Supersonic Speeds, Shock Waves — Question 2

Section 17-9 · Physics pages 2794–2795

Question 2 A ground boom is heard after the source passes overhead A source flies horizontally at altitude h = 1000 m and speed v = 680 m/s in air with c = 340 m/s. Use the steady weak-cone model with sin µ = c/v. A ground observer lies directly under the flight path. O S h x Source position when the cone reaches the ground observer; geometry is schematic. (a) Find the horizontal distance by which the source has passed the observer when the cone arrives. (b) Find the delay after overhead passage and compare with h/c. Explain why the two times need not agree.
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Question 2 - Solution A ground boom is heard after the source passes overhead O S h x (a) Method and calculation At cone arrival, the line from source to observer is a cone generator. Its angle below the backward horizontal direction is µ = 30◦, so tan µ = h x, x = h cot µ = h √ M2 −1 = 1732.05 m. (b) Interpretation and further calculation The overhead-to-arrival delay is tb = x v = h √ M2 −1 Mc = 2.54713 s. By comparison, h/c = 2.94118 s. The cone arrival includes disturbances emitted before the source was overhead, so it is not simply the vertical travel time of the overhead emission. Physical check The source is ahead of the observer when the boom arrives. The result assumes enough prior supersonic flight for the relevant wavefront envelope to exist.

Question 2

A ground boom is heard after the source passes overhead

A source flies horizontally at altitude h=1000mh=1000\,\mathrm m and speed v=680m/sv=680\,\mathrm{m/s} in air with c=340m/sc=340\,\mathrm{m/s}. Use the steady weak-cone model with sin⁡μ=c/v\sin\mu=c/v. A ground observer lies directly under the flight path.

Diagram shown in the original SVG worksheet.

  1. Find the horizontal distance by which the source has passed the observer when the cone arrives.

  2. Find the delay after overhead passage and compare with h/ch/c. Explain why the two times need not agree.

Question 2 - Solution

A ground boom is heard after the source passes overhead

Diagram shown in the original SVG worksheet.

(a) Method and calculation

At cone arrival, the line from source to observer is a cone generator. Its angle below the backward horizontal direction is μ=30∘\mu=30^\circ, sotan⁡μ=hx,x=hcot⁡μ=hM2−1=1732.05m.\tan\mu=\frac hx,\qquad x=h\cot\mu=h\sqrt{M^2-1}=1732.05\,\mathrm m.

(b) Interpretation and further calculation

The overhead-to-arrival delay istb=xv=hM2−1Mc=2.54713s.t_b=\frac xv=\frac{h\sqrt{M^2-1}}{Mc}=2.54713\,\mathrm s.By comparison, h/c=2.94118sh/c=2.94118\,\mathrm s. The cone arrival includes disturbances emitted before the source was overhead, so it is not simply the vertical travel time of the overhead emission.

Physical check

The source is ahead of the observer when the boom arrives. The result assumes enough prior supersonic flight for the relevant wavefront envelope to exist.