Supersonic Speeds, Shock Waves — Question 6

Section 17-9 · Physics pages 2802–2803

Question 6 Finite emission history can remove the nominal cone arrival A source moves along (x, y) = (vte, h), with v = 680 m/s, h = 1000 m and c = 340 m/s. A ground observer is at the origin. A signal emitted at te arrives at tr = te + √ v2t2e + h2/c. O S h x Source position when the cone reaches the ground observer; geometry is schematic. (a) Find the emission time that minimizes arrival time when past emission is unrestricted. (b) If the source begins emitting only at overhead passage te = 0, find the first arrival. Explain the failure of the earlier steady-cone prediction in that case.
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Question 6 - Solution Finite emission history can remove the nominal cone arrival O S h x (a) Method and calculation Differentiate the arrival map: dtr dte = 1 + v2te c √ v2t2e + h2 . The minimum requires te < 0 and gives t∗ e = − h v √ M2 −1 = −0.84904 s. Substitution gives the steady-cone arrival t∗ r = h √ M2 −1/v = 2.54713 s. (b) Interpretation and further calculation If emission begins at te = 0, the minimizer above is unavailable. For all allowed te ≥0, the derivative is positive; the earliest arrival comes from the initial pulse: tr,first = h/c = 2.94118 s. The nominal cone arrival requires a disturbance emitted about 0.849 s before overhead passage. Geometry cannot supply a wavefront that was never emitted. Physical check This is a constraint on source history, not a modification of local sound speed. The unconstrained minimum is valid only if emission began early enough.

Question 6

Finite emission history can remove the nominal cone arrival

A source moves along (x,y)=(vte,h)(x,y)=(vt_e,h), with v=680m/sv=680\,\mathrm{m/s}, h=1000mh=1000\,\mathrm m and c=340m/sc=340\,\mathrm{m/s}. A ground observer is at the origin. A signal emitted at tet_e arrives at tr=te+v2te2+h2/ct_r=t_e+\sqrt{v^2t_e^2+h^2}/c.

Diagram shown in the original SVG worksheet.

  1. Find the emission time that minimizes arrival time when past emission is unrestricted.

  2. If the source begins emitting only at overhead passage te=0t_e=0, find the first arrival. Explain the failure of the earlier steady-cone prediction in that case.

Question 6 - Solution

Finite emission history can remove the nominal cone arrival

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Differentiate the arrival map:dtrdte=1+v2tecv2te2+h2.\frac{dt_r}{dt_e}=1+\frac{v^2t_e}{c\sqrt{v^2t_e^2+h^2}}.The minimum requires te<0t_e<0 and giveste*=−hvM2−1=−0.84904s.t_e^*=-\frac{h}{v\sqrt{M^2-1}}=-0.84904\,\mathrm s.Substitution gives the steady-cone arrival tr*=hM2−1/v=2.54713st_r^*=h\sqrt{M^2-1}/v=2.54713\,\mathrm s.

(b) Interpretation and further calculation

If emission begins at te=0t_e=0, the minimizer above is unavailable. For all allowed te≥0t_e\ge0, the derivative is positive; the earliest arrival comes from the initial pulse:tr,first=h/c=2.94118s.t_{r,\mathrm{first}}=h/c=2.94118\,\mathrm s.The nominal cone arrival requires a disturbance emitted about 0.849s0.849\,\mathrm s before overhead passage. Geometry cannot supply a wavefront that was never emitted.

Physical check

This is a constraint on source history, not a modification of local sound speed. The unconstrained minimum is valid only if emission began early enough.