A warmer medium widens a cone at fixed source speed A source maintains speed v = 680 m/s. The medium changes uniformly from sound speed c1 = 340 to c2 = 374 m/s. Compare the corresponding steady weak Mach cones. S µ v Cross-section of the cone tangent to an earlier spherical wavefront; the source moves right. (a) Calculate Mach number and cone angle in both states. (b) Derive the angle sensitivity dµ/dc at fixed v and explain why the angle becomes especially sensitive near the sonic threshold.
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Question 5 - Solution A warmer medium widens a cone at fixed source speed S µ v (a) Method and calculation The Mach numbers are M1 = 2.00, M2 = 680/374 = 1.81818. µ1 = sin−1(0.50) = 30◦, µ2 = sin−1(0.55) = 33.367◦. The faster sound fronts make the cone wider even though the source speed is unchanged. (b) Interpretation and further calculation Differentiate sin µ = c/v at fixed v: cos µ dµ dc = 1 v, dµ dc = 1 v √ 1 −(c/v)2 . The derivative is in radians per unit speed and grows without bound as c →v−in the ideal geometric model. If c > v, the source is subsonic and no steady Mach cone of this type exists. Physical check Mach number is a speed ratio relative to the medium; constant source speed does not imply constant Mach number when the medium changes.

Question 5

A warmer medium widens a cone at fixed source speed

A source maintains speed v=680m/sv=680\,\mathrm{m/s}. The medium changes uniformly from sound speed c1=340c_1=340 to c2=374m/sc_2=374\,\mathrm{m/s}. Compare the corresponding steady weak Mach cones.

Diagram shown in the original SVG worksheet.

  1. Calculate Mach number and cone angle in both states.

  2. Derive the angle sensitivity dμ/dcd\mu/dc at fixed vv and explain why the angle becomes especially sensitive near the sonic threshold.

Question 5 - Solution

A warmer medium widens a cone at fixed source speed

Diagram shown in the original SVG worksheet.

(a) Method and calculation

The Mach numbers areM1=2.00,M2=680/374=1.81818.M_1=2.00,\qquad M_2=680/374=1.81818.μ1=sin⁡−1(0.50)=30∘,μ2=sin⁡−1(0.55)=33.367∘.\mu_1=\sin^{-1}(0.50)=30^\circ,\qquad\mu_2=\sin^{-1}(0.55)=33.367^\circ.The faster sound fronts make the cone wider even though the source speed is unchanged.

(b) Interpretation and further calculation

Differentiate sin⁡μ=c/v\sin\mu=c/v at fixed vv:cos⁡μdμdc=1v,dμdc=1v1−(c/v)2.\cos\mu\,\frac{d\mu}{dc}=\frac1v,\qquad\frac{d\mu}{dc}=\frac1{v\sqrt{1-(c/v)^2}}.The derivative is in radians per unit speed and grows without bound as c→v−c\to v^- in the ideal geometric model. If c>vc>v, the source is subsonic and no steady Mach cone of this type exists.

Physical check

Mach number is a speed ratio relative to the medium; constant source speed does not imply constant Mach number when the medium changes.