For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Wave Interference

- The total displacement of two or more interfering waves is the sum of all the individual displacements of the waves. This is known as superposition.
- Identical waves traveling in opposite directions in the same medium can create standing wave patterns. Nodes appear to stand still, while antinodes vibrate with maximum amplitude above and below the axis.
- Standing waves in string instruments and open-tube instruments create fundamental frequencies as well as 2nd, 3rd, 4th, … harmonics.
- Standing waves in closed-tube instruments create fundamental frequencies as well as odd-numbered harmonics.
- Two sound waves near-identical frequencies can interfere to create beat patterns.

### Wave Interference

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Superposition
- When More Than One Wave Travels Through the Same Location in the Same Medium
- The Total Displacement is the Sum of All the Individual Displacements of the Waves
- Example 1: Superposition of Pulses
- Types of Interference
- Example 2: Interference
- Example 3: Shallow Water Waves
- Standing Waves
- When Waves of the Same Frequency and Amplitude Traveling in Opposite Directions Meet in the Same Medium
- A Wave in Which Nodes Appear to be Standing Still and Antinodes Vibrate with Maximum Amplitude Above and Below the Axis
- Standing Waves in String Instruments
- Standing Waves in Open Tubes
- Standing Waves in Closed Tubes
- Interference From Multiple Sources
- Beats
- Two Sound Waves with Almost the Same Frequency Interfere to Create a Beat Pattern
- A Frequency Difference of 1 to 4 Hz is Best for Human Detection of Beat Phenomena
- Example 4
- Example 5
- Example 6
- Example 7: Superposition

- Intro 0:00
- Objectives 0:09
- Superposition 0:30
- When More Than One Wave Travels Through the Same Location in the Same Medium
- The Total Displacement is the Sum of All the Individual Displacements of the Waves
- Example 1: Superposition of Pulses 1:01
- Types of Interference 2:02
- Constructive Interference
- Destructive Interference
- Example 2: Interference 2:47
- Example 3: Shallow Water Waves 3:27
- Standing Waves 4:23
- When Waves of the Same Frequency and Amplitude Traveling in Opposite Directions Meet in the Same Medium
- A Wave in Which Nodes Appear to be Standing Still and Antinodes Vibrate with Maximum Amplitude Above and Below the Axis
- Standing Waves in String Instruments 5:36
- Standing Waves in Open Tubes 8:49
- Standing Waves in Closed Tubes 9:57
- Interference From Multiple Sources 11:43
- Constructive
- Destructive
- Beats 12:49
- Two Sound Waves with Almost the Same Frequency Interfere to Create a Beat Pattern
- A Frequency Difference of 1 to 4 Hz is Best for Human Detection of Beat Phenomena
- Example 4 14:13
- Example 5 18:03
- Example 6 19:14
- Example 7: Superposition 20:08

### AP Physics 1 & 2 Exam Online Course

### Transcription: Wave Interference

*Hi everyone and welcome back to Educator.com.*0000

*I am Dan Fullerton and we are going to continue our study of waves today by talking about wave interference.*0003

*Our objectives are going to be to explain how the law of superposition governs the interference of waves, to describe a standing wave pattern and how it is created, to identify nodes and anti-nodes in a standing wave, and determining fundamental and harmonic frequencies in wavelengths for strings, open tubes, and partially closed tubes.*0009

*Super position -- When more than one wave travels through the same location in the same medium at the same time, the total displacement of the medium is governed by what is called the Law of Superposition, which is really a very simple concept.*0030

*All it says is the total displacement is the sum of the individual displacements of the wave.*0044

*The combined effect of the interaction of multiple waves, then is known as wave interference.*0049

*It sounds very complicated, but very, very simple in practice.*0054

*Let us take a look at an example.*0061

*The diagram below shows two pulses approaching each other in a uniform medium. Diagram the superposition of the two pulses.*0063

*That is a fancy way of saying just add them together when they cross each other.*0070

*As this one is going to the right and this one is going to the left and when they hit each other, their displacements are going to add together, so you are going to get something that looks like that.*0073

*If this one is 5 cm and that is 10 cm, then this must be 15 cm; they add together.*0087

*You have the two wave pulses; they are coming together; they get closer and closer and when they start to interfere with each other, they are going to overlap, give you a bigger wave, and then as they move on, they are going to keep going as if they had never met.*0094

*A split second later after they have passed, what this would actually look like then -- You will have the 10 cm one still going to the left and the 5 cm one still going to the right as if they had never met.*0106

*How simple is that?*0120

*Let us talk about types of interference.*0123

*When two or more pulses with displacements in the same direction interact, the effect is known as constructive interference; you get an amplitude that is greater than the individual amplitudes.*0125

*We call this in phase.*0136

*When two or more pulses with displacements in the opposite directions interact -- well now they are going to start to cancel each other out; you are going to get an amplitude that is diminished, and that is called destructive interference or out of phase.*0138

*Now once the pulses have passed each other, let me reiterate, they are going to keep going as if they had never met, even if for a second they completely cancel out and it looks flat, once they pass each other they keep going as if they had never met, whether it is constructive or destructive interference.*0152

*Let us take a look at another example.*0168

*The diagram below over here has two pulses approaching each other from opposite directions in the same medium.*0171

*Which diagram at right best represents the medium after the pulses have passed through each other?*0177

*After they have passed through each other, they keep going as if they had never met, so we need one where we have the smaller pulse traveling to the right and the larger pulse traveling to the left.*0183

*That has to be this one right here, Number 2 -- they do not invert and they do not have anything else fancy going on.*0193

*It is just that simple, they keep going as if they had never met.*0203

*Let us take an example here with shallow water waves.*0208

*The diagram below here shows shallow water waves of constant wavelength passing through two small openings (A) and (B) in the barrier, so the wave velocity must be in that direction.*0211

*When it gets here we are going to see some diffraction around these and we will talk about that in more detail, but what we really notice here is that you have crests and troughs interfering with each other from (A) to (B).*0221

*Which statement best describes the interference, here at point (P)?*0233

*Point (P) is right there, so you will notice that that is occurring where we are on a crest from (A) that is meeting a trough from (B).*0236

*If we have a crest and a trough, that is going to be destructive interference, therefore, our answer must be: It is destructive causing a smaller amplitude because destructive interference reduces the amplitude.*0248

*Standing waves -- When waves of the same frequency and amplitude traveling in opposite directions meet in the same medium, a standing wave is produced.*0264

*A standing wave is a wave in which certain points, which we call nodes, appear to be standing still and other points called anti-nodes over here tend to vibrate with maximum amplitude above and below the axis.*0274

*This is the basis of a bunch of our instruments.*0285

*In this case, as we look here, I see 4 nodes and 3 anti-nodes.*0289

*Now do not get fooled. At any given point the wave is here or here, but you do not double count it.*0305

*This is 1 anti-node, so 4 nodes and 3 anti-nodes and in any standing wave, you always have one more node than anti-node.*0310

*So let us talk about how these occur in instruments.*0335

*You can create a standing wave by holding a string in place at both ends and then introducing a disturbance, plucking it -- think of a guitar, a violin, a cello, a bass, a piano has a hammer that strikes those strings that are held in place.*0338

*That is what is going to give you a sound that sets up a standing wave.*0353

*Now, there is a lot to study about these standing waves though.*0357

*Here we are going to look at a standing wave, the basic standing wave you get if you were to have a node on both ends, the strings held there and you have one anti-node here, so two nodes and one anti-node.*0360

*This is what is known as the first harmonic and we will call that N = 1.*0372

*It is the first harmonic or sometimes referred to as the fundamental frequency.*0376

*You will notice that we only have half of a wave in there, so the length of our string is equal to half a wavelength.*0388

*We could also have at the same time, a second harmonic (N = 2), where we have one full wave within that length (L) and that is called the second harmonic or sometimes referred to as the first overtone.*0396

*In this case (L) is equal to the wavelength of our wave.*0415

*We could have 1 1/2 wavelengths fit in there, and that is our third harmonic, sometimes referred to as the second overtone and now (L) = 1 1/2 wavelengths (3(λ)/2).*0418

*I could take this pattern here and I could try and generalize it by saying that (L) must be equal to N(λ)/2.*0439

*We have 1(λ)/2, 2(λ)/2, 3(λ)/2, for N = 1, 2, 3, and so on.*0449

*Well, if I wanted to solve then for the wavelength, I can rearrange this to say that λ must equal 2L/N.*0459

*If I want the wavelength, it is 2 times the length of my string divided by the harmonic number.*0468

*Or let us rearrange this a little bit more using what we know about the wave equation.*0474

*If V = F(λ), I could rewrite λ as equal to velocity over frequency, so if I replace λ here with velocity over frequency, now I have velocity over frequency is equal to...*0479

*And I still have 2L/N and solving for the frequency, which is often times what we perceive as pitch, and we get that frequency is N (our harmonic number) times the velocity divided by 2 times the length of our string.*0494

*So a couple of equations that can help us analyze what happens when we are talking about a stringed instrument.*0512

*We have that one, we have the wavelength for N = 1, 2, 3 (whole numbers) and the frequency here.*0519

*Now, not all instruments make music just by having a string.*0526

*You can also create standing waves in open and closed types of tubes, like trumpets, pipe organs, flutes, clarinets, obo, and so on.*0532

*Over here we are showing what we had for a stringed instrument and over on the right we are going to look at what happens when we have a tube that is open at both ends.*0541

*You get the same pattern, but note here, we are showing the lines -- these are air pressure, not displacement.*0548

*But over here, we now once again have half a wavelength in this open tube where we could have 1 full wavelength or 3-halfs a wavelength, 1 1/2 wavelengths.*0555

*We have the same basic patterns, but the difference is that these are open at both ends.*0569

*This is showing air pressure and not displacement.*0574

*If we wanted to show the displacement, it would probably look something more like that, which is easy to see is still half a wavelength, but it is just off set a little bit, so open tubes and strings follow the same mathematical sequence.*0576

*Let us take a look at standing waves in a partially closed tube.*0596

*You can close the tube at one end only and you can still set up the standing wave pattern, but closed-tubes only produce odd harmonics.*0600

*Once again, we will show air displacement here.*0608

*We have a closed tube closed at one end and open at the other and now what we have is one-quarter of a wavelength inside this tube.*0611

*Here we have three-quarters of a wavelength and here we have one and a quarter or 5/4 of a wavelength.*0622

*This would be our first harmonic, this would be our third harmonic -- we do not have even harmonics when we have a tube closed at one end -- and likewise, this would be our fifth harmonic.*0630

*I could generalize this with a pattern, that L = N(λ)/4 again, but now (N) must be odd, 1, 3, 5, and so on.*0647

*Or rearranging again for wavelength, λ = 4L/N, where again N = 1, 3, 5 -- we only have odd harmonics.*0660

*Pulling the same trick to find the frequency again, if V = F(λ), then λ = V/F and I can then solve for the frequency to find that frequency if NV/4L and again, N = 1, 3, 5, and so on.*0672

*A lot to take in there, but I think it will get a little bit easier and we will do a sample problem with that here shortly.*0694

*Interference from multiple sources -- this is a pretty cool effect.*0701

*Waves from two or more sources reaching an observation point can lead to constructive or destructive interference depending on the position of the observer.*0707

*Imagine we have a source (A) and a source (B) and if we are standing over here at point (C), because of the distances that these waves are traveling, they are reaching observer at (C), both at crest, so you are going to get constructive interference.*0715

*It is the same sound wave, but an area where it is pretty loud.*0730

*Move just a little bit though, so that now as far as the distances goes, (B) is here at a crest, (A) is here at a trough, which means you are going to get destructive interference, or for a sound wave, it will be quiet.*0734

*Try this sometimes -- Take two speakers and try to put on a continuous tone at a set volume and then stand some distance away with the speakers pointed somewhat toward each other and walk in that sound field.*0746

*As you move your head just a little bit, you may find areas where it is a little bit louder and a little bit softer.*0756

*You can actually perceive that constructive and destructive interference pattern, which is pretty cool.*0762

*Beats -- When you have two sound waves with almost the same frequency, but not quite the same frequency, you can sometimes create what is called a beat pattern, almost a little bit of background ringing.*0770

*You hear a slow, rhythmic change in amplitude.*0781

*It is typically heard by humans when you have a frequency difference of 1-4 Hz and it is very useful for tuning stringed instruments.*0784

*What you do is you try and play the same note on two different strings of a guitar for example.*0792

*If they are exactly matched up, you hear one clear tone at a constant loudness.*0797

*However if they are just off by 1 or 2 Hz, it is sometimes difficult to pick up that difference in pitch at first glance, but what you can hear as you play both notes at the same time is you will hear this rhythmic increase and decrease in the volume...*0804

*...the beat pattern that tells you, you do not quite have them in tune yet, so keep adjusting the frequency of those, the tightness of the strings until you get exactly what you were anticipating.*0820

*By the way, what is happening when you are increasing the tightness of guitar strings -- well you are not changing the wavelength as you are set with that distance, but what you are really doing is you are changing the wave velocity because you are adjusting the medium.*0831

*So you are increasing the tension; you are adjusting the velocity of the wave through that medium and therefore, you can get a different frequency.*0844

*Let us take an example where we are going to look at how we use some of this new information on sound.*0853

*The diagram below show 4 standing sound waves inside a set of organ pipes and we are going to assume the velocity of the sound in air is 343 m/s.*0861

*We want to know the highest frequency for the wave shown, the lowest frequency, the longest wavelength, and the shortest wavelength.*0871

*All right, so let us just do a full analysis of what we have going on here.*0878

*Let us start over here where we have a tube closed at one end and it is pretty easy to see that if this is half a meter -- well, remember λ = 4L/N.*0883

*Well, our (L) is 1/2 m, so that is going to be 4 × 1/2 m, so 4L = 2/N and that is 1 because this is a first harmonic, so our wavelength is 2 m.*0894

*Or you could figure that out by realizing this is one-quarter of a wavelength and one-quarter of a wavelength is 1/2 m and if you want the full thing, you have to multiply by 4, so the wavelength is 2 m.*0907

*Over here you have 1 1/2 waves in there and this would be a wavelength then -- if you have 3/2 in there of 2 m over that third harmonic -- 4 × 0.5 = 2 over the third harmonic -- is going to give you 2/3, so wavelength is going to be 0.667 m.*0918

*Here you have 1 1/4 waves inside this half a meter, so as you go through the math here -- 4 × 0.5 = 2/5 (N) = 0.4 m (wavelength).*0940

*Finally, we have a tube that is opened at both ends and we have one whole wave fit in here, therefore, the wavelength must be 1/2 m.*0960

*Let us write that down -- wavelength equals 0.5 m.*0970

*We found the wavelengths for these, so what is the longest wavelength for the pipe shown?*0974

*We can answer that -- That is 2 m.*0979

*What is the shortest? That must be our 0.4 m.*0981

*Now how about the frequency? To find the frequency, remember V = F(λ), therefore, frequency must equal V/λ, so the frequency over here of our left most tube -- let us call that tube 1, 2, 3, and 4...*0986

*For tube 1, frequency 1 is going to be equal to V/λ1 or 343 m/s/2 m (wavelength) for 172 Hz.*1006

*Frequency 2 is the same idea with velocity divided by wavelength, so we have 343/.667 or about 514 Hz.*1020

*For this third tube, frequency 3, that will be our velocity 343 m/s/.4 m (wavelength) or about 857.5 Hz.*1034

*Our fourth frequency here, F4 = 343/0.5 (wavelength) or 686 Hz.*1049

*So what is the highest frequency for the wave shown? Well, that must be from tube 3, which is 857.5 Hz.*1061

*What is the lowest frequency for the wave shown? Well that is going to be F1 or 172 Hz.*1070

*So we are putting into practice our open and closed tube sorts of problems.*1077

*A musician is designing a custom instrument which utilizes a tube opened at both ends.*1083

*Given the speed of sound in air as 343 m/s, how long should the musician make the tube -- so we are looking for (L) -- to create an (A) or a frequency of 440 Hz.*1089

*As an instruments fundamental frequency, N = 1.*1106

*Well, if you recall, frequency is going to be NV/2L and if we solve for the length, L = NV/2F, our harmonic number is going to be 1 because first harmonic is the fundamental frequency.*1112

*Our velocity of 343 m/s/2 × 440 Hz (frequency) will give us L = 0.39 m.*1131

*If we make that tube 0.39 m long, we are going to get a 440 Hz A.*1146

*A place of maximum displacement on a standing wave is known as...?*1155

*Well, if you recall on the standing waves, we have these areas with minimum displacement, called nodes and the areas with maximum displacement are called anti-nodes.*1159

*In this case we have 3 nodes and 2 anti-nodes and I get in a standing wave it is not at both places at once, it is going to be oscillating back and forth.*1181

*At any given point in time, you might have it look like this and then a second later in time it is going to look like this.*1188

*What are the ones called where we have maximum displacement?*1196

*Maximum displacement is right there and that is an anti-node.*1200

*One last sample problem. The diagram below represents two pulses approaching each other.*1207

*Which diagram below best represents the resultant pulse at the instant the pulses are passing through each other?*1213

*When they interfere they are going to follow the Law of Superposition, which means their amplitudes are going to add.*1219

*We have one positive and one negative and the negative is a little bit bigger than the positive, so what we are going to end up with is a net negative, but smaller than the initial, so 2 must be our correct answer.*1225

*Thanks so much for visiting us with Educator.com. Have a great day everyone!*1240

1 answer

Last reply by: Professor Dan Fullerton

Sat Apr 15, 2017 11:25 AM

Post by James Glass on April 14, 2017

Why can't the answer to number 6 also be amplitude? Thanks

1 answer

Last reply by: Professor Dan Fullerton

Mon Apr 13, 2015 6:38 PM

Post by SeungJoo Han on April 13, 2015

Example 4. I think the equation for wavelength is Wavelength = 2l/n. I want you to explain how dod you get the 4l/n for figuring out the wavelength.

1 answer

Last reply by: Professor Dan Fullerton

Mon Mar 23, 2015 6:29 AM

Post by Wen Nguyen on March 22, 2015

hi

0 answers

Post by Jamal Tischler on September 1, 2014

I found this nice example: http://ro.wikipedia.org/wiki/Und%C4%83_sta%C8%9Bionar%C4%83#mediaviewer/Fi%C8%99ier:Standing_wave_2.gif

3 answers

Last reply by: Professor Dan Fullerton

Sun Apr 6, 2014 7:14 AM

Post by Jerry Liu on April 3, 2014

Example 7 is labeled as Superstition, it should be "Superposition".

Do you not have an editor?

2 answers

Last reply by: varsha sharma

Tue Mar 26, 2013 5:47 AM

Post by varsha sharma on March 24, 2013

Why at B a trough. Please explain.