1608 The Doppler Effect

When a car approaches you with its horn sounding, the pitch seems to drop as the car passes. This phenomenon, first described by the 19th-century Austrian scientist Christian Doppler, is called the Doppler effect.

Moving Listener and Moving Source

Dopper effect for moving listener LL and moving source SS:

fL=v+vLv+vSfS(16.29)\begin{aligned} f_L = \frac{v + v_L}{v + v_S} f_S \quad \bold{(16.29)} \end{aligned}

fLf_L - Frequency heard by listener
fSf_S - Frequency emitted by source
vLv_L - Velocity of listener ( ++ if from LL toward SS, - if opposite)
vSv_S - Velocity of source ( ++ if from LL toward SS, - if opposite)

Examples

16.14 A police car’s siren emits a sinusoidal wave with frequency f S = 300 Hz. The speed of sound is 340340 m/s and the air is still.
(a) Find the wavelength of the waves if the siren is at rest.
(b) Find the wavelengths of the waves in front of and behind the siren if it is moving at 3030 m/s.
Graph

Solution

(a). When the source is at rest.

λ=vfS=340300=1.13 m\begin{aligned} \lambda = \frac{v}{f_S} = \frac{340}{300} = 1.13 \text{ m} \end{aligned}

b. In front of the siren and behind the siren,

λfront=vvSfS=34030300=1.03 mλbehind=v+vSfS=340+30300=1.23 m\begin{aligned} \lambda_{front} &= \frac{v-v_S}{f_S} = \frac{340-30}{300} = 1.03 \text{ m}\\ \lambda_{behind} &= \frac{v+v_S}{f_S} = \frac{340+30}{300} = 1.23 \text{ m} \end{aligned}

16.15 If a listener LL is at rest and the siren in Example 16.14 is moving away from LL at 3030 m/s, what frequency does the listener hear?
Graph

Solution

We have vL=0v_L = 0 and vS=+30v_S = +30 m/s (positive, since the velocity of the source in the direction from listenser to source)

fL=vv+vS=340340+30(300)=276 Hz\begin{aligned} f_L = \frac{v}{v+v_S} = \frac{340}{340 + 30} (300) = 276 \text{ Hz} \end{aligned}

16.16 If the siren is at rest and the listener is moving away from it at 3030 m/s, what frequency does the listener hear?
Graph

Solution

The velocity of the listener is vL=30v_L = - 30 m/s (negative, since the motion
is in the direction from source to listener).

fL=vvLv=340+(30)340(300)=274 Hz\begin{aligned} f_L = \frac{v - v_L}{v} = \frac{340 + (-30)}{340} (300) = 274 \text{ Hz} \end{aligned}

16.17 The siren is moving away from the listener with a speed of 4545 m/s relative to the air, and the listener is moving toward the siren with
a speed of 1515 m/s relative to the air. What frequency does the listener hear?
Graph

Solution

Both the source velocity vS=+45v_S = + 45 m/s and the listener’s velocity vL=+15v_L = + 15 m/s are positive because both velocities are in the direction from listener to source.

fL=v+vLv+vSfS=340+15340+45(300)=277 Hz\begin{aligned} f_L = \frac{v + v_L}{v + v_S} f_S = \frac{340 + 15}{340 + 45}(300) = 277 \text{ Hz} \end{aligned}

16.18 The police car is moving toward a warehouse at 3030 m/s. What frequency does the driver hear reflected from the warehouse?
Graph

Solution

This situation has two Doppler shifts (Fig. 16.34). In the first shift, the warehouse is the stationary “listener.” The frequency of sound reaching the warehouse, which we call fWf_W, is greater than 300 Hz because the source is approaching. In the second shift, the warehouse acts as a source of sound with frequency fWf_W , and the listener is the driver of the police car; she hears a frequency greater than fWf_W because she is approaching the source.
The frequency reaching the warehouse is

fW=vv+vSfS=340340+(30)(300)=329 Hz\begin{aligned} f_W = \frac{v}{v + v_S} f_S = \frac{340}{340 + (-30)} (300) = 329 \text{ Hz} \end{aligned}

Then the frequency heard by the driver is

fL=v+vLvfW=340+30340)(300)=358 Hz\begin{aligned} f_L = \frac{v + v_L}{v} f_W = \frac{340 + 30}{340)} (300) = 358 \text{ Hz} \end{aligned}

Exercises

44, 46, 48

16.44 In Example 16.18 (Section 16.8), suppose the police car is moving away from the warehouse at 2020 m/s. What frequency does the driver of the police car hear reflected from the warehouse?

Solution

Todo

16.46 A railroad train is traveling at 25.025.0 m/s in still air. The frequency of the note emitted by the locomotive whistle is 400400 Hz. What is the wavelength of the sound waves (a) in front of the locomotive and (b) behind the locomotive? What is the frequency of the sound heard by a stationary listener (c) in front of the locomotive and (d) behind the locomotive?

Solution

Todo

16.48 Moving Source vs. Moving Listener. (a) A sound source producing 1.00-kHz waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare to that in part (a)? Explain on physical grounds why the two answers differ.

Solution

Todo

Quiz

(10) If a sound source with a 1000HZ frequency moves away from the listener at 15m/s, and a listener moves a speed of 45m/s away from the source, what is the apparent frequence heard by the listener? (The velocity of sound is 340m/s)?

Solution

Positive direction Listener -> Source

fL=vvLv+vSfS\begin{aligned} f_L = \frac{v-v_L}{v+v_S} f_S \end{aligned}