   Chapter 14, Problem 35P

Chapter
Section
Textbook Problem

A supersonic jet traveling at Mach 3.00 at an altitude of h = 2.00 × 104 m is directly over a person at time t = 0 as shown in Figure P14.35. Assume the average speed of sound in air is 335 m/s over the path of the sound. (a) At what time will the person encounter the shock wave due to the sound emitted at t = 0? (b) Where will the plane be when this shock wave is heard?

(a)

To determine
The time when the shock wave reaches the observer.

Explanation

Given Info: velocity of the plane is Mach 3.00

Formula to calculate the half-angle of the conical wave front is,

θ=sin1(vsoundvplane)

• vsound is the velocity of the sound
• vplane is the velocity of the plane

Mach 1.00 speed of aircraft in air is equal to the speed of sound in air. So, Mach 3.00 equals 3 times the speed of sound in air vplane=3vsound .

Substitute 335m/s for vsound and 3 vsound for vplane in the above expression to get θ

θ=sin1(335m/s(3)(335m/s))=19.47°

The below figure 1 shows the position of the Jet above the observe.

The below figure 2 shows the distance covered by the jet plane when the observer hears the shock wave.

(b)

To determine
The distance of the plane when the shock is heard by the person.

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