Doppler Effect/Shift• By the end of the lecture students will beable to calculate frequencies of sound asthey apply to the Doppler effect/shift.
Doppler EffectA tone is not always heard at the same frequency at which it isemitted. When a train sounds its horn as it passes by, the pitch ofthe horn changes from high to low. Any time there is relativemotion between the source of a sound and the receiver of it, there isa difference between the actual frequency and the observedfrequency. This is called the Doppler effect.The Doppler effect applied to electromagnetic waves helpsmeteorologists to predict weather, allows astronomers to estimatedistances to remote galaxies, and aids police officers catch youspeeding.The Doppler effect applied to ultrasound is used by doctors tomeasure the speed of blood in blood vessels, just like a cop’s radargun. The faster the blood cell are moving toward the doc, the greaterthe reflected frequency.
DopplerEquationfD = fSv ± vDv ± vS)(fD = frequency as heard by a listenerfS = frequency produced by the sourcev = speed of sound in the mediumvD = speed of the listenervS = speed of the sourceThis equation takes into account the speed of the source of thesound, as well as the listener’s speed, relative to the air (orwhatever the medium happens to be). The only tricky part is thesigns. First decide whether the motion will make the observedfrequency higher or lower. (If the source is moving toward thelistener, this will increase fD, but if the listener is moving awayfrom the source, this will decrease fD.) Then choose the plus orminus as appropriate. A plus sign in the numerator will make fDbigger, but a plus in the denominator will make fD smaller.Examples are on the next slide.
Which way + or - ?D SNegative Direction Positive Direction
Doppler Set-ups fD = fSv ± vDv ± vS)(still10 m/sf D = 1000343343 -10 )(= 1030 HzThe horn is producing a pure 1000 Hz tone. Let’s find the frequency asheard by the listener in various motion scenarios. The speed of soundin air at 20 °C is 343 m/s.still10 m/sfD = 1000343 + 10343 )(= 1029 HzNote that these situation are not exactly symmetric.Also, in real life a horn does not produce a single tone.More examples on the next slide.
3 m/sDoppler Set-ups(cont.)fD = fSv ± vDv ± vS)(10 m/sfD = 1000343 - 3343 -10 )(= 1021 HzThe horn is still producing a pure 1000 Hz tone. This time both thesource and the listener are moving with respect to the air.fD = 1000343 + 3343 - 10 )(= 1039 Hz10 m/s 3 m/sNote the when they’re moving toward each other, the highestfrequency possible for the given speeds is heard. Continued . . .
10 m/s10 m/s3 m/sDoppler Set-ups(cont.)fL = fSv ± vDv ± vS)(fD = 1000343 - 3343 + 10 )(= 963 HzThe horn is still producing a pure 1000 Hz tone. Here are the finaltwo motion scenarios.fD = 1000343 + 3343 + 10 )(= 980 Hz3 m/sNote the when they’re moving toward each other, the highestfrequency possible for the given speeds is heard. Continued . . .
Doppler ProblemMr. Magoo & Betty Boop are heading toward each other. Mr. Magoodrives at 21 m/s and toots his horn (just for fun; he doesn’t actually seeher). His horn sounds at 650 Hz. How fast should Betty drive so thatshe hears the horn at 750 Hz? Assume the speed of sound is 343 m/s.21 m/s vDfD = fSv ± vDv ± vS)( 750 = 650343 + v D343 - 21 )(vD = 28.5 m/s