Illustration 18.5: The Location of a Supersonic Airplane

Please wait for the animation to completely load.

If an airplane is flying faster than the speed of sound, it will produce a shock wave called a sonic boom.  In the animation, consider an airplane that is flying from point A to point B.  A listener, the ear, is located at point C.  We consider how the airplane's speed and the position of the listener affect when the sound from the airplane's engines is heard by the listener.

In the animation, you can change the airplane's speed, which we call v.  The speed of sound is fixed at 343 m/s and we will represent it as v_s (shown as v_s in the animation).  Press "play" to begin the animation.  In addition,

• You can drag the ear across the screen to change its location.

• The program draws sound wave paths to the listener.

• The animation pauses when the sonic boom arrives at the listener; the animation can be resumed by clicking the right mouse button.

• The color of the paths of the sound waves changes to blue when those sound waves reach the listener.  The order in which the sound from different paths arrives at the listener is shown as numbers located at the point that the sound was produced.

• Press "reset" for default values.

Consider a sound generated by the airplane when it is at some point A traveling toward some point B along the straight path AB.  The listener hears the sound as the airplane flies toward point B (AB > AC).  DC is the path of a subsequent sound generated at some point D and traveling to the listener at point C. 

Consider a few time intervals (Δt = |Δx|/v).  The time it takes sound to travel from A to C is AC/v_s, the time it takes the plane to move from A to D is AD/v, and the time it takes sound to travel from D to C is DC/v_s.

Now, how does the time interval AC/v_s compare to the time interval AD/v + DC/v_s?  In other words, which event happens first: the sound emitted at A reaches C or the sound emitted at D reaches C? 

First consider an airplane traveling at less than or equal to the speed of sound.  AD/v + DC/v_s > AC/v_s because the path ADC is longer than the path AC.  The best you can do is when the time interval for AD is the smallest it can be, which is when v = v_s.  In this case comparing the two time intervals is equivalent to comparing the two paths.  Clearly, ADC > AC.  When v << v_s, the situation is worse and the time interval for the path ADC is even longer.  Therefore, you would hear the sound from the airplane when it was at A before you heard it from when it was at point D.

Now consider what you will hear if a supersonic airplane flies over you (v > v_s).  Again, what you hear is dependent on whether AD/v + DC/v_s is greater than, less than, or the same as AC/v_s.  If v is large enough, the extra path difference, AD, accounts for a smaller and smaller time interval, and since DC < AC we may hear the sound emitted at D before hearing the sound emitted at A.  Try it in the applet above.  Set v and move the ear around.  Notice when you "hear" the sounds from the airplane by looking at the numbers that show the ordering of the events.

Illustration authored by Fu-Kwun Hwang and Mario Belloni.
Applet authored by Fu-Kwun Hwang, National Taiwan Normal University.
© 2004 by Prentice-Hall, Inc. A Pearson Company