The Doppler effect is the shift in frequency of a wave that occurs when the wave source, or the detector of the wave, is moving. Applications of the Doppler effect range from medical tests using ultrasound to radar detectors and astronomy (with electromagnetic waves).

We've all observed the Doppler effect with our ears. A good example is when an emergency vehicle passes with its siren going. When the vehicle comes toward you the siren frequency is higher than usual. The frequency shifts abruptly to lower than usual when the vehicle passes you and moves away. A similar effect occurs if the sound source is stationary and you move toward it or away from it.

We will focus on sound waves in describing the Doppler effect, but it works for other waves too.

The wave speed = 1.0

Consider a stationary source of sound broadcasting a single frequency sound wave. You are the observer of the sound wave, and you are also stationary. The usual relationship between frequency, speed, and wavelength is:

f = v/λ

v represents the speed of sound through the medium.

Let's say you, the observer, now move toward the source with velocity v_{O}. You encounter more waves per unit time than you did before. Relative to you, the waves travel at a higher speed:

v^{/} = v + v_{O}

The frequency of the waves you detect is higher, and is given by:

f^{/} = v^{/}/λ = (v + v_{O})
/λ

If you moved away from the source the observed frequency is lower. In general the observed frequency when the observer moves is:

f^{/} = (v +/- v_{O})
/λ = (v +/- v_{O})* f/v
= f * (v +/- v_{O})/v

Use the first sign (+) when the observer moves toward the source and the second sign (-) when the observer moves away.

What happens when the source of the waves moves toward you, a stationary observer? Again, you encounter more waves per unit time than you did before so the frequency is shifted up. This time, though, the shift occurs because the wavelength has been lowered by the movement of the source.

When nothing moves the wavelength is equal to vT, where T is the period, or v/f, because T = 1/f. When the source moves at speed v_{s}, the wavelength is different by the distance traveled by the source in one period:

Change in wavelength = v_{s}T = v_{s}/f

The effective wavelength is λ
^{/} = v/f -/+ v_{s}/f = [v -/+ v_{s}]/f

Use the first sign (-) when the source moves toward the observer, and the second sign (+) when it moves away.

The detected frequency is:

f^{/} = v/λ
^{/} = f v/[v -/+ v_{s} ]

In some situations both the source and the observer move. We can write out a general Doppler equation for the observed frequency by simply combining the previous results.

The general equation accounting for any motion is:

f^{/} = f (v +/- v_{O}) / (v -/+ v_{s} )

For both sets of signs use the first sign when the motion is toward the other thing, and the second sign when the motion is away.

A particular bat emits ultrasonic waves with a frequency of 56.00 kHz. The bat is traveling at 20.00 m/s toward a moth, which is flying away from the bat at 8.00 m/s. The speed of sound is 340.0 m/s.

(a) Assuming the moth could detect the waves, what frequency waves would it observe?

(b) The waves reflect off the moth and are detected by the bat. What frequency are the waves detected by the bat?

Part (a). We use the general Doppler equation:

f^{/} = f (v +/- v_{O}) / (v -/+ v_{s} )

where f = 56.0 kHz and v = 340 m/s

v_{O} = 8.00 m/s (use the minus sign - moving away)

v_{s} = 20.0 m/s (use the minus sign - moving toward)

Combining this gives:

f^{/} = 56.0 kHz (340 - 8.00) / (340 - 20.0)

f^{/} = 58.1 kHz

**Part (b).** Once again we use the general Doppler equation, but this time the bat is the observer and the moth acts as the source.

f^{//} = f^{/} (v +/- v_{O}) / (v -/+ v_{s} )

where f^{/} = 58.1 kHz and v = 340 m/s

v_{O} = 20.0 m/s (use the plus sign - moving toward)

v_{s} = 8.00 m/s (use the plus sign - moving away)

These give:

f^{//} = 58.1 kHz (340 + 20.0) / (340 + 8.00)

f^{//} = 60.1 kHz

The bat detects a 60.1 kHz wave, which it could use to figure out how fast the moth is flying.

Sonic booms occur when the source travels at, or faster than, the speed of sound. Why is this?

Assuming a stationary observer and a source moving at the speed of sound, the Doppler equation predicts an infinite frequency. Why is this?

If the source is traveling at the speed of sound, the waves pile up and move along with the source. All the peaks are at the same place, so the wavelength is zero and the frequency is infinite. This overlay of all the waves produces a shock wave known as a sonic boom.

When the source moves faster than the wave speed the source outruns the wave. The equation can give negative frequency values, but -500 Hz is pretty much the same as +500 Hz as far as an observer is concerned.

Now the waves pile up along a particular angle, again producing a shock wave (sonic boom). The angle at which the shock wave moves away from the path of the source depends on the speed of the source relative to the speed of sound.

Does the pattern look familiar? It looks a lot like the wake left behind by a boat moving through water. What does this tell you about how a boat's speed compares to the speed of water waves?