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0 votes
1.7k views
in Wave Motion by (50.5k points)

Describe Doppler’s effect for sound waves and calculate the formula for experienced frequency when:
(a) Source is moving towards the stationary observer.
(b) Observer is moving towards the stationary source.

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2 Answers

+2 votes
by (1.7k points)

You, the person hearing the sounds, are called the observer or listener and the thing emitting the sound is called the source. As mentioned in the introduction, there are two situations which lead to the Doppler effect:

  1.  

    When the source moves relative to a stationary observer.

  2.  

    When the observer moves relative to a stationary source.

In points 1 and 2 above there is relative motion between the source and the observer. Both the source and the observer can be moving at the same time but we won't deal with that case in this chapter.

Doppler effect

The Doppler effect is the change in the observed frequency of a wave when the source or the detector moves relative to the transmitting medium.

The Doppler effect occurs when a source of waves and/or observer move relative to each other, resulting in the observer measuring a different frequency of the waves than the frequency that the source is emitting. The medium that the waves are travelling through, the transmitting medium, is also stationary in the cases we will study.

The question that probably comes to mind is: "How does the Doppler effect come about?". We can understand what is happening by thinking through the situation in detail.

Case 1: Moving source, stationary observer (ESCMP)

Let us consider a source of sound waves with a constant frequency and amplitude. The sound waves can be represented as concentric circles where each circle represents a crest or peak as the wavefronts radiate away from the source. This is because the waves travel away from the source in all directions and the distance between consecutive crests or consecutive troughs in a wave is constant (the wavelength as we learnt in Grade 10). In this figure, the crests are represented by the black lines and the troughs by the orange lines.

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Figure 6.1: Stationary sound source as more wavefronts are emmitted.

The sound source is the police car in the middle and is stationary. For the Doppler effect to take place (manifest), the source must be moving relative to the observer.

Let's consider the following situation: The source (represented by the black dot) emits one wave (the black circles represent the crests of the sound wave) that moves away from the source at the same rate in all directions. The distance between the crests represents the wavelength (\(\lambda\)) of the sound. The closer together the crests, the higher the frequency (or pitch) of the sound according to \(f=\frac{v}{\lambda}\), where \(v\) (speed of sound) is constant.

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As this crest moves away, the source also moves and then emits more crests. Now the two circles are not concentric any more, but on the one side they are closer together and on the other side they are further apart. This is shown in the next diagram.

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If the source continues moving at the same speed in the same direction, then the distance between crests on the right of the source is constant. The distance between crests on the left is also constant. The distance between successive crests on the left is constant but larger than the distance between successive crests on the right.

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When a car approaches you, the sound waves that reach you have a shorter wavelength and a higher frequency. You hear a sound with a higher pitch. When the car moves away from you, the sound waves that reach you have a longer wavelength and lower frequency. You hear a sound with a lower pitch.

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Figure 6.2: Moving sound source as more wavefronts are emitted.

Case 2: Moving observer, stationary source (ESCMQ)

Just as we did before, let us consider a source (a police car) of sound waves with a constant frequency and amplitude. There are two observers, one on the left that will move away from the source and one on the right that will move towards the source. We have three diagrams:

  1. shows the overall situation with the siren starting at time \(t_1\);
  2. shows the situation at time \(t_2\) when the observers are moving; and
  3. shows the situation at \(t_3\) after the observers have been moving for a time interval, \(\Delta t=t_3-t_2\).

The crests and troughs are numbered so you can see how they move further away and so that we can track which ones an observer has measured.

image

The observers can hear the sound waves emitted by the police car and they start to move (we ignore the time it takes them to accelerate).

image

The frequency of the wave that an observer measures is the number of complete waves cycles per unit time. By numbering the crests and troughs we can see which complete wave cycles have been measured by each of the observers in time, \(\Delta t\). To find the frequency we divide the number of wave cycles by \(\Delta t\).

In the time interval that passed, the observer moving towards the police car observed the crests and troughs numbered 1 through 5 (the portion of the wave is highlighted below). The observer moving away encountered a smaller portion of the wavefront, crest 3 and trough 4. The time interval for each of them is the same. To the observers, this will mean that the frequency they measured is different.

image

The motion of the observer will alter the frequency of the measured sound from a stationary source:

  • An observer moving towards the source measures a higher frequency.
  • An observer moving away from the source measures a lower frequency.

It is important to note that we have only looked at the cases where the source and observer are moving directly towards or away from each other and these are the only cases we will consider.

We didn't actually need to analyse both cases. We could have used either explanation because of relative motion. The case of a stationary source with a moving observer is the same as the case of the stationary observer and the moving source because the relative motion is the same. Do you agree? Discuss with your friends and try to convince yourselves that this is the case. Being able to explain work to each other will help you understand it better. If you don't understand it, you won't be able to explain it convincingly.

For a real conceptual test, discuss what you think will happen if the source and the observer are both moving, in the same direction and at the same speed.

The formula that provides the relationship between the frequency emitted by the source and the frequency measured by the listener is:

\[\boxed{{f}_{L}=\left(\frac{v\pm{v}_{L}}{v\pm{v}_{S}}\right){f}_{S}}\]

  • where \({f}_{L}\) is the frequency perceived by the observer (listener),
  • \({f}_{S}\) is the frequency of the source,
  • \(v\) is the speed of the waves,
  • \({v}_{L}\) the speed of the listener and
  • \({v}_{S}\) the speed of the source.

Note: The signs show whether or not the relative motion of the source and observer is towards each other or away from each other:

Source moves towards listener

\({v}_{S}\): negative

Source moves away from listener

\({v}_{S}\): positive

Listener moves towards source

\({v}_{L}\): positive

Listener moves away from source

\({v}_{L}\): negative

We only deal with one of the source or observer moving in this chapter. To understand the sign choice you can think about the pictures of the motion. For the listener/observer we are using the numerator in the equation. A fraction gets larger when the numerator gets larger so if we expect the frequency to increase we expect addition in the numerator (\({f}_{L}=(\frac{v+{v}_{L}}{v}){f}_{S}\)). If the numerator gets smaller the fraction gets smaller so if we expect the frequency to decrease then it is subtraction in the numerator (\({f}_{L}=(\frac{v-{v}_{L}}{v}){f}_{S}\)).

For the denominator, the reverse is true because of the fact that we divide by the denominator. The larger the denominator the smaller the fraction and vice versa. So if we expect the motion of the source to increase the frequency we expect subtraction in the denominator (\({f}_{L}=(\frac{v}{v-v_{S}}){f}_{S}\)) and if we expect the frequency to decrease we expect addition in the denominator (\({f}_{L}=(\frac{v}{v+v_{S}}){f}_{S}\)).

+1 vote
by (34.7k points)

Therefore Doppler Effect is a phenomena, it is not only applicable on sound waves but also works for electro-magnetic waves. Here we will only study sound waves and the changes in frequency at different situations of the source and observer.

Suppose sound source is S and observer is O and source S is vibrating with a frequency n and υ is the velocity of sound in stable medium then:

The most important use of Doppler’s Effect in sound is to calculate the velocity of submarines. Sound wave are sent from the seashore towards the sea; they are reflected back from the enemy submarines and are gained back at sea-shore (SONAR station). Now by calculating the change in the wavelength of the reflected wave the velocity of submarine is calculated as follows.

There is no change in the wavelength of the waves reaching the submarine from the SONAR station because in this case source is stationary but there is change in the wavelength of the reflected waves because for the reflected waves submarine will act as sound source. If submarine (sound source) is coming towards the SONAR station then the apparent wavelength will be:

Where υs is the velocity of source means that of submarine and λ is the velocity of sound in water and X is the original wavelength of the sound waves. Negative sign means submarine is coming closer and the wavelength of the reflected waves from it decreases. If submarine is going away from SONAR station then there would be positive sign. Hence, simple, formula for change in wavelength will be:

\(\Delta \lambda = \lambda' - \lambda = \pm[\frac{v_s}{v}]\lambda\)

By using this formula, the velocity of submarine (υs) is calculated and the position of the submarine relative to the SONAR station is also calculated.

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