How can the Doppler effect be explained both qualitatively and quantitatively?
Why are there differences when applying the Doppler effect to different types of waves?
What are some practical applications of the Doppler effect?
Understandings
Guidance:
How can the Doppler effect be explained both qualitatively and quantitatively?
What is the frequency of sound emitted by the source?
Doppler effect
Let’s derive the equation for the moving source!
Subtract for objects getting closer and add for objects getting farther apart
Add for objects getting closer and subtract for objects getting farther apart
Rule: Choose the operation which gives a higher observed frequency for objects approaching; choose the operation which gives a lower observed frequency for objects separating.
Equations for Doppler Effect
A car horn has a frequency of 520 Hz. The car travels to the right at 25 m s-1 and the speed of sound on this day is 340 m s-1.
(a) What frequency is heard by the driver of the car?
(b) What frequency is heard by an observer ahead of the car?
(c) What frequency is heard by an observer behind the car?
(a) The driver hears the horn at 520 Hz. The car horn is not moving relative to the driver. Therefore there is no Doppler effect
(b) Objects are getting closer together – use subtraction
(c) Now objects getting farther apart; use addition
A car horn has a frequency of 520 Hz. The car is stationary while an observer travels toward it at 25 m s-1. What frequency is heard by the moving observer if the speed of sound is 340 m s-1?
Approaching the source, a higher observed frequency is expected;
Notice how this is a different result from when the source was moving toward the observer
A bat at rest sends out ultrasonic waves of 50.0 kHz. The waves reflect from an owl moving directly at it with a speed of 25.0 m/s. At what frequency does the bat receive the reflected ultrasonic signal (use v = 340 m/s).
Why is the change in sound so sudden?
The abruptness you hear (e.g., a car horn as it zooms past) happens at the moment of closest approach.
The speed of a water wave is c. A boat moving at speed v makes waves. How fast will those waves move relative to the water?
The speed of the wave is independent of the motion of the source or the observer (only depends on the medium)
When the boat moves through the water, it generates waves. However, the speed of the waves generated by the boat depends primarily on the properties of the water (depth, wavelength, etc.), and not on the speed of the boat itself. This is because the wave speed c is a characteristic of the medium (water in this case).
The speed of the waves relative to the water will remain c, even though the boat itself is moving at v. The boat's motion doesn’t alter the inherent speed at which waves propagate through the water.
(We will do this explanation when we do Relativity by analyzing the motion from three perspectives: the water’s frame of reference (stationary relative to the water), the boat's frame of reference, an observer on land.)
Clearing up common misconceptions:
The Doppler effect is about perceived frequency (and so wavelength). It relates to the change in the pitch of the sound.
At the same time, a change in intensity will be observed because of the change in distance – but the Doppler effect is NOT about intensity!
Why Do Boats Make This Shape?
Explain how doppler effect is used in medical physics
Doppler Ultrasound
Why are there differences when applying the Doppler effect to different types of waves?
Doppler effect for light
Start at 19:25!
source
Speed of light is the same in all frames of reference!
The Doppler effect – light
The Doppler effect also applies to electromagnetic waves where v = c.
Thus f ’ = f [ v / ( v ± uS ) ] becomes
f ’ = f [ c / ( c ± uS ) ] = f [ 1 /( 1 ± uS / c)] = f ( 1 ± uS / c )-1.
And f ’ = f [ ( v ± uO ) / v ] becomes
f ’ = f [ ( c ± uO )/ c ] = f [ 1 ± uO / c)] = f ( 1 ± uO / c ) 1.
Perhaps you recall the binomial theorem.
Then because x = uO / c or x = uS / c, clearly |x| < 1 (a requirement for convergence). Both formulas reduce to
f ’ = f (1 ± 1×u / c) = f ± fu / c → ∆f / f = ±u / c.
binomial theorem
(1 ± x) n = 1 ± nx / 1! + n(n –1)x2 / 2! + …
(1 ± x)-n = 1 ± nx / 1! + n(n +1)x2 / 2! + …
Derivation not required!
The u in ∆f / f = ±u / c can be either the speed of the source or the speed of the observer, or the relative speed between both. The IBO changes the u to a v, where v is not the speed of the wave in the medium. ∙Because c = λf = λ’f ’we see that
∆f / f = (f ’ – f )/ f
= (c / λ’ – c / λ)/ (c / λ)
= (λ / λ’ – 1)
= (λ – λ’ )/ λ’
= ∆λ / λ
Doppler effect for light
∆f / f = ∆λ / λ = v / c
v is the relative speed between source and observer
EXAMPLE: A star in a galaxy is traveling away from us at a speed of 5.6×106 ms-1. It has a known absorption spectrum line that should be located at 520 nm on an identical stationary star. Where is this line located on the moving star?
SOLUTION: Use ∆λ / λ = v / c:
∆λ / λ = v / c = 5.6×106/ 3.00×108 = 0.01867
∆λ = 0.01867 λ = 0.01867 (520 nm) = 9.7 nm.
The star is moving away from us so that λ’ will be bigger than λ. Then
λ’ – λ = λ’ – 520 nm = 9.7 nm.
λ = 520 nm + 9.7 nm = 530 nm
PRACTICE: The absorption spectra of stars of varying distances from Earth are shown here. The sun is the bottom spectrum (D).
What is the approximate speed at which star A is receding from Earth?
SOLUTION: Use ∆λ / λ = v / c → v = c ∆λ / λ.
We can estimate that λ = 390 nm (bottom arrow).
We can estimate that λ’ = 490 nm (top arrow).
Thus ∆λ = 490 nm – 390 nm = 100 nm.
v = c ∆λ / λ
= 3.00×108 ×100 / 390
= 7.69×107 ms-1.
A
B
C
D
This is the so-called “redshift” and it is used as evidence for an expanding universe.
Which star has the highest velocity relative to Earth? Is it moving towards us, or away from us?
SOLUTION:
From ∆f = (v / c) f we see that the higher the relative velocity v the greater the shift ∆f. Thus A is our candidate.
Observing the heavy black line in D we see that it is shifting to the red region. Thus A is receding.
A
B
C
D
This is the so-called “redshift” and it is used as evidence for an expanding universe.
Redshift > moving away
Blueshift > moving towards