C.5 Doppler effect – IB Physics exam questions with mark scheme

1StandardHL only6 marks

An ambulance siren emits sound of frequency $900\ \mathrm{Hz}$. The ambulance travels at a constant $25\ \mathrm{m\,s^{-1}}$ along a straight road, directly towards and then directly away from a stationary observer at the side of the road. The speed of sound is $340\ \mathrm{m\,s^{-1}}$.

(a) Calculate the frequency heard by the observer as the ambulance approaches. [2]

(b) Calculate the frequency heard as the ambulance moves away. [1]

(c) A spectral line from hydrogen has wavelength $656.3\ \mathrm{nm}$ in the laboratory but $659.6\ \mathrm{nm}$ in the light from a distant galaxy. Estimate the speed of the galaxy. Take $c=3.00\times10^8\ \mathrm{m\,s^{-1}}$. [2]

(d) State whether the galaxy is moving towards or away from the Earth. [1]

$f'=f\,v/(v-u_s) = 900\times340/(340-25)$(1)
$f'=971$ Hz (accept 970 Hz)(1)
$f'=900\times340/(340+25) = 838$ Hz (accept 840 Hz)(1)
$v\approx c\,\Delta\lambda/\lambda = 3.00\times10^8\times3.3/656.3$ (accept dividing by 659.6)(1)
$v=1.5\times10^{6}\ \mathrm{m\,s^{-1}}$ (accept $1.51\times10^6$)(1)
Away from the Earth (reason not required; e.g. wavelength increased / redshift)(1)
Answer: (a) 971 Hz (b) 838 Hz (c) $1.5\times10^6$ m s$^{-1}$ (d) away from Earth
Examiner tip: For a moving source use minus in the denominator when approaching and plus when receding.

2Core4 marks

The diagram, drawn to scale, shows four successive wavefronts of sound emitted by a small source S that moves at constant velocity. The wavefronts were emitted at equal time intervals. A and B are two stationary observers on the line of motion of S.

(a) State the direction in which S is moving. [1]

(b) Explain, with reference to the diagram, why the frequency heard by B is greater than the frequency emitted by S. [2]

(c) By taking measurements from the diagram, determine the ratio $\dfrac{\text{frequency heard by B}}{\text{frequency heard by A}}$. [1]

towards B / to the right (wavefronts are bunched on B's side)(1)
the wavefronts are closer together on B's side, so the wavelength reaching B is shorter (than when S is at rest)(1)
the speed of sound is unchanged / the same, so (since $f=v/\lambda$) more wavefronts reach B per second and the frequency is higher; do not accept "the waves travel faster towards B"(1)
spacing behind S (towards A) is three times the spacing in front of S (towards B), so ratio = 3; accept 2.8 to 3.2(1)
Answer: (a) towards B (b) wavefronts closer together on B's side: shorter wavelength at the same wave speed, so higher frequency (c) 3
Examiner tip: The wave speed does not change: the explanation must say the wavelength is shorter at the same speed, not that the sound travels faster.

3Core4 marks

The calcium K absorption line has a wavelength of $393.4\ \mathrm{nm}$ when measured in a laboratory. In the spectrum of a nearby star the same line is observed at $393.1\ \mathrm{nm}$.

(a) State and explain whether the star is moving towards or away from the Earth. [1]

(b) Calculate the speed of the star relative to the Earth along the line of sight. Take $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$. [2]

towards the Earth, because the observed wavelength is shorter / the line is blue-shifted; reason needed for the mark(1)
$\Delta\lambda = 0.3\ \mathrm{nm}$ and use of $v = c\,\dfrac{\Delta\lambda}{\lambda} = 3.00\times10^{8}\times\dfrac{0.3}{393.4}$(1)
$v = 2.3\times10^{5}\ \mathrm{m\,s^{-1}}$; accept $2.29\times10^5$; ignore a negative sign(1)
the speed of the star is much less than the speed of light ($v \ll c$); allow "non-relativistic speeds"(1)
Answer: (a) towards, blue shift (b) $2.3\times10^{5}\ \mathrm{m\,s^{-1}}$ (c) $v \ll c$
Examiner tip: Students often subtract the wrong way and then say "away": a shorter observed wavelength always means approach.

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4Standard5 marks

A roadside speed camera uses radar. It emits microwaves of frequency $24.15\ \mathrm{GHz}$ towards a car that is driving directly towards the camera. The microwaves reflected from the car are received back at the camera, and the camera measures the difference between the received and emitted frequencies as $3.50\ \mathrm{kHz}$.

(a) State whether the received frequency is higher or lower than the emitted frequency. [1]

(b) Explain why the frequency shift for the reflected wave is twice the shift that the car itself would detect. [2]

higher (the car is approaching)(1)
the car acts as a moving observer/receiver: it receives microwaves at a frequency shifted (by $\approx fv/c$)(1)
the car then re-emits / reflects the waves as a moving source, so the frequency is shifted again by the same amount (total $\approx 2fv/c$)(1)
$\Delta f = \dfrac{2fv}{c}$ so $v = \dfrac{c\,\Delta f}{2f} = \dfrac{3.00\times10^{8}\times3.50\times10^{3}}{2\times24.15\times10^{9}}$(1)
$v = 21.7\ \mathrm{m\,s^{-1}}$; accept 22 m s⁻¹; award 1 mark only for 43.5 m s⁻¹ (factor of 2 missed)(1)
Answer: (a) higher (b) shifted once on reception by the moving car and again on reflection (c) $21.7\ \mathrm{m\,s^{-1}}$
Examiner tip: Forgetting the factor of 2 for a reflected wave doubles the speed and costs the final mark.

5StandardHL only5 marks

A fire alarm fixed to the wall of a building emits a steady note of frequency $1800\ \mathrm{Hz}$. A cyclist rides at a constant speed of $9.0\ \mathrm{m\,s^{-1}}$ along a straight road, first directly towards the alarm and later directly away from it. The air is still and the speed of sound is $343\ \mathrm{m\,s^{-1}}$.

(a) Calculate the frequency heard by the cyclist while riding towards the alarm. [2]

(b) Determine the change in the frequency heard by the cyclist as she passes the alarm. [1]

(c) The cyclist is a moving observer. Explain why the frequency she hears changes even though the wavelength of the sound in the air is not changed by her motion. [2]

$f' = f\,\dfrac{v+u_o}{v} = 1800\times\dfrac{343+9.0}{343}$(1)
$f' = 1850\ \mathrm{Hz}$; accept 1847 Hz(1)
receding: $f' = 1800\times\dfrac{334}{343} = 1753\ \mathrm{Hz}$, so change $= 94\ \mathrm{Hz}$ (decrease); accept 90 to 95 Hz; allow ECF from (a)(1)
the speed of the waves relative to the cyclist is $v + u_o$ (towards) / $v - u_o$ (away)(1)
so she meets more (or fewer) wavefronts per second than if at rest, and the frequency she hears is (speed relative to her)/$\lambda$; do not accept "the wavelength gets shorter"(1)
Answer: (a) $1850\ \mathrm{Hz}$ (1847 Hz) (b) falls by about $94\ \mathrm{Hz}$ (c) relative speed of the waves changes, so wavefronts are met at a different rate
Examiner tip: For a moving observer the wavelength in the air is unchanged: credit goes to the change in relative speed, not to a change in wavelength.

6HardHL only6 marks

A train sounds its horn continuously while travelling at constant speed along a straight track. An observer stands very close to the track. The graph shows how the frequency heard by the observer varies with time. The train passes the observer at $t = 5\ \mathrm{s}$ and is far from the observer at $t = 0$ and $t = 10\ \mathrm{s}$. The speed of sound is $340\ \mathrm{m\,s^{-1}}$.

(a) Use the graph to determine the speed of the train. [3]

(b) Calculate the frequency emitted by the horn. [2]

reads $f_1 = 530\ \mathrm{Hz}$ (approaching) and $f_2 = 470\ \mathrm{Hz}$ (receding); allow ±2 Hz(1)
uses $f_1 = f\,\dfrac{v}{v-u_s}$ and $f_2 = f\,\dfrac{v}{v+u_s}$ and eliminates $f$, e.g. $\dfrac{f_1}{f_2} = \dfrac{v+u_s}{v-u_s}$(1)
$u_s = v\,\dfrac{f_1-f_2}{f_1+f_2} = 340\times\dfrac{60}{1000} = 20\ \mathrm{m\,s^{-1}}$; accept 20.4 m s⁻¹(1)
$f = f_1\,\dfrac{v-u_s}{v} = 530\times\dfrac{340-20.4}{340}$ OR $f = f_2\,\dfrac{v+u_s}{v}$(1)
$f = 498\ \mathrm{Hz}$; accept 500 Hz; allow ECF from (a); do not accept the mean of the two readings without working(1)
the observer is (slightly) to the side of the track, so only the component of the train's velocity along the line to the observer counts, and this component changes gradually as the train passes(1)
Answer: (a) $20\ \mathrm{m\,s^{-1}}$ (20.4) (b) $498\ \mathrm{Hz}$ (c) the velocity component along the line of sight changes gradually
Examiner tip: Two unknowns need two equations: write both Doppler expressions and divide them, rather than guessing the emitted frequency as the mean.

7Hard6 marks

A spiral galaxy is viewed edge-on from Earth. A spectral line emitted by oxygen ions has a laboratory wavelength of $500.7\ \mathrm{nm}$. The same line is measured at the two opposite edges, X and Y, of the galaxy:

edge X: $504.1\ \mathrm{nm}$
edge Y: $503.3\ \mathrm{nm}$

Take $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$.

(a) Determine the speed at which the centre of the galaxy is moving away from the Earth. [3]

(b) Determine the speed at which the edges of the galaxy rotate about its centre. [2]

wavelength from the centre = mean of the two edges = $503.7\ \mathrm{nm}$, so $\Delta\lambda = 3.0\ \mathrm{nm}$(1)
$v = c\,\dfrac{\Delta\lambda}{\lambda} = 3.00\times10^{8}\times\dfrac{3.0}{500.7}$(1)
$v = 1.8\times10^{6}\ \mathrm{m\,s^{-1}}$(1)
shift of each edge relative to the centre $= \pm 0.4\ \mathrm{nm}$ (half of the 0.8 nm difference)(1)
rotational speed $= 3.00\times10^{8}\times\dfrac{0.4}{500.7} = 2.4\times10^{5}\ \mathrm{m\,s^{-1}}$; award 1 mark only for $4.8\times10^5$ (full difference used); accept dividing by 503.7(1)
Y (its line is less red-shifted than the centre's)(1)
Answer: (a) $1.8\times10^{6}\ \mathrm{m\,s^{-1}}$ (b) $2.4\times10^{5}\ \mathrm{m\,s^{-1}}$ (c) Y
Examiner tip: The whole galaxy is receding, so both edges are red-shifted: rotation shows up only as the difference from the centre's shift.

8StandardHL only5 marks

A source of sound emits waves of frequency $f$ and period $T$. It moves at constant speed $u_s$ directly towards a stationary observer. The speed of sound in still air is $v$.

(a) By considering the distance between two wavefronts emitted one period apart, show that the frequency heard by the observer is $f' = f\,\dfrac{v}{v-u_s}$. [3]

(b) A model aircraft flies directly towards the observer at one-tenth of the speed of sound. Calculate the ratio $f'/f$. [1]

in time $T$ the first wavefront travels $vT$ while the source moves $u_sT$ towards the observer(1)
so the wavelength in front of the source is $\lambda' = (v-u_s)T$(1)
the wave speed is still $v$, so $f' = \dfrac{v}{\lambda'} = \dfrac{v}{(v-u_s)T}$ and with $T = 1/f$, $f' = f\,\dfrac{v}{v-u_s}$; must state that the speed of sound is unchanged(1)
$\dfrac{f'}{f} = \dfrac{v}{v-0.1v} = \dfrac{10}{9} = 1.11$(1)
the air is still / there is no wind OR the speed of sound is the same in all directions / unaffected by the source; allow uniform temperature(1)
Answer: (a) derivation (b) $1.11$ ($10/9$) (c) still air (no wind)
Examiner tip: In a 'show that' every step must appear: state that the wave speed is unchanged before writing $f' = v/\lambda'$.

9Hard6 marks

Sound of a single frequency is produced in still air. In situation 1 the source moves at constant speed towards a stationary observer. In situation 2 the source is stationary and the observer moves at the same constant speed towards it.

Using wavefront diagrams, explain why the observer hears a higher frequency in both situations, and compare what happens to the wavelength and to the speed of the sound relative to the observer in each situation. [6]

situation 1: wavefronts drawn as non-concentric circles, centres shifted in the direction of motion, bunched in front of the source(1)
situation 1: the source moves a distance $u T$ between emissions, so the wavelength in front of it is shorter(1)
situation 1: the speed of the sound relative to the (stationary) observer is unchanged ($v$), so a shorter wavelength gives a higher frequency(1)
situation 2: wavefronts drawn as concentric, evenly spaced circles centred on the source(1)
situation 2: the wavelength in the air is unchanged(1)
situation 2: the speed of the waves relative to the observer increases to $v + u$, so the observer meets more wavefronts per second(1)
comparison: source motion changes the wavelength; observer motion changes the relative speed(1)
the medium (air) is the reference: the two situations are not equivalent and give (slightly) different frequencies for the same speed(1)

Level	Marks	What it looks like
Level 3	5–6	Correct wavefront diagrams (or clear descriptions of them) for both situations. Explains the higher frequency in each case correctly: shorter wavelength at unchanged speed for the moving source, and greater relative speed at unchanged wavelength for the moving observer. Makes a clear, explicit comparison.
Level 2	3–4	Correct explanation for one situation with a suitable diagram, and a partly correct account of the other. Some comparison is made, but it may be incomplete or contain a minor error (e.g. implies the wavelength also changes for the moving observer).
Level 1	1–2	States that the frequency rises in both cases, with a limited explanation or diagram for one situation. Physics is confused, e.g. the speed of sound is said to increase for a moving source.
Level 0	0	No relevant physics.

Answer: Moving source: wavelength ahead shortened, wave speed unchanged. Moving observer: wavelength unchanged, relative speed increased. Both give more wavefronts received per second.
Examiner tip: A diagram showing evenly spaced circles for the moving observer is essential: drawing bunched wavefronts in situation 2 is the most common error.

10StretchHL only6 marks

A bat flies at constant speed $u$ directly towards a flat vertical wall. It emits ultrasound of frequency $42.0\ \mathrm{kHz}$ and hears the echo from the wall at a frequency of $43.6\ \mathrm{kHz}$. The air is still and the speed of sound $v$ is $343\ \mathrm{m\,s^{-1}}$.

(a) Show that the frequency $f''$ of the echo heard by the bat is related to the emitted frequency $f$ by $f'' = f\,\dfrac{v+u}{v-u}$. [2]

(b) Determine $u$. [3]

wall receives $f' = f\,\dfrac{v}{v-u}$ (moving source, stationary observer)(1)
wall reflects at $f'$ as a stationary source and the bat is a moving observer: $f'' = f'\,\dfrac{v+u}{v} = f\,\dfrac{v+u}{v-u}$(1)
rearranges: $f''(v-u) = f(v+u)$ OR $u = v\,\dfrac{f''-f}{f''+f}$(1)
substitution: $u = 343\times\dfrac{1.6}{85.6}$(1)
$u = 6.4\ \mathrm{m\,s^{-1}}$; accept 6.41 m s⁻¹; 6.5 m s⁻¹ from the approximation $\Delta f/f \approx 2u/v$ scores max 2(1)
$f' = 42.0\times\dfrac{343}{343-6.41} = 42.8\ \mathrm{kHz}$; allow ECF from (b); accept $f' = \sqrt{f f''}$ approach giving 42.8 kHz(1)
Answer: (b) $6.4\ \mathrm{m\,s^{-1}}$ (c) $42.8\ \mathrm{kHz}$
Examiner tip: Treat the echo as two separate Doppler shifts: moving source to the wall, then the wall as a stationary source and the bat as a moving observer.

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