Doppler Effect PDF

C.5 Doppler eect What are some practic al applic ations of the Doppler eect? How c an the Doppler eect be explained both qualitatively and quantitatively? Why are there dierences when app...

C.5 Doppler eect What are some practic al applic ations of the Doppler eect? How c an the Doppler eect be explained both qualitatively and quantitatively? Why are there dierences when applying the Doppler eect to dierent types of waves? The Doppler effect is a f re quency shift that is detected when a source of w av e s is moving relative to the observer of the w av e s. It was originally a curiosity when it was f i rs t identified in 1842, even though the hy p o t h e s i s was tested and found to be corre ct within a f ew y e a rs. N ow a d ays , techniques using the Doppler effe ct are used ex t e n s i v e l y in m e d i c i n e, a s t r o n o my, f l ow me asurements and in m a ny other applic ations. The D opple r effe ct is de sc ribe d in ter ms of the moveme nt of the w ave f ronts thro ugh a me dium as pe rceive d by the o bserver. These qual itati ve desc r iptio ns are use d as a bas is for qu antitative e quations that c an pre dict the f re que ncy s hift. Topic A.5 showed that, when the speed of an object approaches c, we need to apply the theory of special ▴ Figure 1 An ultrasound image of the heart. The Doppler relativity rather than Newtonian mechanics to the situation. eect enables the speed of blood ow to be measured. The This applies to the Doppler effect too, whatever the type dierent colours represent the dierent ow speed. of wave under consideration. The Doppler effect for light is 6 always relativistic bec ause c is invariant. relativistic eects are of the order of 10 %. For this reason, relativity c an be ignored when evaluating how the Doppler The speed of sound, whether in uid or solid, is never aects sound and other mechanic al waves. However, close to c. The speed of a sound wave through diamond, when considering Doppler shis for electromagnetic 1 at 12 km s , is one of the largest values known and this is radiation, we must be more c areful to establish the still only 0.004% of the speed of light in a vacuum. The approximations being used. In this topic, you will learn about: the Doppler effect for sound waves and shifts in spectral lines that provide information about the electromagnetic waves motion of astronomic al objects visualizing the Doppler effect when either the source or quantitative Doppler shifts in sound and LHA the observer is moving mechanic al waves when either the source or the observer is moving. the relative change in wavelength for a light wave due to the Doppler effect 456  C. Wave behaviour Introduction An ambulance travels along the road at speed, sounding its siren. You notice a change in the frequency of the siren — high to low — as the vehicle passes you. This is the Doppler effect. It was first suggested by the Austrian scientist Christian Doppler, who showed mathematic ally that an observed wave frequency depends on the relative speed between the sound source and an observer. In France, the effect is ascribed to Doppler–Fizeau bec ause the Frenchman Amand Fizeau extended the work to the spectral shifts in light from stars shortly before Doppler ’s death. ▴ Figure 2 An observer will hear a The nature of the Doppler eect change in the frequency of an ambulance's siren as it passes. As you saw in Topic C.2, wavefronts from a point source spread out in a sphere. Figure 3 is a wavefront diagram showing the effect in two dimensions, but this is easily extended to 3D in your imagination. S is the point source and the observer of the waves is O. In this and subsequent wavefront diagrams, you are acting as a second observer, at rest relative to the medium, and you c an see the wavefronts together with the movement of both source and observer. This description and the later derivation of the Doppler equations on page 459 ignore relativistic effects. In Topic A.5 you met the theory of special relativity. This c theory has practic al importance when speeds are greater than about (c here is 5 The changes to the Doppler the speed of electromagnetic radiation in a vacuum). When we are dealing with equations under relativistic speeds of sound or with observer/source speeds close to c, then we must allow conditions are described on for time dilation and length contraction effects. Wave speeds are invariably very page464. much less than c, so this approximation is a good one. S O S O S O S O S O ▴ Figure 3 A spheric al wave expands. Both source (S) and observer (O) are stationary. O observes the same wavelength and frequency as emitted by S. Figure 3 shows what happens when neither the source S nor observer O move. In this diagram the source is emitting a wavefront at a regular rate (the frequency of the wave). The red circles show the position of these wavefronts at equal time intervals. Time increases from left to right in the figure. The circular wavefront increases in radius by one square between diagrams. The first wavefront reaches the observer three time intervals after it was generated. The second wavefront crosses the observer one time interval later. The source and the observer agree on the rate at which the wavefronts are being generated and detected. 457  Topic C.5 Doppler effect What happens when the observer or source move relative to the wave as it spreads through the medium? There are two c ases: source moving and observer stationary, and source stationary with observer moving. These are analysed separately. For both c ases, the source and observer move along the straight line that joins them. S O The Doppler eect analysed (a) Moving source — Stationary observer When the Doppler shift is c aused by source movement, there is a change in the wavelength of the waves relative to the stationary medium. This is observed source starting position as a shifted frequency bec ause the wavefronts sweep across the observer at a ▴ Figure 4 O is stationary and S is different rate from that at which they were emitted. moving towards O at half a square per time In Figure 4, the source S is moving to the right. It moves half a square during the interval. The wavefronts in the medium are time that a wavefront advances by a full square. M athematic ally, the source speed compressed into a smaller volume bec ause is half that of the wave speed. of the movement of S. Five time intervals elapse over the course of the diagram. There are five wavefronts shown and the source has moved 2.5 squares to the right. The wavefront emitted earliest is about to cross the observer. Bec ause the wavefront moves at the wave speed in the medium, it will cross the observer twice as quickly as it would without the source movement. (Look again at Figure 4, remembering that the radius of each wavefront is constantly expanding by one S O square every time interval.) The observed frequency increases for the stationary observer (in fact, doubling in this c ase). When the source is moving away from the observer, the frequency change is in the opposite direction. The observed frequency is now less than the emitted source starting position frequency (Figure 5). The sc ale distances between wavefronts as they move across the observer are ▴ Figure 5 When S moves away from O, larger than for the original static situation (Figure 3), showing that the wavefronts the wavelengths observed by O are further apart. pass over the observer less frequently than before. (b) Stationary source — Moving observer When the observer moves but the source is stationary, the wavefronts move symmetric ally through the medium, expanding as concentric spheres. However, bec ause the observer moves towards or away from them, they are detected more often or less frequently than they were emitted. The waves that are created in the medium have the usual spheric al pattern S O (circular in 2D). When O is moving towards S, the wavefronts are crossed more quickly than they were emitted. An increased frequency is observed with an unchanged wavelength (bec ause that wavelength is determined by the medium, not by the observer). Relative to the observer, the speed of the wave has changed. ▴ Figure 6 O is now moving towards S. O When O is moving away from S, then the wavefronts take longer to c atch the crosses the wavefronts more oen than they observer up, so that the time period is longer and the apparent frequency are emitted, so observes a higher frequency. (according to the observer) is lower. 458  C. Wave behaviour Worked example 1 A source of sound S is moving at a constant speed along the line joining two stationary observers A and B. The diagram shows the wavefronts emitted by S in equal time intervals. a. Explain, with reference to wavelength and wave speed, why A S B observer B will detect sound of a higher frequency than observer A. 1m frequency observed by B b. Determine the ratio frequency observed by A Solutions a. B observes a shorter wavelength of sound than A, bec ause at the position of B the wavefronts are closer to each other than at the position of A. The speed of the waves is the same according to both observers bec ause they c are both stationary relative to the medium. Since f = , the shorter wavelength observed by B results in a higher λ frequency. λ A b. The neighbouring wavefronts are 2.5 m apart at A and 1 m apart at B. Hence = 2.5. λ B c frequency observed by B λ λ A B = = = 2.5. c frequency observed by A λ B λ A LHA C alculating the observed frequency due to the Doppler eect (a) Moving source — Stationary observer The change in frequency c an be related to the speed of the wave v, the frequency emitted by the source f and the source speed u s v v f f u s A B S S u s f ▴ Figure 7 Modelling Doppler shi when the source is moving and the observer is stationary. Figure 7 shows the source position S as the point source emits its first wave. One 1 time period T later, the source has moved to position S′. T is. The distance SS′ f u s is the source speed × one wave period, which is. The radius of the wavefront f v after time period T is. During this first time period, the wave emitted when the f source was at S has expanded to become a sphere of diameter AB. Focus on the wavefront at B which is the position of the observer. At the point S′, the source is just emitting another wavefront, so that from the observer ’s point of view the apparent wavelength of the wave is the distance from S′ to B. This is v u (v – u ) s s , which c an be written as. ( ) f f f 459  Topic C.5 Doppler effect LHA The observed (apparent) frequency f ′ as detected by the observer is wave speed in medium v = apparent wavelength (v u ) s f This c an be rearranged to v f ′ = f ( ) v u s This equation predicts that, when the source is moving towards an observer at B, the observed frequency will increase (bec ause the denominator in the fraction is smaller than the numerator). An observer at A on the other side of the source will see the opposite effect. S is v u s moving away and the apparent wavelength is now increased to +. When the f f equation is rearranged as before, the observed frequency becomes v f ′ = f ( ) v + u s and f ′ is smaller than f Worked example 2 1 An ambulance moves at a constant speed of 30.0 m s towards a stationary observer. The siren 1 of the ambulance emits a sound of frequency 1600 Hz. The speed of sound in air is 340 m s. C alculate: a. the frequency of the sound heard by the observer b. the observed wavelength. Solutions 1 1 a. We have v = 340 m s and u = 30.0 m s. The observed frequency is higher than the emitted s frequency, so we must use the minus sign in the Doppler eect equation: v 340 f ′ = f = 1600 × = 1750 Hz. v u 340 30 s 1 b. The observed speed of sound is 340 m s , so the wavelength becomes v 340 λ′ = = = 0.194 m. f ′ 1750 Worked example 3 A train moving at a speed u along a straight track sounds a whistle of frequency 1300 Hz. The frequency heard by an observer standing next to the track is 1250 Hz. a. State the direction of motion of the train relative to the observer. u b. C alculate the ratio of the speed of the train to the speed of sound. v Solutions a. The frequency heard is less than that emitted. Hence the train is moving away from the observer. b. The observer is stationary, and we c an nd the speed of the train u by solving the Doppler eect v f u f 1300 equation: f ′ = f ⇒ u = − 1 v. From here, = − 1 = − 1 = 0.040. The train is ( ) v + u f ′ v f ′ 1250 moving at 4% of the speed of sound. 460  C. Wave behaviour LHA Practice questions In questions 1 and 2, assume that the speed of sound in air 3. A police c ar moves along the line joining two 1 is 340 m s stationary observers P and Q. The speed of the c ar is 10% of the speed of sound in still air. The siren of the 1. A loudspeaker emitting sound with a frequency of c ar emits a sound detected by both observers. 850 Hz moves in a circle at a constant linear speed of 1 12 m s. An observer, at rest relative to the centre of direction of motion P Q the circle, is some distance away and in the plane of the circle. C alculate the maximum and the minimum The frequency of the sound measured by observer P is f. frequency of the sound heard by the observer. What is the frequency measured by observer Q? 2. A train approaching a station sounds a horn of 11 10 12 11 A. f B. f C. f D. f frequency 900 Hz. A stationary observer on the station 10 9 10 9 platform measures a frequency of 970 Hz. C alculate the speed of the train. (b) Stationary source — Moving observer Figure 8 shows the arrangement. The observer moves towards the source with a speed u. The speed of the wave in the medium is v o P u o u × T’ v × T’ o ▴ Figure 8 Modelling Doppler shi when the observer is moving towards the stationary source. The observer at O begins exactly one wavelength (in the medium) away from the source at S. At this instant, another wavefront is emitted by the source just as the observer crosses the previous wavefront. After a time T ′, the observer is at point P on Figure 8 and observes the second wavefront which has travelled from S in the time T ′. This means that the true wavelength λ in the medium is λ = u × T ′ + v × T ′ o The wavelength λ is equal to the product of the speed v in the medium and the time period T as emitted by the source: λ = v × T 1 1 Therefore v × T = u × T ′ + v × T ′and, bec ause f ′ = and f = , this leads to o T ′ T v u v o = + f f ′ f ′ Simplifying this equation gives v + u o f ′ = f ( ) v When the observer moves away from the source the sign is negative: v – u o f ′ = f ( ) v 461  Topic C.5 Doppler effect LHA Worked example 4 1 Waves of frequency 0.50 Hz travel across the surface of a lake at a constant speed of 2.5 m s. A boat is moving at right angles to the wavefronts. A passenger in the boat observes that the boat crosses the wavefronts with a frequency of 0.70 Hz. C alculate the speed of the boat relative to the lake. Solution v + u f ′ o We need to solve the Doppler equation for the unknown speed u of the boat. f ′ = f ⇒ u = − 1 v o o ( ) v f 0.70 1 1 We substitute v = 2.5 m s , f = 0.50 Hz and f ′ = 0.70 Hz. u = − 1 × 2.5 = 1.0 m s o ( ) 0.50 Worked example 5 1 A bat is ying in a c ave at a constant velocity of 6.00 m s towards a at vertic al wall. The bat emits an ultrasound pulse of frequency 45.0 kHz towards the wall. The ultrasound is reected o the wall and returns to the bat. C alculate the frequency of the ultrasound that the bat will hear, assuming that the 1 speed of sound in air is 340 m s Solution The bat is a moving source of waves travelling towards the stationary wall, and the frequency of the 340 ultrasound reaching the wall is 45.0 × = 45.8 kHz. The wall reects the ultrasound with no 340 6.00 change in frequency, and the bat becomes a moving observer of the approaching wave. The frequency 340 + 6.00 heard by the bat is therefore 45.8 × = 46.6 kHz. 340 Practice questions In questions 4 and 5 assume that the speed of sound in air 5. A stationary motion sensor emits a sound pulse of −1 is 340 m s. frequency 80.0 kHz towards a c art approaching the −1 sensor at a speed of 18.0 m s. The sound is reected 4. A stationary siren emits sound of frequency 800 Hz. o the front of the c art and returns to the sensor. a. A cyclist is moving towards the siren with a −1 Determine the frequency of the returning sound. constant speed of 9.0 m s. C alculate the Hint: consider the c art as (1) a moving observer of frequency heard by the cyclist. the emitted pulse and (2) a moving source of the b. Passengers of a c ar moving away from the siren reected pulse. hear a frequency of 750 Hz. C alculate the speed of the c ar. Predictions — When both source and observer move The equations c an be combined for the general c ase where the source has speed u and the observer has speed u : s o v ± u o f ′ = f ( ) v ∓ u s The upper sign is used when the source and observer are approaching; the lower sign when they are moving apart. You will not be asked questions involving the simultaneous movement of source and observer in the IB Diploma Programme physics examinations. 462  C. Wave behaviour Doppler eect and light What are the similarities The nature of electromagnetic radiation means that the Doppler analysis above and differences between c annot be used for light or any other part of the electromagnetic spectrum. This is light and sound waves? bec ause, as you saw in Topics A.5 and C.2: Light waves and sound waves Electromagnetic radiation does not require a medium through which to travel. have many similarities. Both types A postulate of special relativity states that the velocity of light waves is of wave demonstrate the wave constant for all inertial frames (page 164). This is not the c ase for waves that properties described in earlier require a medium such as sound. For example, when a sound source moves, in Theme C. Reflections of light the wavefronts in the medium are closer than they should be. may be re-named as echoes in sound, but the basic description The motion of a source and an observer c annot be distinguished and the of the phenomenon is the same. If concepts of source speed and observer speed have no meaning in special you fill a toy balloon with c arbon relativity. They must be replaced by a relative velocity. dioxide gas (with its high “optic al” However, when the relative speed v between source and observer is very density), it acts as a very effective much less than the speed of light c (v ≪ c), the equations derived earlier are lens for sound waves. Both sets approximately correct. However, modific ations to the equations are required: of waves c an undergo diffraction and interference with diffracting The wave speed symbol v changes to c apertures and source separations The source speed or observer speed becomes v. (There is now no need to of an appropriate size. distinguish between them.) However, the mechanisms that v The moving source–stationary observer equation f ′ = f changes lead to these effects in light and ( ) v + u s sound are very different. Light c c to f ′ = f. The equation f ′ = f c an be re-written as ( ) ( ) reflection involves absorption c + v c + v and re-emission of photons, 1 v f ′ = f = f 1 +. The binomial theorem allows this to be ( ) whereas sound reflection involves c + v c ( ) compressions and rarefactions ( ) c 2 3 at a solid boundary. Also, v v v … expanded as f ′ = f 1 − + + +. As v ≪ c, only the rst ( ( ) ( ) ( ) ) electromagnetic radiation in a c c c vacuum has a universal speed v two terms are signic ant so that f ′ ≈ f 1 −. (You are only required to ( ) (which underpins the theory of c know this result, not its derivation.) A further simplific ation is possible: the special relativity in TopicA.5) which leads to differences fv change in observed frequency Δf is (f ′ f), so that Δf ≈ , which leads between the Doppler behaviour c Δf v oflight and sound waves. to ≈ providing an equation for the fractional change in frequency. f c It is straightforward to extend the equation to the fractional change in You will only need to use the wavelength, giving result for the fractional changes Δf Δλ v = ≈ in frequency and wavelength. f λ c The derivation given here is an Re m e m b e r that this set of equations is only true when the relative explanation and is not required as speed v between source and observer is much l e ss than that of part of the course. ele ctromagnetic ra d i a t i o n c 463  Topic C.5 Doppler effect What happens if the speed of light is not much larger than the relative speed between the source and the observer? The Doppler effect for sound c an seem quite different from that for electromagnetic waves both algebraic ally and in its origin. A relativistic treatment should really be used for both. However, as mentioned earlier in this topic, the speed of Δf Δλ v sound waves is so much less than c, so = ≈ is a good approximation. f λ c When v and c are closer, then a full relativistic treatment is required. The ideas of Topic A.5 lead to a full relativistic expression for the Doppler effect which applies to all waves: 2 v S v 1 – R 1 – ( ) ( ) c c S f R 2 = × v R f v 1 – S S 1 + ( ) c ( ) c S Here v and v are the speed of receiver and source, respectively. It is no longer appropriate to talk in terms of source R S and observer. The speed of sound in the medium is c. When v , v , and c are much smaller than c, then the value of the S R S S term within the square root is close to 1 and v R 1 – ( ) c S f R ≈ f S v S 1 + ( ) c S which is the same as the approximate expression earlier. Another special c ase is c = c, which gives the relativistic Doppler equation for light. S Worked example 6 A sodium lamp emits light of wavelength 588.995 nm. Calculate the wavelength according to an 1 observer moving towards the lamp at a speed of 25 km s Solution The change in the wavelength c an be c alculated from the Doppler equation: 3 v 25 × 10 2 Δλ ≈ λ = 588.995 × = 4.9 × 10 nm. The distance between the source and the 8 c 3.00 × 10 observer decreases, so the observed wavelength is shorter than the emitted wavelength. λ = 588.995 – 0.049 = 588.946 nm. obs Worked example 7 An ultraviolet wavelength of 85.0 nm is observed in the spectrum of a distant galaxy. A corresponding stationary source on E arth emits a wavelength of 78.0 nm. a. Explain the direction of motion of the galaxy relative to the E arth. b. C alculate the relative speed of the galaxy. Solutions a. The observed wavelength (85.0 nm) is longer than the emitted wavelength (78.0 nm). Hence the galaxy is moving away from E arth. Δλ 85.0 78.0 2 b. v ≈ c = c = 8.97 × 10 c. The galaxy is receding at approximately 9% of the λ 78.0 speed of light. 464  C. Wave behaviour Practice questions 6. The frequency of light reaching E arth from a 7. A spaceship approaches a space station at a relative distant galaxy is 1.2% lower than the frequency of a speed of 0.050 c. The space station emits a navigation corresponding stationary source. signal of frequency 2.80 MHz towards the spaceship. C alculate the frequency of the signal received by a. C alculate, in terms of c, the relative speed of the the spaceship. galaxy. State whether the galaxy is approaching or moving away from E arth. b. A stationary source emits a spectral line of wavelength 527.0 nm. C alculate the wavelength of the same line observed in the light from the galaxy. Applic ations of the Doppler eect Medic al Ultrasound consists of sound waves that have a frequency greater than 20 kHz transmitter–receiver ultrasound beam and that are inaudible to most people. It has several uses in medicine. One of blood these is the non-invasive measurement of the speed of blood flow in blood cells θ vessels. An assessment of the speed and quality of flow is important in the v diagnosis of some heart and tissue diseases. direction