A police car is parked by the side of the highway, sounding its 1000 Hz siren. If you are also parked by the highway, you will hear that same frequency. However, if there is relative motion between you and the police car, either toward or away from each other, you will hear a different frequency. For example, if you are driving toward the police car at 120 km/h (about 75 mi/h), you will hear a higher frequency (1096 Hz, an increase of 96 Hz). If you are driving away from the police car at that same speed, you will hear a lower frequency (904 Hz, a decrease of 96 Hz).
These motion-related frequency changes are examples of the Doppler effect. The effect was proposed (although not fully worked out) in 1842 by Austrian physicist Johann Christian Doppler. It was tested experimentally in 1845 by Buys Ballot in Holland, “using a locomotive drawing an open car with several trumpeters.”
The Doppler effect holds not only for sound waves but also for electromagnetic waves, including microwaves, radio waves, and visible light. Here, however, we shall consider only sound waves, and we shall take as a reference frame the body of air through which these waves travel. This means that we shall measure the speeds of a source S of sound waves and a detector D of those waves relative to that body of air. (Unless otherwise stated, the body of air is stationary relative to the ground, so the speeds can also be measured relative to the ground.) We shall assume that S and D move either directly toward or directly away from each other, at speeds less than the speed of sound.
If either the detector or the source is moving, or both are moving, the emitted frequency f and the detected frequency f′ are related by
where ν is the speed of sound through the air, νD is the detector’s speed relative to the air, and νs is the source’s speed relative to the air. The choice of plus or minus signs is set by this rule:
When the motion of detector or source is toward the other, the sign on its speed must give an upward shift in frequency. When the motion of detector or source is away from the other, the sign on its speed must give a downward shift in frequency.
In short, toward means shift up, and away means shift down.
Here are some examples of the rule. If the detector moves toward the source, use the plus sign in the numerator of Eq. 17-47 to get a shift up in the frequency. If it moves away, use the minus sign in the numerator to get a shift down. If it is stationary, substitute 0 for νD. If the source moves toward the detector, use the minus sign in the denominator of Eq. 17-47 to get a shift up in the frequency. If it moves away, use the plus sign in the denominator to get a shift down. If the source is stationary, substitute 0 for νs.
Next, we derive equations for the Doppler effect for the following two specific situations and then derive Eq. 17-47 for the general situation.
1. When the detector moves relative to the air and the source is stationary relative to the air, the motion changes the frequency at which the detector intercepts wavefronts and thus changes the detected frequency of the sound wave.
2. When the source moves relative to the air and the detector is stationary relative to the air, the motion changes the wavelength of the sound wave and thus changes the detected frequency (recall that frequency is related to wavelength).
Detector Moving, Source Stationary
In Fig. 17-19, a detector D (represented by an ear) is moving at speed νD toward a stationary source S that emits spherical wavefronts, of wavelength λ and frequency f, moving at the speed ν of sound in air. The wavefronts are drawn one wavelength apart. The frequency detected by detector D is the rate at which D intercepts wavefronts (or individual wavelengths). If D were stationary, that rate would be f, but since D is moving into the wavefronts, the rate of interception is greater, and thus the detected frequency f′ is greater than f.
Fig. 17-19 A stationary source of sound S emits spherical wavefronts, shown one wavelength apart, that expand outward at speed ν. A sound detector D, represented by an ear, moves with velocity 
toward the source. The detector senses a higher frequency because of its motion.
Fig. 17-20 The wavefronts of Fig. 17-19, assumed planar, (a) reach and (b) pass a stationary detector D; they move a distance νt to the right in time t.
Let us for the moment consider the situation in which D is stationary (Fig. 17-20). In time t, the wavefronts move to the right a distance νt. The number of wavelengths in that distance νt is the number of wavelengths intercepted by D in time t, and that number is νt/λ. The rate at which D intercepts wavelengths, which is the frequency f detected by D, is
In this situation, with D stationary, there is no Doppler effect—the frequency detected by D is the frequency emitted by S.
Now let us again consider the situation in which D moves in the direction opposite the wavefront velocity (Fig. 17-21). In time t, the wavefronts move to the right a distance νt as previously, but now D moves to the left a distance νDt. Thus, in this time t, the distance moved by the wavefronts relative to D is νt + νDt. The number of wavelengths in this relative distance νt + vDt is the number of wavelengths intercepted by D in time t and is (νt + vDt)/λ. The rate at which D intercepts wavelengths in this situation is the frequency f′, given by
From Eq. 17-48, we have λ = ν/f. Then Eq. 17-49 becomes
Note that in Eq. 17-50, f′ must be greater than f unless νD = 0 (the detector is stationary).
Similarly, we can find the frequency detected by D if D moves away from the source. In this situation, the wavefronts move a distance νt − νDt relative to D in time t, and f′ is given by
In Eq. 17-51, f′ must be less than f unless νD = 0.
We can summarize Eqs. 17-50 and 17-51 with
Source Moving, Detector Stationary
Let detector D be stationary with respect to the body of air, and let source S move toward D at speed νs (Fig. 17-22). The motion of S changes the wavelength of the sound waves it emits and thus the frequency detected by D.
To see this change, let T (= 1/f) be the time between the emission of any pair of successive wavefronts W1 and W2. During T, wavefront W1 moves a distance νT and the source moves a distance νsT. At the end of T, wavefront W2 is emitted. In the direction in which S moves, the distance between W1 and W2, which is the wavelength λ′ of the waves moving in that direction, is νT − νST. If D detects those waves, it detects frequency f′ given by
Note that f′ must be greater than f unless νS = 0.
Fig. 17-21 Wavefronts traveling to the right (a) reach and (b) pass detector D, which moves in the opposite direction. In time t, the wavefronts move a distance νt to the right and D moves a distance νDt to the left.
Fig. 17-22 A detector D is stationary, and a source S is moving toward it at speed νs. Wavefront W1 was emitted when the source was at S1, wavefront W7 when it was at S7. At the moment depicted, the source is at S. The detector senses a higher frequency because the moving source, chasing its own wavefronts, emits a reduced wavelength λ′ in the direction of its motion.
In the direction opposite that taken by S, the wavelength λ′ of the waves is νT + νsT. If D detects those waves, it detects frequency f′ given by
Now f′ must be less than f unless νs = 0.
We can summarize Eqs. 17-53 and 17-54 with
General Doppler Effect Equation
We can now derive the general Doppler effect equation by replacing f in Eq. 17-55 (the source frequency) with f′ of Eq. 17-52 (the frequency associated with motion of the detector). The result is Eq. 17-47 for the general Doppler effect.
That general equation holds not only when both detector and source are moving but also in the two specific situations we just discussed. For the situation in which the detector is moving and the source is stationary, substitution of νS = 0 into Eq. 17-47 gives us Eq. 17-52, which we previously found. For the situation in which the source is moving and the detector is stationary, substitution of νD = 0 into Eq. 17-47 gives us Eq. 17-55, which we previously found. Thus, Eq. 17-47 is the equation to remember.
CHECKPOINT 5 The figure indicates the directions of motion of a sound source and a detector for six situations in stationary air. For each situation, is the detected frequency greater than or less than the emitted frequency, or can’t we tell without more information about the actual speeds?
A rocket moves at a speed of 242 m/s directly toward a stationary pole (through stationary air) while emitting sound waves at frequency f = 1250 Hz.
(a) What frequency f′ is measured by a detector that is attached to the pole?
Solution: We can find f′ with Eq. 17-47 for the general Doppler effect. The Key Idea here is that, because the sound source (the rocket) moves through the air toward the stationary detector on the pole, we need to choose the sign on νS that gives a shift up in the frequency of the sound. Thus, in Eq. 17-47 we use the minus sign in the denominator. We then substitute 0 for the detector speed νD, 242 m/s for the source speed νS, 343 m/s for the speed of sound ν (from Table 17-1), and 1250 Hz for the emitted frequency f. We find
which, indeed, is a greater frequency than the emitted frequency.
(b) Some of the sound reaching the pole reflects back to the rocket as an echo. What frequency f″ does a detector on the rocket detect for the echo?
Solution: Two Key Ideas here are the following:
1. The pole is now the source of sound (because it is the source of the echo), and the rocket’s detector is now the detector (because it detects the echo).
2. The frequency of the sound emitted by the source (the pole) is equal to f′ from part (a), the frequency of the sound the pole intercepts and reflects.
We can rewrite Eq. 17-47 in terms of the source frequency f′ and the detected frequency f″ as
A third Key Idea here is that, because the detector (on the rocket) moves through the air toward the stationary source, we need to use the sign on νD that gives a shift up in the frequency of the sound. Thus, we use the plus sign in the numerator of Eq. 17-56. Also, we substitute νD = 242 m/s, νs = 0, ν = 343 m/s, and f′ = 4245 Hz. We find
which, indeed, is greater than the frequency of the sound reflected by the pole.
CHECKPOINT 6 If the air in this sample problem is moving toward the pole at 20 m/s, (a) what value for vS should be used in the solution of part (a), and (b) what value for νD should be used in the solution of part (b)?
Leave a Reply