A.3 (11.2) Doppler Effect
A.3.1 The Doppler Effect/A.3.2 Explaining the Doppler Effect



Doppler Effect -» The phenomenon in which a sound source moving towards an observer
will result in an increase of observed frequency, while a sound source moving away from an
observer will result in a decrease of observed frequency.
The same phenomenon occurs when the source is stationary and the observer is moving.
In the stationary case, shown below, the frequency f produced by the source is the same as
that observed by A and B.

If the source is moving towards point A, the resulting wave fronts are shown below.

Clearly, the waves are bunched together in front of the source, and spread out behind the
source. How much they will bunch together or spread out depends on the speed of the
source. This difference in wavelength means that the observers at A and B will detect
different frequencies. In general, this follows the formula
𝑣
𝑓𝐴 = 𝜆
𝐴
and
𝑣
𝑓𝐵 = 𝜆
𝐵
where v is the speed of waves in the medium, and λA and λB are the wavelengths at
A and B, respectively.

In the phenomenon above, the important thing is the relative velocity of the observer and
the source, so the situation will be the same if the observers are moving and the source is
stationary.
Question: What will happen if the source or observer are accelerating?
Answer: The frequency will be constantly changing.
A.3.3 The Doppler Equations for Sound

Let’s say there is an ambulance driving towards an observer who is standing on the sidewalk.
Normally, the ambulance’s siren produces a sound of frequency f. The speed of sound is v,
and the speed of the ambulance is vs.
vs
-v

v
O
Let’s determine a simple formula for wavelength, λ. In some time, ∆t, how many
wavelengths will the observer receive? Remembering that frequency is the time for one
period, we can see that the answer is f∆t. The distance that these waves have travelled in
this time is v∆t. The wavelength will be the total distance travelled divided by the number
of wavelengths, so
𝑣∆𝑡
𝑣
𝜆 = 𝑓∆𝑡 = 𝑓
or
𝑣 = 𝑓𝜆
Please note this equation as a general rule for waves.

Now we will determine the wavelength of the waves in front of the ambulance, λ’. When
we discuss the velocity, v, the important thing is really the velocity relative to the source. In
this case, the effective velocity of the sound waves will really be v – vs.
𝜆′ =
(𝑣 − 𝑣𝑠 )∆𝑡 𝑣 − 𝑣𝑠
=
𝑓∆𝑡
𝑓
The speed of the sound waves relative to the observer is still v, so the frequency that the
observer hears, f’ is
𝑣
𝑓 ′ = 𝜆′
or
𝑣
𝜆′ = 𝑓′
Equating the two equations for λ’, we have
𝑣 𝑣 − 𝑣𝑠
𝜆′ = =
𝑓′
𝑓
So,
𝑓′ =
𝑣
𝑣−𝑣𝑠
𝑓
or
𝑓′ =
1
𝑣 𝑓
1− 𝑠
𝑣
Please remember that for this equation, a source moving away from the observer will have a
negative velocity. Knowing that, we can see that the frequency heard by an observer will be
greater if the source is moving toward them, and smaller if the source is moving away.

Now we will look at the situation where the observer is moving and the source is stationary.
v
vo
O
We will start with the original equation, 𝑣 = 𝑓𝜆.
Remember, we care about the velocity relative to the source. In this case, that velocity is
simply v. So, the wavelength of the waves that reach the observer is
𝑣
𝑓
The speed of the sound waves relative to the observer is v + vo, so the observed frequency is
given by the equation
𝜆′ =
𝑓′ =
𝑣 + 𝑣𝑜 𝑣 + 𝑣𝑜 𝑣 + 𝑣𝑜
= 𝑣 =
𝑓
𝜆′
𝑣
𝑓
This can be simplified to the form
𝑓 ′ = (1 +
𝑣𝑜
)𝑓
𝑣
Remember, if the observer is moving toward the source, vo is positive. If the observer is
moving away from the source, vo is negative.

The change in frequency is defined by the equation
𝑣𝑜
𝑣𝑜
∆𝑓 = 𝑓 ′ − 𝑓 = (1 + ) 𝑓 − 𝑓 = 𝑓
𝑣
𝑣
A.3.4 Solve Doppler Effect Problems for Sound
(ex 1) A source emits a sound of frequency 440 Hz. It moves in a straight line towards a
stationary observer with a speed of 30 m s-1. The observer hears a sound of frequency
484 Hz. Calculate the speed of sound in air.
Use the equation,
𝑓′ =
1
𝑣 𝑓
1 − 𝑣𝑠
Solving for v, we have
𝑣=
𝑣𝑠
1−
𝑓
𝑓′
Substituting the values gives
𝑣=
30 𝑚 𝑠 −1
= 330 𝑚 𝑠 −1 = 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 𝑖𝑛 𝑎𝑖𝑟
440𝐻𝑧
1 − 484𝐻𝑧
(ex 2) Oh no, the school is on fire! If I am riding away from the school on my bicycle at 25 m s-1,
and the normal frequency of a fire alarm is 500 Hz, what will be the frequency that I hear
as I ride away?
Use the equation,
𝑓 ′ = (1 +
𝑣𝑜
)𝑓
𝑣
Remembering that moving away from the source means negative velocity, and
substituting the answer from (ex 1), we have
𝑓 ′ = (1 +
−25 𝑚 𝑠 −1
) (500 𝐻𝑧) = 462 𝐻𝑧
330 𝑚 𝑠 −1
A.3.5 The Doppler Equations for Light

When we talked about the Doppler effect for sound, we looked at two things:
o The wavelength, λ, where we considered the velocity of sound relative to the source.
o The observed frequency, f’, where we considered the velocity of sound relative to the
observer.
Light, however, is very special. Light has the same speed for all observers. This means that,
even if the source is moving and the observer is stationary, they will both measure light as
having the same speed, c = 3 x 108 m s-1.
Therefore, when we talk about the Doppler effect for light, we are mostly concerned with
the velocity of the source relative to the observer, v.

Because light is so unique, the equations for the Doppler effect are very difficult, and are
part of Einstein’s Theory of Special Relativity. In this class, however, we will only look at the
simple situation where the source is moving at a speed v, which is much less than the speed
of light, c.

For v«c,
𝑣
𝑓
𝑐
Remember that for a source moving toward the observer, v is positive. For a source moving
away from an observer, v is negative.
∆𝑓 = 𝑓 ′ − 𝑓 =

The Doppler effect of light is usually seen in stars. A star moving away from us will have
negative ∆f, which means that the frequency we see will be shifted down. The lowest
frequency visible light is red, so this phenomenon is called red-shift. If the star is moving
toward us, it is known as blue-shift.
(ex) A particular radio signal from a galaxy is measured as having a frequency of 1.39 × 109 Hz.
The same signal from a stationary source in a laboratory has a frequency of 1.42 × 109 Hz.
Suggest why the galaxy is moving away from Earth and calculate its recession speed (i.e.
the speed with which it is moving away from Earth).
The frequency of the moving galaxy is less than that of a stationary source, so this
suggests that the galaxy is moving away from us.
𝑣
Rearranging the equation, ∆𝑓 = 𝑐 𝑓, we have
𝑐∆𝑓 3𝑥108 𝑚 𝑠 −1 𝑥 (1.42 − 1.39)𝑥109 𝐻𝑧
𝑣=
=
= 6.34𝑥106 𝑚 𝑠 −1
𝑓
1.42𝑥109
This can also be written as 𝑣 = 0.021𝑐. Note that for speeds greater than this, we
would probably need to use Einstein’s more complicated equations.
A.3.6 Using the Doppler Effect

The Doppler effect can be used to measure speed, as seen in the previous example. Here is
another example:
(ex) We have a car travelling toward a large mirror. The car is equipped with a light transmitter
and a light receiver. We can determine the speed of the car based on the change in
frequency between the transmitter and receiver.
Basically, the Doppler effect will happen twice. We can imagin the mirror as a stationary
receiver, with the source moving towards it, so the light that hits the mirror will follow
the equation
𝑣
𝑓′ − 𝑓 = 𝑓
𝑐
When these waves are reflected and go back to the car, it will be like the mirror is a
stationary source, and the car is an observer moving towards it. This second Doppler
effect will be defined by
𝑣
𝑓 ′′ − 𝑓 ′ = 𝑓′
𝑐
So, the total Doppler shift will be
𝑣
𝑣
∆𝑓 = 𝑓 ′′ − 𝑓 = 𝑓′ + 𝑓
𝑐
𝑐
𝑣
But, from the first equation, we see that 𝑓 ′ = (1 + 𝑐 ) 𝑓 , so
𝑣 𝑣 𝑣
𝑣
𝑣
𝑣2
∆𝑓 = 𝑓 (1 + ) + 𝑓 = 𝑓 + 𝑓 + 2
𝑐 𝑐 𝑐
𝑐
𝑐
𝑐
For v«c,
𝑣2
𝑐2
≈ 0 , which gives us
∆𝑓 =
2𝑣
𝑓
𝑐
We only consider cases where v«c. Note that this would be the same if we were using
sound waves instead of light waves, with c being the speed of sound.
HOMEWORK: Physics 3rd ed. Pg 299-301 #11.2 (a), (b)