Moving Observer, Stationary Source

Case 2: Moving Observer, Stationary Source

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.

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).

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.

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.

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:

We only deal with one of the source or observer moving in this section. 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 (\({f}_{L}=(\frac{v}{v+v_{S}}){f}_{S}\)).

Optional Video: Introduction to the Doppler Effect by Khan Academy

See some examples below:

Example: Ambulance Siren

Question

The siren of an ambulance emits sound with a frequency of \(\text{700}\) \(\text{Hz}\). You are standing on the pavement. If the ambulance drives past you at a speed of \(\text{20}\) \(\text{m·s$^{-1}$}\), what frequency will you hear, when

1. the ambulance is approaching you

2. the ambulance is driving away from you

Take the speed of sound in air to be \(\text{340}\) \(\text{m·s$^{-1}$}\).

Step 1: Analyse the question

The question explicitly asks what frequency you will hear when the source is moving at a certain speed. This tells you immediately that the question is related to the Doppler effect. The values given in the question are all in S.I. units so no conversions are required.

Step 2: Determine how to approach the problem based on what is given

We know that we are looking for the observed frequency with a moving source. The change in frequency can be calculated using:

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

To correctly apply this we need to confirm that it is valid and determine what signs we need to use for the various speeds. You (the listener) are not moving but we have to consider two different cases, when the ambulance is moving towards you (a) and away from you (b). We have been told that if the source is moving towards the observer then we will use subtraction in the denominator and if it is moving away, addition. This means:

\begin{align*} {f}_{S}& = \text{700}\text{ Hz}\\ v& = \text{340}\text{ m·s$^{-1}$}\\ {v}_{L}& = 0 \text{ because you. the observer. are not moving}\\ {v}_{S}& = -\text{20}\text{ m·s$^{-1}$} \text{ for (a) and}\\ {v}_{S}& = +\text{20}\text{ m·s$^{-1}$} \text{ for (b)} \end{align*}

Step 3: Determine \({f}_{L}\) when ambulance is approaching

\begin{align*} {f}_{L}&=\left(\frac{v\pm{v}_{L}}{v\pm{v}_{S}}\right){f}_{S} \\ {f}_{L}& = \left(\frac{340+0}{340-20}\right)\left(700\right)\\ & = \text{743.75} \text{Hz} \end{align*}

Step 4: Determine \({f}_{L}\) when ambulance has passed

\begin{align*} {f}_{L}&=\left(\frac{v\pm{v}_{L}}{v\pm{v}_{S}}\right){f}_{S} \\ {f}_{L}& = \left(\frac{340+0}{340+20}\right)\left(700\right)\\ & = \text{661.11} \text{Hz} \end{align*}

Step 5: Quote the final answer

When the ambulance is approaching you, you hear a frequency of \(\text{743.75}\) \(\text{Hz}\) and when it is going away you hear a frequency of \(\text{661.11}\) \(\text{Hz}\)