Ch 16: Waves & SoundWorksheetSee all chapters
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Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
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Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
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Sections
What is a Wave?
The Mathematical Description of a Wave
Waves on a String
Wave Interference
Standing Waves
Sound Waves
Standing Sound Waves
Sound Intensity
The Doppler Effect
Beats

Concept #1: The Doppler Effect

Transcript

Hey guys, in this video we're going to talk about the Doppler Effect which is an interesting thing that happens to waves when the source and the listener to the source or the detector are moving relative to one another. Alright, let's get to it. The measurement of the frequency of a sound will always be the same whenever you measure it at rest. If the source is at rest and you are at rest no matter where you are in relation to the source the frequency of the sound emitted will always be the same. The frequency will not be the same if there is a relative motion between you and the source when you measure it. The Doppler Effect is the change in the frequencies of sound when in motion relative to a source.

The frequency increases when moving towards the source, when you're coming together, which is sometimes called approaching and the frequency decreases when moving away from the source sometimes when you two are moving away sometimes called receding. I want to show two figures here, I'm going to get out of the way for this, at the left we have a source at rest. This isn't physical space each of these circles is a wavefront, it's a peak of a sound wave and they are emitted with the distance being the wavelength and they are emitted at different times. A pulse, a pulse, a beat, a beat, a beat. But they are emitted at particular intervals of time and no matter where you happen to be you are going to hear the same frequency of sound because no matter where you are the wavelength between these sounds is the same. You guys see that? That wavelength is the same no matter where you measure it but what if it's in motion? This outermost ring is measured, is produced at the initial position of the source then the source moves to the right and the second ring is emitted at the second position of the source then it moves to the right and this third ring is emitted at the third position of the source then it moves to the right again and this littlest ring is emitted where it is right now. So this is with the source moving to the right. Look at what this does to the frequency, if an observer's over here it sees a huge wavelength. These waves are very very very far apart. Over here these wavelengths stack up, it's too small for even for me to even draw, so if you see when they stack up it hears wavefront, wavefront, wavefront, wavefront very very very fast. That means that the frequency is up on this side because of how rapidly those wavefronts are coming and the frequency is down on this side because of how slowly it takes the wavelengths to reach them because of how far apart those wavelengths are. So the frequency increases when moving towards the source or when the source is moving to you when you guys are approaching one another and the frequency decreases when moving away from the source or when the source is moving away from you, when you are receding. Now the equation for the Doppler effect gives us the frequency detected due to the Doppler effect, due to the relative motion between the source and the detector and it's given by V plus VD which is the speed of the detector divided by V Plus VS which is the speed of the source times the frequency of the source in minutes sometimes called the natural frequency. Alright, and now in order to use this equation with these particular signs you need to use a very particular coordinate system where your detector is at the origin and the positive axis is drawn from the detector to the source.

Let's do a quick example. A speaker emits a sound at 20 Hertz while at rest. If the speaker were moving away from you at 45 meters per second, what frequency of sound would you hear? Consider the speed of sound to be 343 meters per second. So first what's the natural frequency of the speaker? 20 Hertz. It says that's what it emits while at rest, that's its natural frequency. Next what's the speed of the source? It's 45 meters per second but what sign should I give it? Notice that it's moving away from you, you are at the detector, so the source is moving away from you so the source is moving in a positive direction so this is positive 45 meters per second. What about your speed? It doesn't mention anything about you moving so don't assume you are, assume it's zero and the speed of sound is 343 meters per second. So all we have to do is plug and chug with the Doppler Effect equation. So the frequency detected is V plus VD over V plus the VS times FS which is 343 plus, sorry not 20 20 is the frequency, this is 45 divided by the detector is zero divided by 343 plus the source is moving at 45 and the source frequency is 20 Hertz and this whole thing becomes 17.7 Hertz and it should be lower, it should have decreased because the source is moving away from you so the frequency detected should definitely be lower than the natural frequency. Alright guys, that wraps up our discussion on the Doppler effect. Thanks for watching.

Practice: A police siren emits a sound somewhere around 700 Hz. If you are waiting at a red light, and a police car approaches you from behind and passes you, moving at a constant 30 m/s, what is the frequency you hear from the siren as it approaches you from behind? What about once it’s passed you? Assume the air temperature to be 20°C.

Example #1: Two Submarines Approaching One Another

Transcript

Hey guys, let's do an example. 2 submarines approach 1 another underwater. Sub 1 emits a sonar pulse at 1 times 10 to the 3 Hertz which travels from sub 1 towards sub 2 is then reflected off of sub 2 travelling back to sub 1. If the speed of sound in water is 1500 meters per second, sub 1 is approaching sub 2 at a speed of 18 meters per second and sub 2 is approaching sub 1 at a speed of 15 meters per second, what frequency to sub 2 detect the sonar pulse at? What frequency does sub 1 detect the reflected sonar pulse at? So I'll break these up into questions A and questions B. Question A is what frequency does sub 2 detect the pulse emitted by sub 1 at? So here's 1 sub, here's the seconds sub my crappy subs, sub 1 is moving towards sub 2 at 18 meters per second. Sub 2 is moving towards sub 1 at 15 meters per second and sub 1 is emitting sound towards sub 2. So we have the source, we have the detector. The source is sub 1, the detector is sub 2. So the detected frequency is going to be V plus VD over V plus VS FS. Now the speed of sound is just going to be 1500 just like the problem says 1500 meters per second, the question is what are the signs going to be and it's emitted at 100 Hertz. What are the signs and values going to be for our velocities for the detector and source? The source is sub 1 so clearly the source should be 18 but is it plus or minus? The detector is sub 2 so clearly this should be 15 for the detective speed but is it plus or minus? Don't forget the coordinate system. The detector's at the origin and you draw a line from the detector to the source and that's the positive direction. So the detector is moving towards the source so that's positive, the source is moving towards the detector so that is negative and all of this equals a detected frequency of 1022 Hertz. So this is the frequency that sub 2 detects the sonar pulse from sub 1 at.

Now part B, same problem but now sub 2 sub 1, this one's moving at 18 still this one's moving at 15 still, but now sub 2 is acting like the source because those sonar waves are bouncing off of sub 2 and heading back towards sub 1. Remember that sound that any wave was reflected off of any surface at the same frequency so this is exactly like this sub emitting a sound at 1022 Hertz that's exactly the set up to the problem. So the detected sound is V plus VD, V plus VS FS but our detector and our source have flipped. Now sub 2 is the source and sub 1 is the detector. So this is still going to be 1500 plus some speed, some velocity. This is still going to be 1500 plus some velocity but now the source frequency is 1022 Hertz. And our coordinate system has sub 2 on the right, sub 1 on the left and remember that the coordinate system's from the detector to the source is positive so the detector's velocity is positive this is plus 15, oh sorry, this is the source I got these backwards. I had also gotten this one backwards but it still worked because the mirror image looks the same. In this case, the detector is truly on the right and the source is on the left if I were to flip this around you would see sub 2 is the detector, sub 1 is the source. This is positive they're moving towards each other the detector speed is still positive, the source speed is still negative so it worked out the exact same way the detector is moving towards the source, source is moving towards the detector so you get that same thing but now the detector sub 1 so this is plus 18. The source is sub 2 so this is minus 15 and plugging this all in we get 1045 Hertz. This is how sonar works. Sub 1 emitted a pulse at 1000 Hertz and received the pulse back at 1045 Hertz because that sub knows its own speed it can use this exact equation to find the speed of the other sub based on how large that Doppler shift is. So this is how sonar for submarines work. Alright guys, that wraps up this problem. Thanks for watching.