Basic Physics III Lecture 8
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Basic Physics III8 / 27
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JointCaliberMicrowaveSpare partInterference (wave propagation)Interference (wave propagation)LoudspeakerPhase (matter)Separation processBrillouin zoneCosmic distance ladderAmmeterWavelengthAudio frequencyWind waveSoundSpare partSignal (electrical engineering)VideoCartridge (firearms)AmplitudeNegativer WiderstandMultiplizitätEnvelopeStock (firearms)DefecationSolidFuelTypesettingRulerBird vocalizationOrder and disorder (physics)Power (physics)SizingSummer (George Winston album)Speckle imagingYearSunlightToolFahrgeschwindigkeitScoutingRoll formingPackaging and labelingDampfbügeleisenBlanking and piercingMondayLecture/Conference
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CaliberParticle physicsAudio frequencyWind waveEnvelopeDirect currentComputer animationDiagram
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Source (album)CaliberOptischer HalbleiterverstärkerSpare partMeasuring instrumentSource (album)FuelAudio frequencyJuneCartridge (firearms)FlightActive laser mediumBending (metalworking)Effects unitCrest factorWavelengthCosmic distance ladderHot isostatic pressingSoundOrbital periodDoppler-VerschiebungBird vocalizationFire apparatusString theoryElectric guitarRing strainTone (linguistics)Car tuningWind waveSirenLightPitch (music)VakuumphysikSpeed of soundComputer animationLecture/Conference
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Audio frequencyLambda baryonSpeed of soundOrbital periodWavelengthLecture/Conference
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Source (album)RankingGoldsmithCaliberGentlemanSmart meterHull (watercraft)ElektronentheorieActive laser mediumParticle physicsFahrgeschwindigkeitDoppler-VerschiebungSource (album)SoundWavelengthAudio frequencyCartridge (firearms)TypesettingSpeed of soundDirect currentWind waveKickstandCrest factorTruckRailroad carApparent magnitudeActive laser mediumCommand-line interfaceAtmosphere of EarthLambda baryonOrbital periodFACTS (newspaper)Faraday cageZirkulatorElectronic mediaEisengießereiRulerSpare partSirenProzessleittechnikYearCar dealershipWinterCold inflation pressureBird vocalizationSchwimmbadreaktorContinuous trackEffects unitComputer animationLecture/Conference
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Active laser mediumCaliberParticle physicsBrigantineMinuteSource (album)FahrgeschwindigkeitMeasurementInkSonic boomChandrasekhar limitLarge Hadron ColliderKopfstützeRailroad carAudio frequencyCartridge (firearms)SensorSonic boomSpeed of soundReflexionskoeffizientInterface (chemistry)GunRadarVolumetric flow rateWind waveFahrgeschwindigkeitSource (album)Weather frontWindSirenField-effect transistorCrest factorWavelengthDirect currentSoundDoppler-VerschiebungInterference (wave propagation)PhysicistPlane (tool)Intensity (physics)Shock waveAngle of attackGasPressure vesselFunkgerätEffects unitFACTS (newspaper)Bird vocalizationProzessleittechnikAmmeterKey (engineering)Hot isostatic pressingMint-made errorsHammerFood storageHeat exchangerHitzeschildSchubvektorsteuerungSpare partRoll formingFighter aircraftLecture/Conference
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Sonic boomJet (brand)SoundScabiesCaliberParticle physicsOptischer HalbleiterverstärkerLuggerEffects unitSonarWavelengthDiffractionEcho <Ballonsatellit>Field-effect transistorString theorySpeckle imagingAudio frequencyHarmonicAtmosphere of EarthDensityCartridge (firearms)Vacuum tubeTransducerEcho (phenomenon)Amateur radio repeaterWater vaporPitch (music)Cosmic distance ladderJet aircraftSpeed of soundFighter aircraftWind waveLongitudinal waveSound pressureDoppler-VerschiebungSoundMultiplizitätShock wavePlane (tool)Weather frontFundamental frequencyAbsolute threshold of hearingFahrgeschwindigkeitSonic boomSpare partCrest factorSource (album)MotorjachtAstronomisches FensterFinishing (textiles)Electronic mediaActive laser mediumIntensity (physics)MaterialDirect currentBe starReflexionskoeffizientCrystal structureSensorStromschlagBleisatzBird vocalizationJet (brand)AmmeterLastComputer animation
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Lecture/ConferenceComputer animation
Transcript: English(auto-generated)
00:04
You may remember the phenomenon of interference of waves in general. And like any other wave, sound waves also interfere. For example, if you have two loudspeakers separated
00:20
by some distance d that produce waves that are in phase with each other, then depending on where you receive the sound, where you listen, you will observe constructive, destructive, or partially destructive interference. For example, at point c, the distance between loudspeaker
00:44
b and loudspeaker a is the same. And therefore, both waves will constructively interfere and produce a bigger signal. However, if you move to some other position, for example, point d, the distances are now different.
01:02
If I look at the triangle between loudspeaker a, point d, and point b, and mark off the shorter distance a, d along the line of d, e, and count how many wavelengths is the difference in path length between loudspeaker
01:20
a and loudspeaker b, then if this difference is, that is, distance b, e equals a half a wavelength or three halves of a wavelength or five halves of the wavelength and so on, then destructive interference will result. However, if it's an integral multiple of the wavelengths,
01:44
then constructive interference will result. So let me demonstrate that with another of Professor Casper's programs. So here we have our two loudspeakers, the blue and the red.
02:01
And now we start the sound from both of them. And we choose some position. And what you see in this little rectangle here is the oscillations as a function of time
02:20
due to wave a, the red one, or the blue loudspeaker and the sum. And you can see that in this case we have constructive interference, because the sum, the green, is roughly twice as big as the red or the blue. However, if I move to a different place,
02:44
for example, over here, then we have almost complete destructive interference. You can see that the red and the blue are out of phase with each other, and therefore the sum is almost exactly equal to zero.
03:03
You can also change the separation between the loudspeakers. And then the distances between constructive interference here and destructive interference get smaller.
03:22
So we get more zones of constructive and destructive interference than if they are closer together. Or we can change the frequency, and then we
03:42
get even more zones of constructive and destructive interference next to each other. So let's calculate an example. So let's assume that the two loudspeakers are one meter apart, and a person stands four meters away
04:02
from one of the speakers. The question is, how far away from the other speaker does he have to be to detect destructive interference when the loudspeakers emit an 1,150 hertz sound? So the first problem we have is to determine the wavelength.
04:23
And we know that the wavelength is the wave velocity divided by the frequency. And the wave velocity is the sound velocity, or 343 meters per second. And we divide that by the 1,150 hertz, or 1,150
04:42
per second of the frequency, and we get the wavelength as 0.298 meters, or roughly 30 centimeters. So that we hear destructive interference, the difference in distance must be equal to half a wavelength, or 3 halves of a wavelength,
05:02
or 5 half of a wavelength, and so on. And that means that we have several possible distances. For example, we can have 15 centimeters less, or 15 centimeters more to the 4 meters, or 3.85 meters,
05:22
and 4.15 meters. Or we can have 45 centimeters more or less than the 4 meters, which is 3.55 meter, or 4.45 meters. Or we can have 75 centimeters more or less than the 4 meters.
05:41
And that's the 3.25 meter and the 4.75 meter. The next set, which would be roughly 1 meter and 5 centimeter difference, that cannot occur anymore because the loudspeakers are only 1 meter apart. So if you are 4 meters away from one of the loudspeakers,
06:04
you can't be 5.04 meter away from the other loudspeaker, or 2.96 meter away from the other loudspeaker. Let me now talk about a different type of interference. The interference that we discussed so far basically
06:23
happens in space. We look at different positions in space and see whether we observe constructive or destructive interference. But if you have two waves of slightly differing frequencies and you look at a fixed position in space,
06:41
then you get constructive and destructive interference at different points in time, different instances in time. As usual with interference, we simply sum the two waves. So we have those two waves, which have slightly different frequencies.
07:00
And at one particular point, we know that the equation of motion is given by a simple harmonic motion with the frequency of the first wave, in this case, and the second wave, in that case. And this is the amplitude of both waves, which we assume to be the same.
07:20
And we use this formula from trigonometry, the sum of sine x and sine y equals twice the sine of the average of x and y times the cosine of one half of the difference. And if you apply that to this sum,
07:41
then we get twice the amplitude times cosine of the difference divided by 2 times the time multiplied with the sine of the average frequency times the time. And we can consider this as a time-dependent amplitude.
08:05
This will be relatively slowly varying compared to this. And so when we look at this sum, which is indicated here of those two waves, the blue and the pink, and we get this red sum wave, and we
08:22
can define an envelope, which is this first part of the result, which is also time-dependent but varies much more slowly. And this is how this envelope will look like.
08:40
So one more of Professor Kasper's demonstration program. So first, let's just look at two waves. And let's now detune them slightly in frequency, both traveling in the same direction.
09:00
And if I allow them to interfere, then I get the green sum of the two frequencies. And we can now look at the envelope as a function of time. And we can try what happens if we detune the frequencies further.
09:26
And we see that the envelope starts to vary faster and faster. And eventually, it is kind of difficult to see it as an envelope at all.
09:42
So this kind of phenomenon is usually referred to as a beat. So the frequency A is beating against frequency B. So let's illustrate this with an example as well. So let's say we have a tuning fork, which
10:02
produces a steady 400 hertz tone. And this tuning fork is struck and then held near a vibrating guitar string. And if you do that, then you can hear distinct beats. And if you count them, you find out
10:21
that there are 20 beats for every five seconds. And the question is, what are the possible frequencies produced by the guitar string? And the answer is, the beat frequency is 20 beats divided by 5 seconds, or that's about 4 hertz.
10:42
And the beat frequency, as shown on the previous slide, the beat frequency is related to the difference of the individual frequencies. And therefore, the difference between the tuning fork and the guitar string must be 4 hertz.
11:02
Or it is either two possibilities, either 396 hertz or 404 hertz. So one can make use of these beat frequencies to tune instruments very precisely. Let's now talk about another complex phenomenon
11:20
that occurs for sound waves, but also for any other kind of waves. And it's called the Doppler effect. The observation is as follows. If you have a fire truck driving by with its siren going, then as it approaches you, you hear a higher pitch.
11:42
Then when it passes you and when it goes away from you, you hear a lower pitch. So this phenomenon is called Doppler effect. And it occurs when either the sound source or the listener or the medium is moving. In the case of light, a Doppler effect exists also,
12:02
but the only motion that is relevant there is the source relative to the observer, at least if the light is moving in vacuum. If the source is moving towards the listener, the frequency is increased. And if the source is moving away from the listener,
12:20
then the frequency is decreased. Let's first consider just a moving source. So in the case of a moving source, let's first consider the source not moving. So if the source is not moving, it emits wave crests. And we can count those and we can determine the wavelength.
12:41
For example, this distance here is supposed to be exactly one wavelength. If you compare that situation to that where the source is moving, we would observe a slightly different wavelength. Because by the time we hear this crest,
13:00
the source has moved a little bit. And when you now measure the distance to the next crest, it's somewhat closer than it was before. And it is closer by the speed of the source times the time between two crests, which is the period of the wave. So in other words, the lambda,
13:21
the wavelength, appears to be shorter by the distance v source times period of the wave. Since we know that the time period is inversely related to the frequency, and we know that the frequency times
13:41
wavelength equals sound speed, that doesn't change. So time period is one over the frequency. And lambda divided by the sound speed. So we have the sound speed is frequency times wavelength,
14:05
which is one over the period times wavelength. And so if I multiply by t over v, then we get t equals lambda over the sound speed.
14:30
If we now enter this expression into the time period t in this formula, we get that lambda prime,
14:42
the changed wavelength, is lambda minus lambda times the source velocity divided by the sound velocity. And we can pull out the wavelength lambda out of the expression and get that lambda prime equals lambda times one minus v source over v sound.
15:04
However, we also know that the frequency that is associated with that wavelength must be the sound speed divided by that wavelength. And if we substitute this into that equation,
15:22
then we get that the changed frequency f prime equals the sound speed divided by that expression, or the frequency prime equals the frequency divided by the factor one minus v source over v sound.
15:42
And indeed, if f prime is bigger than f, since the denominator gets smaller, the more you increase the speed of the source. Similarly, we could derive an expression if we are moving away from the listener. And in that case, the lambda prime gets longer by this amount.
16:04
So we basically replace this minus sign, which comes from that minus sign, by a plus sign. And so in that case, f prime equals f over one plus v source over v sound. In that case, this frequency is smaller
16:22
since the denominator is getting bigger. So unfortunately, Professor Kasper doesn't have a program for this case. But I found one on the web. So if you type fizzled into Google,
16:43
you get to this web page, and it runs this Java application where we can play with the Doppler effect as far as moving sources are concerned. So right now, the source isn't moving at all.
17:00
So we observe nice circular waves going in all directions. But I can now start to move the source with a higher and higher velocity. And now you see that the wavelength
17:23
in the direction of motions becomes smaller while the wavelength away from the direction of motion becomes bigger. And perpendicular to the direction of motion, it is about the same as it was before.
17:42
And so if you now imagine standing here and this being the fire truck, you hear a bigger frequency, meaning more crests per second. Then as the truck passes you by, you hear the normal frequency of the siren. And when the truck has passed you,
18:02
then you hear a lower frequency than before because you have fewer crests per second. So the next question we ask ourselves is what happens if not the source is moving but the listener? So let's say you're sitting in a car yourself and you're listening to some sound that is stationary.
18:24
What happens in that case? What happens in that case is that the wavelength is the same because the source isn't moving but the apparent sound velocity is increased. If you're moving towards the source, then the two velocities add.
18:43
So the changed sound velocity, v prime, is the sum of the normal sound velocity plus the motion of the listener. And if you plug that into our usual expression relating frequency and wavelength,
19:02
then we get that the changed frequency is f times one plus the velocity of the listener over the velocity of sound. And in this case, f prime is also bigger than f because now this expression is in the numerator.
19:21
And the numerator is getting bigger, therefore f prime is bigger than f. And similarly, if we move away from the source and the two velocities subtract, and we get this expression. These two expressions are different
19:40
than what we got from the moving source. Even if we change the sign of the velocity, what we have here is that it matters whether you move towards the source or the source is moving towards you or you moving away from the source
20:01
and the source is moving away from you. So in other words, this is a manifestly non-relativistic effect, meaning that it doesn't just matter what the relative velocity of u with respect to the source is, but it also matters how fast
20:21
are you moving against the medium, which in this case is the air. So what happens if both the source and the listeners are moving and we can combine both formulas, or all four formulas into one. So the change frequency is f times
20:42
one plus or minus v listener over v sound divided by one minus or plus v source over v sound. The plus for the listener means that the listener moves towards the source and the minus for the listener means that it moves away.
21:00
For the source, the plus means that the source moves towards the listener and the minus means that it moves away from the listener. You basically just have to remember this one formula if you can keep those signs straight.
21:22
So there's a third option and that is if neither the source nor the listener is moving, but the medium is moving. So far we have always considered that the air is still, which transports the sound.
21:40
If there is some wind, then there's also a Doppler effect. And there we can use the principle of relativity that physics must be equivalent whether we move with an absolute motion or not.
22:01
And so we can replace the moving medium with a moving source and a moving listener. For example, let's say source and listeners are still, but there's a wind of some velocity vwind that's going towards the listener. We can replace that scenario
22:21
with one where the listener approaches the source with the same velocity and the source is moving away from him with that same velocity. Then we can use the previous formula. Let me now calculate two examples of the Doppler effect. The first problem is
22:43
we hear the siren of a police car at rest and it emits a predominant frequency of 1600 Hz. The question is what frequency will you hear if you are at rest and the police car moves at 25 m per second
23:02
towards you or away from you. For the A part, we assume that it's moving towards you and in this case we have a moving source and so f prime equals f divided by one minus v source over v sound.
23:21
The source velocity is 25 m per second. The sound velocity is 343 m per second. And if you plug in the numbers, then the frequency is... If the police car is moving away from us, then the minus sign gets replaced by a plus sign
23:42
and if you plug in the numbers, then we get 1490 Hz. Let's now consider a more complicated case. Let's say we have a stationary source of 5000 Hz and the sound wave reflects from an object which moves with 3.5 m per second
24:02
towards the source. The question is what is the frequency of the wave that is reflected by the moving object as detected by a detector at rest near the source. What happens here? We have a stationary source. It emits a wave
24:22
and as far as the reflection on that interface is concerned, that is kind of like this object is the listener. The listener is moving and then it emits the same frequency as this moving listener back again but that is to the detector
24:43
a moving source. We first calculate what is the frequency that the object receives. That frequency is f' equals f times 1 plus v listener over v sound where v listener is the speed
25:01
of the object. If we plug in the 3.5 m per second, we get 3.5 m per second over 343 m per second and the 5000 Hz changes to 5050 Hz. Now the frequency that is then emitted by the object
25:22
to a non-moving listener so the detector that is addressed with the source is then this frequency divided by 1 plus v source over v sound where v source is again the 3.5 m per second. In this case the source is moving
25:42
towards us and so we get a minus sign. If we do that, the 5050 Hz changes to 5100 Hz. In summary, instead of 5 kHz reflection which we would have gotten from a stationary object,
26:03
we get a 5.1 kHz reflection. If we have a frequency dependent detector, we can actually measure the speed of the object. This is exactly how a radar gun works for police cars. Only that in this case it's radar waves rather
26:22
than sound waves. One can also use this technique with sound to detect things like blood flows or heartbeat in the human body. Let's take the Doppler effect to an extreme. Let's go back to
26:40
our moving source but let's increase the speed by a lot. In fact, so much so that the speed of the object is faster than the speed of the sound. What happens in that case is that the object passes its own
27:01
sound wave crests that it produced. There's no more sound in front of the object. All the sound is behind the object. There is constructive interference of a lot of wave crests just at this
27:21
boundary here. Those are all the wave crests overlapping each other. This is constructive interference and that means that right there it's very loud. This is usually referred as the sonic boom. To recap,
27:40
if the object is at rest we get circular wave crests surrounding it. The wavelength doesn't change in any of the directions. We start moving and the wavelengths get shorter and shorter in the direction of motion. If we are at the sound speed then all the
28:01
wave fronts meet just in front of the object. If we go past that speed then there is a conical region where sound is emitted. Since there's constructive interference of all those wave crests here, there's very high intensity
28:21
just at this boundary. That is also called a shock wave. This comes about because sound waves can't keep up with the motion of the object. We can actually define a so-called Mach number after the physicist Mach, which is
28:42
the ratio of the object speed divided by the sound speed. Plane moving with Mach number 2 means it moves twice as fast as the speed of sound. This Mach number is related to the opening angle of this cone. Opening angle means half the angle
29:02
of this conical shape. The sine of this opening angle is V object over V sound. It should be V sound over V object. Therefore, this is inversely related to the Mach number.
29:20
A typical example of such supersonic booms is the crack of a gunshot. That's not to be confused with the explosive release of gases from the gun, which you can shield or dampen with a muffler. The actual shot itself produces
29:41
a snap or a crack and that is such a shockwave produced by the supersonic bullet. A more familiar example perhaps is a supersonic jet fighter. In this case multiple shockwaves are emitted. Basically, every part of the plane is producing such a shockwave.
30:02
In this picture the shockwaves are made visible. You see, for example, the shockwave of the nose here. There is also one produced here. In general, one can distinguish those that are produced by the front of the plane and those that are produced by the tail of the plane.
30:21
The reality is more complicated even than just two shockwaves. Or we can look at water waves and a speedboat. Then the same phenomenon happens. The speedboat if it goes very fast, it passes over the crest of its own wake
30:41
and then produces a shockwave that sort of moves in a somewhat conical shape away from it. In the case of the jet plane the shockwave since it is very intense can cause considerable damage. For example, it can burst windows.
31:01
Let me finish today with various applications of sound. The first one that comes to mind are sonars. Sonars detect underwater objects by using the time delay of an echoing sound pulse. In a sonar you emit a sound pulse and then you wait
31:22
for the reflection coming back to you. You judge from the reflection what kind of objects are near you. Or, alternatively with the same principle, one can probe the eternal structure of Earth using waves
31:41
travelling through the Earth. Often these waves are produced intentionally with explosions. But one can also use earthquake waves that happen naturally. To maximize resolution in the case of
32:01
sonars, one usually uses ultrasound waves. That is because the wavelength is shorter and therefore the diffraction effects are smaller. A kind of sonar is also used in
32:21
medical imaging. So-called sonograms are images used for diagnostic purposes. It uses the same principle as the same pulse echo technique as a sonar. Typically the frequencies used are between 1 and 10 MHz.
32:40
Echoes are observed from any kind of boundary in the body where the density changes. For example, organs or lesions or tumours or fluid pockets or anything else. In that case you have a transducer that you
33:00
place outside the human being and then it produces a sound, it reflects off those various ingredients that you have in the human body and for each one of them you receive an echo. Then you repeat this procedure by placing the transducer at a slightly different position.
33:20
Like in this picture here. So if you make many such traces, you observe different kind of echoes and you can form a kind of image of what is happening inside the body. What you need to know of course is the sound velocity in the human body which is roughly the same as that of water or 1540
33:42
m per second. The time delay between emitting the pulse and observing it is related to the distance to the boundary, twice the velocity times delta t. The intensity of the echo tells you something
34:00
about the difference in density of the various media. The most famous example probably is to look at pictures of a human fetus. For example here is one. One can see that it is possible to make very detailed pictures. We can then make use of the Doppler effect
34:22
to detect whether there is a heartbeat in the fetus or if it's used for other diagnostic purposes you can detect the motion of the blood stream etc. Let me now summarize chapter 16. First of all, sound means both a sensation in our
34:42
consciousness as well as a physical phenomenon that originates that sensation. Sound waves are longitudinal waves which means that air molecules for example vibrate along the direction of propagation. Sound is transported in air or other materials.
35:01
The sensation of pitch is related to the frequency. Human beings can hear frequencies between 20 Hz and 20,000 Hz. The sensation of loudness is related to the intensity and the way we perceive loudness is proportional to the
35:21
logarithm of the intensity. Or we can define the sound level as 10 decibels times the logarithm of the intensity I divided by the intensity I0 where I0 is taken as the human hearing threshold or 10 to the minus
35:41
12 watts per meter square. The sources of sound are vibrating objects and typically standing waves are produced in those vibrating objects. The frequency of the sound wave in air is the same as the frequency of the vibrating
36:01
object but the wavelength is typically different. If the vibrating object is a string of some length or if it's an open air column in a tube that is open on both sides then the fundamental wavelength of the fundamental mode is given by twice the length
36:21
and then odd and even higher harmonics exist. And for closed air tubes which means that one end is closed the other one is open, the fundamental wavelength is four times the length and only odd harmonics exist. The wavelength of the nth harmonic is given by
36:42
the fundamental wavelength divided by n and the frequency is given by the fundamental frequency multiplied with n. Finally the Doppler effect changes the pitch of sound or the frequency if the source or the listeners are moving using
37:01
this formula. If the medium moves one can replace that motion with an appropriate motion of source and listener. Objects moving faster than the sound velocity emit shock waves called sonic booms and sonars and medical
37:20
imaging devices use ultrasound echoes to determine distances to objects and their speed. Thank you very much.