Describe what is meant by the Doppler effect Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations Apply the Doppler effect equations for sound Solve problems on the Doppler effect for sound Solve problems on the Doppler effect for electromagnetic waves using the approximation  f = vf/c Outline an example in which the Doppler effect is used to measure speed. Topic 11: Wave phenomena 11.2 Doppler effect

Describe what is meant by the Doppler effect.  The Doppler effect has to do with moving sound sources (or moving observers).  For example if a sound source of frequency f approaches you at a speed u s its wave fronts will bunch together and you will hear a frequency f’ which is higher than f. ( f’ > f ).  On the other hand if a sound source recedes from you at a speed u s its wave fronts will stretch out and you will hear a frequency f’ which is lower than f. ( f’ < f ). Topic 11: Wave phenomena 11.2 Doppler effect usus

Describe what is meant by the Doppler effect. Topic 11: Wave phenomena 11.2 Doppler effect PRACTICE: Who claims he is wearing a Doppler effect costume? SOLUTION: Sheldon makes this claim. Leonard Raj Sheldon

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  Consider the ice-cream truck that parks in the hood playing that interminable music.  Since the ice cream truck is not yet moving, the wave fronts are spherically symmetric. Topic 11: Wave phenomena 11.2 Doppler effect Planar cross- section The cross- section is symmetric. u s = 0

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  Whether Dobson is in front of the truck, or behind it, he hears the same frequency. Topic 11: Wave phenomena 11.2 Doppler effect U s = 0 FYI  Even though the sound waves are drawn as transverse, they are in actuality longitudinal.

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  Suppose the ice-cream truck is now moving, but ringing the bell at the same rate as before.  Note how the wave fronts bunch up in the front, and separate in the back.  The reason this happens is that the truck moves forward a little bit during each successive spherical wave emission. FYI  Don’t forget, the actual speed of the wavefronts through the stationary medium of the air is the speed of sound v (bunched or otherwise). Topic 11: Wave phenomena 11.2 Doppler effect usus

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  Dobson will hear a different frequency now, depending on his position relative to the truck.  If Dobson is in front of the moving truck, f’ > f.  If he is behind, he will hear f’ < f. Topic 11: Wave phenomena 11.2 Doppler effect usus f’ > f f’ < f

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  Now we want to look at TWO successive pulses emitted by a source that is MOVING at speed u s :  is the “real” wavelength of the sound.  Since the speed of sound in air is v, = vT where T is the period of the sound source. Note that Pulse 1 has traveled = vT in the time t = T. Topic 11: Wave phenomena 11.2 Doppler effect uSuS Pulse 1 t = 0 Pulse 1 t = T Pulse 2 t = T = vT d = u s T ’

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  ’ is the wavelength detected by the observer.  The distance d between the emission of the two pulses is simply the velocity of the source u s times the time between emissions T. Thus d = u s T.  Since = d + ’ then ’ = – d = vT – u s T. So ’ = (v – u s )T Topic 11: Wave phenomena 11.2 Doppler effect uSuS Pulse 1 t = 0 Pulse 1 t = T Pulse 2 t = T = vT d = u s T ’

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  From v = ’f’ and ’ = (v – u s )T we see that f’ = v/ ’ = v/[ ( v – u s )T ] f’ = ( 1/T )v/( v – u s ) f’ = f[ v / ( v – u s ) ]. Topic 11: Wave phenomena 11.2 Doppler effect uSuS Pulse 1 t = 0 Pulse 1 t = T Pulse 2 t = T = vT d = u s T ’ FYI  If the source is receding, replace u s with –u s. [ f = 1/T ]

Apply the Doppler effect equations for sound. Topic 11: Wave phenomena 11.2 Doppler effect Doppler effect moving source f’ = f[ v / ( v  u s ) ] v is the speed of sound. Approach (-u s ), recede (+u s ). EXAMPLE: A car horn has a frequency of 520 Hz. The car is traveling to the right at 25 ms -1 and the speed of sound is 340 ms -1. (a) What frequency is heard by the driver? SOLUTION: (a) The driver has a relative speed of u s = 0. Thus f’ = 520 Hz.

Apply the Doppler effect equations for sound. Topic 11: Wave phenomena 11.2 Doppler effect Doppler effect moving source f’ = f[ v / ( v  u s ) ] v is the speed of sound. Approach (-u s ), recede (+u s ). EXAMPLE: A car horn has a frequency of 520 Hz. The car is traveling to the right at 25 ms -1 and the speed of sound is 340 ms -1. (b) What frequency is heard by Observer 1? SOLUTION: (b) For approach use -u s. f’ = 520[ 340/( ) ] = 560 Hz.  Be sure f’ is higher since car is approaching.

Apply the Doppler effect equations for sound. Topic 11: Wave phenomena 11.2 Doppler effect Doppler effect moving source f’ = f[ v / ( v  u s ) ] v is the speed of sound. Approach (-u s ), recede (+u s ). EXAMPLE: A car horn has a frequency of 520 Hz. The car is traveling to the right at 25 ms -1 and the speed of sound is 340 ms -1. (c) What frequency is heard by Observer 2? SOLUTION: (c) For approach use +u s. f’ = 520[ 340/( ) ] = 480 Hz.  Be sure f’ is lower since the car is receding.

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  Suppose the source is stationary.  If the observer is not moving, he will hear the true frequency of the source. EXAMPLE: A sound source with wavefronts is shown. A stationary observer (the blue circle) is immersed in the sound. Observing the clock and the wavefronts, determine the frequency f of the source (and the frequency f’ detected by the observer). SOLUTION: Both frequencies are the same.  Thus f = f’ = 5 cycles / 12 s = 0.42 Hz. Topic 11: Wave phenomena 11.2 Doppler effect o 5 wavefronts pass observer in 12 seconds.

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  The diagram shows which wavefronts pass the observer in the time interval ∆t (red):  Because the wave speed is v, the length of the region in question is v∆t.  If we divide the length v∆t by the wavelength we get the number of cycles (or wavefronts). Thus #cycles detected = v∆t/. Topic 11: Wave phenomena 11.2 Doppler effect o v∆tv∆t FYI  Recalling the definition of f’ we have f’ = #cycles detected /∆t = (v∆t/ )/∆t = v/.  But v/ = f so that f’ = f. (EXPECTED)

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  Suppose the source is stationary.  If the observer is moving TOWARD the source at u o, he will hear a higher frequency. EXAMPLE: A moving observer (the blue circle) is now immersed in the sound. Observing the clock and the wavefronts, determine the frequency f of the source (and the frequency f’ detected by the observer). SOLUTION: f = 5 cycles / 12 s = 0.42 Hz.  Now 8 wavefronts pass the observer in 12 s.  Thus f’ = 8 cycles / 12 s = 0.66 Hz. Topic 11: Wave phenomena 11.2 Doppler effect o

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  The diagram shows which wavefronts pass the moving observer in the time interval ∆t:  The red region v∆t is still due to the wave speed itself (still 5 cycles in our example).  The blue region u o ∆t is due to the oberver’s speed u o (3 cycles in our example).  Now for the observer #cycles detected = (v∆t + u o ∆t)/.  Thus f’ = #cycles detected /∆t = [(v∆t + u o ∆t) / ]/∆t = v/ + u o /. Topic 11: Wave phenomena 11.2 Doppler effect o v∆tv∆tuo∆tuo∆t

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.  From f’ = v/ + u o / and the relation v = f we can write f’ = v/ + u o / f’ = f + u o /(v/f) f’ = fv/v + fu o /v f’ = f[ ( v + u o ) / v ].  If the observer is moving AWAY from the source, substitute –u o for u o. Topic 11: Wave phenomena 11.2 Doppler effect Doppler effect moving observer f’ = f[ ( v  u o ) / v ]. v is the speed of sound. Approach (+u o ), recede (-u o ). FYI  Pick the sign that makes f’ do what is expected.

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations. Topic 11: Wave phenomena 11.2 Doppler effect Doppler effect moving observer f’ = f[ ( v  u s ) / v ]. v is the speed of sound. Approach (+u s ), recede (-u s ). EXAMPLE: A car horn has a frequency of 520 Hz. The car is stationary and the speed of sound is 340 ms -1. (a) An observer approaches the car at 25 ms -1. What frequency is heard by the observer? SOLUTION: (a) For approach use +u s. f’ = 520[ ( ) / 340 ] = 560 Hz.  Be sure f’ is higher since the observer is approaching the car.

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations. Topic 11: Wave phenomena 11.2 Doppler effect Doppler effect moving observer f’ = f[ ( v  u s ) / v ]. v is the speed of sound. Approach (+u s ), recede (-u s ). EXAMPLE: A car horn has a frequency of 520 Hz. The car is stationary and the speed of sound is 340 ms -1. (b) An observer recedes from the car at 25 ms -1. What frequency is heard by the observer? SOLUTION: (b) For receding use -u s. f’ = 520[ ( ) / 340 ] = 480 Hz.  Be sure f’ is lower since the observer is receding from the car.

Solve problems on the Doppler effect for electromagnetic waves using the approximation  f = vf/c.  We have discussed the Doppler effect for sound waves, but we can also discuss it for electromagnetic waves, with this major difference: The speeds of source u s and observer u o are very much less than the speed of light c.  We define v to be the relative velocity between source and observer. Thus v = u s – u o.  Then f’, the frequency detected by the observer, is related to f, the frequency of the light source, by the formula f’ = f + (v/c)f. Topic 11: Wave phenomena 11.2 Doppler effect Doppler effect for light ∆f = (v/c)f v is the relative speed between the source and the observer.

Solve problems on the Doppler effect for electromagnetic waves using the approximation  f = vf/c. Topic 11: Wave phenomena 11.2 Doppler effect EXAMPLE: A star in another galaxy is traveling away from us at a speed of 5.6  10 6 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 c = f to find f = c/ :  f = c/ = 3  10 8 / 520  = 5.8  Hz.  The star is moving away from us so the effect is to decrease the frequency. Thus v = -5.6  10 6 ms -1. f’ = f + (v/c)f = f[ 1 + (v/c)] f’ = 5.8  (  10 6 / 3  10 8 ) = 5.7  Hz. ’ = c/f’ = 3  10 8 / 5.7  = 5.3  = 530 nm.

Outline an example in which the Doppler effect is used to measure speed. PRACTICE: The absorption spectra of stars of varying distances from Earth are shown here. The sun is the bottom spectrum (D). Which star has the highest relative velocity 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.  As a reference look at the heavy black line in D.  It is shifting to the red region. Thus its frequency is less, and thus A is receding. Topic 11: Wave phenomena 11.2 Doppler effect A B C D

Outline an example in which the Doppler effect is used to measure speed. 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 D is receding from Earth? SOLUTION: Use c = f ( f = c/ ):  From A, f = 3  10 8 /390  = 7.7  Hz.  From D, f’ = 3  10 8 /490  = 6.1  Hz.  Then ∆f = f’ – f = (6.1–7.7)  = -1.1  Hz.  Finally from ∆f = (v/c)f we get v = c∆f/f = (3  10 8 )(-1.1  ) /7.7  =  10 7 ms -1. Topic 11: Wave phenomena 11.2 Doppler effect A B C D This is the so-called “redshift” and it is used as evidence for an expanding universe. Receding…

Outline an example in which the Doppler effect is used to measure speed. Topic 11: Wave phenomena 11.2 Doppler effect EXAMPLES: The following are all examples of the Doppler effect. 1. The radar gun used by police. It uses the form v = c∆f/f to find the velocity of the car. It actually measures the difference in frequency between the emitted radar beam, and the reflected-and- returned one.

Outline an example in which the Doppler effect is used to measure speed. Topic 11: Wave phenomena 11.2 Doppler effect EXAMPLES: The following are all examples of the Doppler effect. 2. A Doppler ultrasound test uses reflected sound waves to see how blood flows through a blood vessel. It helps doctors evaluate blood flow through major arteries and veins, such as those of the arms, legs, and neck. It can show blocked or reduced blood flow through narrowing in the major arteries of the neck that could cause a stroke. It also can reveal blood clots in leg veins (deep vein thrombosis, or DVT) that could break loose and block blood flow to the lungs (pulmonary embolism). During pregnancy, Doppler ultrasound may be used to look at blood flow in an unborn baby to check the health of the fetus.

Outline an example in which the Doppler effect is used to measure speed. Topic 11: Wave phenomena 11.2 Doppler effect EXAMPLES: The following are all examples of the Doppler effect. 3. A Doppler radar is a specialized radar that makes use of the Doppler effect to produce velocity data about objects at a distance. It does this by beaming a microwave signal towards a desired target and listening for its reflection, then analyzing how the frequency of the returned signal has been altered by the storm’s motion. This variation gives accurate measurements of the radial component of a storm’s velocity.

Outline an example in which the Doppler effect is used to measure speed. Topic 11: Wave phenomena 11.2 Doppler effect EXAMPLES: The following are all examples of the Doppler effect. 4. If you look at a rotating luminous object like the sun, you see that one side is moving away from the observer while the other side is approaching the observer. Once again, the formula v = c∆f/f comes into play. The sun’s right side will be red shifted, whereas the left will be blue shifted. We can then find the speed of rotation of the sun (or any star, even if it is quite distant).

Solve problems on the Doppler effect for sound. Topic 11: Wave phenomena 11.2 Doppler effect  Since the source is receding ’ had better increase.  It will increase by how far d = VT the source has traveled.  Note that the source is NOT traveling at v, the speed of sound!

Solve problems on the Doppler effect for sound. Topic 11: Wave phenomena 11.2 Doppler effect  Consider both the stationary and moving source S.  By placing a scale from the first diagram into the second…

Solve problems on the Doppler effect for sound. Topic 11: Wave phenomena 11.2 Doppler effect  A is wrong because f goes negative. Beware!

Solve problems on the Doppler effect for sound. Topic 11: Wave phenomena 11.2 Doppler effect

Solve problems on the Doppler effect for sound. Topic 11: Wave phenomena 11.2 Doppler effect  For Doppler light, the v in the formula ∆f = (v/c)f is the relative velocity.  Because the speed of is a constant c regardless of the speed of the source of the light, its frequency cannot change.

Solve problems on the Doppler effect for sound. Topic 11: Wave phenomena 11.2 Doppler effect  For moving away the wavelength INCREASES. FYI  Regardless of who is observing, the speed of the wave is determined by the air, and nothing else. Thus the speed is V.

Solve problems on the Doppler effect for sound. Topic 11: Wave phenomena 11.2 Doppler effect  For moving observer we use f’ = f(v+u o )/v.  In terms of the given symbols f = f 0 (v + 0.1v)/v.  Thus f = 1.1f 0 IF the observer O and the source S were perfectly IN-LINE. They are not. Thus B is the answer.