A collection of Doppler effect problems

Here is a set of four problems on the Doppler shift of sound waves. The solutions to the problems are given at the end. If you need a refresher on the Doppler effect, the following note might be useful: How does Doppler effect work?

Question 1

While standing at a crosswalk, you hear a frequency of 560 Hz from an approaching police car. After the police car passes and is moving away from you, you hear a frequency of 480 Hz. What is the speed of the police car (take speed of sound to be v=343 m/s)?

  1. 13.1 m/s
  2. 21.1 m/s
  3. 28.8 m/s
  4. 26.4 m/s

Question 2

The Doppler shift of ultrasonic waves is used to measure the speed of blood in an artery. The velocity of sound in the blood is 1.5 m/s. If the frequency of the stationary source is 100 Hz and the reflected sound is Doppler shifted to a frequency of 150 Hz, what is the speed of the blood flow?

  1. 0.3 m/s
  2. 0.7 m/s
  3. 1.5 m/s
  4. 2.2 m/s
  5. 3.3 m/s

Question 3

A bat flying toward a wall at 5.0 m/s emits a chirp of frequency 40 kHz. If this sound pulse is reflected, what is the frequency of the echo received by the bat? (Assume the speed of sound in air is 330 m/s.)

  1. 40.2 kHz
  2. 40.6 kHz
  3. 41.2 kHz
  4. 41.8 kHz
  5. 42.0 kHz

Question 4

A bat, moving at 5 m/s, is flying toward an insect. If the bat emits a 40.0 kHz chirp and receives back an echo at 40.4 kHz, at what speed is the insect moving (with respect to the ground)? Take the speed of sound in air to be v = 343 m/s.

  1. 3.29 m/s toward the bat
  2. 3.29 m/s away from the bat
  3. 1.62 m/s toward the bat
  4. 1.62 m/s away from the bat

Question 5

On a windy day, a 1200 Hz warning siren on the town hall sounds. The wind is blowing at 55 m/s in a direction from the siren toward a person standing 1 km away. With what frequency does the sound wave reach the person? (The speed of sound in air is 330 m/s.)

  1. 1440 Hz
  2. 1000 Hz
  3. 1030 Hz
  4. 1200 Hz
  5. 1400 Hz


Question 1: When the police car is approaching you, the observed frequency is

(1)   \begin{equation*}   f_1 = \frac{v}{v-v_{\rm src}} f_{\rm src} . \end{equation*}

When the car is moving away from you, you hear a frequency

(2)   \begin{equation*}   f_2 = \frac{v}{v+v_{\rm src}} f_{\rm src} . \end{equation*}

Taking the ratio of the above two equations, we obtain

(3)   \begin{equation*}   \frac{f_2}{f_1} = \frac{v-v_{\rm src}}{v+v_{\rm src}}   \implies   v_{\rm src} = \frac{f_1-f_2}{f_1+f_2} v \end{equation*}

The problem tells you that f_1=560 Hz, f_2=480 Hz, and v = 343 m/s. Plugging in these values we obtain v_{\rm src} = 26.4 m/s. The answer is (D).

Question 2: We will solve the problem in two stages:

(i) The ultrasound device (source) is at rest and the blood (observer) is moving. The sound as received by the blood (observer) has a frequency

(4)   \begin{equation*}   f_1 = \frac{v+v_b}{v} f_{\rm src} \end{equation*}

(ii) The blood then acts as the source for the reflected sound, which is received by the device with a frequency

(5)   \begin{equation*}   f_2 = \frac{v}{v-v_b} f_1 = \frac{v}{v-v_b} \left( \frac{v+v_b}{v} f_{\rm src} \right ) = \frac{v+v_b}{v-v_b} f_{\rm src} \end{equation*}

We know that v_b = 1.5 m/s, f_{\rm src} = 100 Hz and f_2 = 150 Hz. Solving, we find v_b = 0.3 m/s. The answer is (A).

Question 3: We solve the problem in two stages. First, the bat is the source and the wall is a stationary observer. The wall receives sound of frequency

(6)   \begin{equation*}   f_1 = \frac{v}{v-v_{\rm bat}} f_{\rm bat} . \end{equation*}

Next, the wall acts as a source emitting a frequency f_1 and the bat is the observer moving toward it. It hears an echo with frequency

(7)   \begin{equation*}   f_2     = \frac{v+v_{\rm bat}}{v} f_1     = \frac{v+v_{\rm bat}}{v-v_{\rm bat}} f_{\rm bat}     = 41.2 \ {\rm kHz} . \end{equation*}

So the answer is (C).

Question 4: Let us assume that the bat is moving with a velocity v_b = 5 m/s in the +x direction and that the insect is moving with a velocity v_i in the -x direction. We will solve the problem in two phases:

(i) the bat (source) emits a signal with frequency f_{\rm src} = 40 kHz which is received by the insect at a frequency of

(8)   \begin{equation*}   f_1 = \frac{v+v_i}{v-v_b} f_{\rm src} \end{equation*}

(ii) Next the sound reflects off the insect (who is now the source) and is received by the bat (observer) at a frequency

(9)   \begin{equation*}   f_2 = \frac{v+v_b}{v-v_i} f_1 = \frac{v+v_b}{v-v_i} \left( \frac{v+v_i}{v-v_b} f_{\rm src} \right ) \end{equation*}

Substituting v_b = 5 m/s, f_{\rm src} = 40 kHz and f_2 = 40.4 kHz we find that v_i = -3.29 m/s. The negative sign for the answer means that the insect is moving in a direction opposite to the one we originally assumed. Therefore the insect is moving away from that bat at 3.29 m/s (+x direction). The answer is (B).

Question 5: The velocity of the medium has no bearing on the observed frequency. To see why, recall the formula for the observed frequency, with source and observer moving toward one another,

(10)   \begin{equation*}   f_{\rm obs} = \frac{v + v_{\rm obs}}{v-v_{\rm src}} f_{\rm src} . \end{equation*}

This formula is derived under the assumption that the medium is not moving. In the present problem we can bring the medium to rest by switching to a reference frame F' that is moving with the wind, at velocity v_{\rm w}. In this frame, the source and observer, which were both at rest in the original reference frame, will both move with velocity -v_{\rm w}. We may therefore apply (10) with v_{\rm obs}=v_{\rm w} and v_{\rm src}=-v_{\rm w}, since the observer is moving towards the source but the source is moving away from the observer in F'. This gives us f_{\rm obs} = f_{\rm src}, so the answer is (D).

We can also understand this by pictorial arguments similar to the ones used in this note. This is left as an exercise.

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