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Doppler Problem

Posted by Ross Gualco
Grade level: nonaligned
School: No school given.
City: Woodacre State/Province: CA
Country: U.S.A.
Area of science: Physics


Here's a Doppler problem I've been pondering. How fast must an object be travelling so the sound I perceive as it directly passes me will drop in pitch exactly one octave? Thanks in advance. Ross.

When I first looked at this question, I thought it would be easy. Unfortunately, sometimes physics can through us for a loop! This explanation is kind of long, and involves a bit of algebra. Hold on tight!

We need to know two things to solve this problem: 1) How (quantitatively) does the Doppler shift work, and 2) What does a one octave change in pitch mean physically?

Pretty much everyone is familiar with the Doppler shift. When an object is moving past another object, any sound emitted by the first object is changed in pitch due to the relative motion. Imagine an object sitting still a few feet away from you. Any sound it emits is really a pulse of waves, like waves hitting a beach. The distance between the waves determines the pitch you hear: the longer the distance, the lower the pitch. If the waves are emitted at a steady rate, the sound you hear has a steady pitch. If the object moves toward you, the distance between waves decreases, and the pitch goes up: the object has moved a bit toward you since the last wave was emitted, and so the wavelength decreases. The opposite is true if it moves away from you; the waves get stretched out, and the pitch drops.

There is a simple equation that can be used to calculate the Doppler shift when the velocity of the object is small. Unfortunately, to drop a whole octave the source must be moving pretty quickly. That means you have to use a more complicated formula. As it turns out, the only formula I could find in textbooks was the Doppler shift for light, which is also a wave. I see no reason why we cannot use this equation for sound, substituting the speed of sound for the speed of light. Here is the equation:

   wl(heard) - wl(emitted)
   -----------------------  + 1 =
                    /------     /   /--------
                   /     v     /   /     v
                  /  1 + -    /   /  1 - -
                \/       c   /  \/       c

"wl(emitted)" is the wavelength of sound emitted by the object at rest, "wl(heard)" is the wavelength you hear from the object, "v" is the velocity of the object, and "c" is the speed of sound.

Now we're almost ready to answer your question, but we have to look at number (2) now: what wavelength change corresponds to a one octave pitch change? It turns out that this is simply a factor of two! For example, middle A on a piano has a frequency of 440 Hertz (or 440 cycles per second). To raise it an octave, the frequency would have to go to 880 Hertz, and to drop an octave the frequency must drop to 220 Hertz. Now, wavelength and frequency are closely related: a drop in two in frequency means an increase in wavelength by the same amount. So a drop of pitch of one octave is exactly a factor of two in wavelength. Let's plug-and-chug!

If the wavelength heard is twice the emitted wavelength, we can replace "wl(heard)" in the equation with 2 x wl(emitted). You have to do some algebra to solve this, which is tough to show on a web page! However, when you finish, the solution says that the velocity of the object must be 0.6 times the speed of sound away from you. The speed of sound is about 1100 km/hour (700 miles per hour), so the object must really be moving! It is receding at about 700 km/hr (400 mph).

At first I though this must be wrong. It seems too fast! But then I thought about race cars. They move at a pretty good clip; maybe 350 km/hr. You hear the engine whine as it heads towards you, then VRRROOOOMMMM, it passes you and the pitch drops dramatically. It heads toward you at 350 km/hr, then away at 350 km/hr, which is a total change of speed of 700 km/hr, or pretty much the speed we calculated. Sure enough, when I think about it, the pitch change is about an octave, so I think this calculation is accurate!

©2008 Phil Plait. All Rights Reserved.

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