Visit ilectureonline.com for more math and science lectures! In this video I will show you how to calculate I=? when there is sound (or wave) interference.
Honestly, as much as I like khanacademy, freelance, brightstorm...this professor has by far the best grasp in what high school/college physics students need. I apologize for spamming you with comments here and there but you need way more subscribers, viewers, and public recognition. Please continue to do what you do and hope you know that there are people who sincerely appreciates your content.
Nvrloptimism, No need to apologize. All comments favorable and unfavorable are welcome. I appreciate the feedback so I can adjust to the needs of the students out there. Your words are encouraging and it has given me great pleasure that thousands of students around the world have been helped by these videos.
It is certainly very comforting to see that no matter what problems I have in math and physics courses, you always have a bunch of videos on that exact topic.
Thank you, professor Michel Van Biezen for provide us practical problems/examples. You're my first source when I want to practice some problems/questions after read the theory.
You're video's are great and they really help me practice for my exams, so thank you. This example, however, has me a bit confused. The intensity of the two waves won't be exactly the same, because one has traveled a greater distance and thus has lost some of it's magnitude - right?
+Triston McLean That is a very good point. We often "ignore" such things in order to concentrate on the basic principles being presented. In more advanced classes we increase the complexity of the material and begin to include these additional details.
Hi Michael, first of all thank you for your lectures/videos. Just a quick comment. In the snapshot of the video you have a typo: "distructive" instead of "destructive". Thanks!
Professor I have a question... Is we consider our two ears as one point?..Because our two ears are little bit far away from each other..According to this reason is it consider to avoid the complexity?? Please reply....
And another question professor...How do we able to hear the sound in our left ear if the sound source is in right??How ear particles enter in the left ear if the head blocked the dusturbance from right? What is the reason of it?? Please reply
Sound waves are essentially pressure waves and when you look at the diffraction principles, you will see that pressure waves are able to bend around corners. (That is why we can hear another conversation held in an adjacent room with the open door).
when the person stand somewhere that the wave are exactly same, i.e. constructive interference happens, shouldn't the intensity be 4 times higher? as you have A+A=2A and I =(2A)^2=4A^2. cannot remember how much db that would be though, need a calculator.
@@MichelvanBiezen I mean time 4:10-4:25. Seems that you were talking the intensity is double, I think intensity is 4 times higher, that about 6 dB i guess.
You are adding some good questions. We have videos that describe that in more detail. See the following 2 videos: th-cam.com/video/zd_QBB3O0Vk/w-d-xo.html and th-cam.com/video/Jq1xz_DAA5Y/w-d-xo.html
@@MichelvanBiezen Yes, I learned a lot of stuff from these two videos. I am not sure whether you listened to this video at 4:10-4:25. I really appreciate you did such fantastic job to explain the scientific problem, hopefully you won't be offended to correct a small error here.
How would I calculate the lowest frequency obtained by the person, if I only have the velocity of sound and the distances of the speakers and to the person, assuming that the frequency on each of the speakers are the same?
John, The frequency is fixed by the speakers, so that cannot change. But if the distance to each speaker is different, then there will be interference caused be the path length difference from each speaker to the listener. The closer this path length difference is to n + (1/2) wavelength where n is an integer number then the closer the sound intensity will go to zero.
If I have 2 speakers emitting 1250hz, unknown distance apart. And a person stands 12m directly in front of one of them. How would I calculate the distance between speakers so the person would hear no sound? I've been stuck on this problem for awhile now.
Redgie,I suspect that there is something wrong with the information given because that makes it impossible to solve the problem. Maybe the frequency is 1250 MHz? (1250 million Hz). In that case the wavelength would be 0.24 m (instead of 240,000 m). Then the xtra distance would be a half wavelength = 0.12 m. If the diagonal distance to the other speaker is h = sqrt (x^2 + 12^) then h - 12 = 0.12 and solving for x gives you 1.7 m
Thank you for the reply. The exact wording of the question is: "Two loudspeakers (S&S) are a certain distance apart, and person P stands 12m in front of one of them on a line perpendicular to the baseline of the loudspeakers. If both loudspeakers emit identical 1250hz tones, calculate the distance between the two loudspeakers so that the person: i. Hear no sound. ii. Hears maximum sound. "
My mistake, we are talking about sound waves here, so we need to assume the velocity of sound (since it wasn't given). If we take the velocity of sound to be 340 m/sec then the wavelength will be 0.272 m. Half a wavelength will be 0.136 m. Thus for no sound we take xdis to be 0.136 m and for sound we take xdis to be 0.272 m. Use the method above and you'll get for no sound x = 1.81 m and for sound you'll get x = 2.57 m (h - 12) = 0.272 m while h = sqrt (x^2 + 12^)
This is confusing, where is the original question? what is the question asking? you go over this in your video however i would like to see the question written down :)
The video is just an illustration of wave interference. The concept is: determine the path length difference between the two waves. If the path length difference is an integer multiple of a wavelength, then the waves will experience constructive interference. If the path length difference is an integer multiple of a wavelength + (1/2) wavelength, then the waves will experience destructive interference.
how comes we never experience this ? you never find a spot in a room with two speaker where there's no music.. and also you never find a spot where music get louder.. it sounds the same everywhere in the room
Honestly, as much as I like khanacademy, freelance, brightstorm...this professor has by far the best grasp in what high school/college physics students need. I apologize for spamming you with comments here and there but you need way more subscribers, viewers, and public recognition. Please continue to do what you do and hope you know that there are people who sincerely appreciates your content.
chAndLer .bong
Nvrloptimism,
No need to apologize. All comments favorable and unfavorable are welcome. I appreciate the feedback so I can adjust to the needs of the students out there. Your words are encouraging and it has given me great pleasure that thousands of students around the world have been helped by these videos.
It is certainly very comforting to see that no matter what problems I have in math and physics courses, you always have a bunch of videos on that exact topic.
This guy has legit carried me from high-school to computer engineering in university 🙏
Your are going in the right direction. Glad we can be part of it!
same, computer engineering too btw
Thank you, professor Michel Van Biezen for provide us practical problems/examples. You're my first source when I want to practice some problems/questions after read the theory.
A really straightforward and easy to understand explanation
Michel, you're my hero
You're video's are great and they really help me practice for my exams, so thank you.
This example, however, has me a bit confused. The intensity of the two waves won't be exactly the same, because one has traveled a greater distance and thus has lost some of it's magnitude - right?
+Triston McLean
That is a very good point.
We often "ignore" such things in order to concentrate on the basic principles being presented.
In more advanced classes we increase the complexity of the material and begin to include these additional details.
You just saved my assignment mark 😅🙏🏼
Thank you very much from Iraq, you are a great teacher 💗.
Glad you found our videos! Welcome to the channel!
Hi Michael, first of all thank you for your lectures/videos. Just a quick comment. In the snapshot of the video you have a typo: "distructive" instead of "destructive". Thanks!
Thanks for letting us know.
Thankfully the sound from the speaker that is 2 meters away will have loss some energy. But how many dB will the remaining sound be?
Totally agree with nvrlopyimism! Thanks so much Michel!
Thank you! Great explanation!
love this channel helps alot
Professor I have a question...
Is we consider our two ears as one point?..Because our two ears are little bit far away from each other..According to this reason is it consider to avoid the complexity??
Please reply....
That is correct. We first learn the principles with simplified models. Then in more advanced classes we introduce this type of additional complexity.
Ok..Please introduce this type of questions.I am eagerly waiting for it...
Thank you professor for your reply...
Love from india😊❤️
And another question professor...How do we able to hear the sound in our left ear if the sound source is in right??How ear particles enter in the left ear if the head blocked the dusturbance from right?
What is the reason of it??
Please reply
Sound waves are essentially pressure waves and when you look at the diffraction principles, you will see that pressure waves are able to bend around corners. (That is why we can hear another conversation held in an adjacent room with the open door).
great thanks!
You're welcome!
when the person stand somewhere that the wave are exactly same, i.e. constructive interference happens, shouldn't the intensity be 4 times higher? as you have A+A=2A and I =(2A)^2=4A^2. cannot remember how much db that would be though, need a calculator.
No, the amplitude is only twice as high.
@@MichelvanBiezen I mean time 4:10-4:25. Seems that you were talking the intensity is double, I think intensity is 4 times higher, that about 6 dB i guess.
You are adding some good questions. We have videos that describe that in more detail. See the following 2 videos: th-cam.com/video/zd_QBB3O0Vk/w-d-xo.html and th-cam.com/video/Jq1xz_DAA5Y/w-d-xo.html
@@MichelvanBiezen Yes, I learned a lot of stuff from these two videos. I am not sure whether you listened to this video at 4:10-4:25. I really appreciate you did such fantastic job to explain the scientific problem, hopefully you won't be offended to correct a small error here.
Not at all. Yes sometimes I make mistakes or I misspeak.
Very nice explination
Thank you.
i would like to know. what is the reference you are using on this topic sir,
thanks.
I use a lot of my notes from teaching the topics.
@@MichelvanBiezen is there any suggestion for book reference that i can read about this topic sir?
thanks
/sorry about my english hehe
مردانه خوب توضیح داد
Thank you.
How would I calculate the lowest frequency obtained by the person, if I only have the velocity of sound and the distances of the speakers and to the person, assuming that the frequency on each of the speakers are the same?
John,
The frequency is fixed by the speakers, so that cannot change.
But if the distance to each speaker is different, then there will be interference caused be the path length difference from each speaker to the listener.
The closer this path length difference is to n + (1/2) wavelength where n is an integer number then the closer the sound intensity will go to zero.
If sound intensity is literally 0 W/m^2, does that mean -inf decibels?
That is correct. However when decibels are used, the minimum intensity used = 1 x 10^-12 W/m^2
so does that mean that it can still be perfectly destructive even if you'd move a bit further away or even closer?
+David Colegado just as long as the extra distance the sound waves is .47?
+David Colegado I hope my question is understandable hehehe
+David Colegado
As long as the extra distance = (1/2) * wavelength
Sir, are the wavelength (λ) for both wave always same?
Typically in these types of problems the wavelengths are the same.
@@MichelvanBiezen thank you!!
is the extra distance called path difference?
I like to call it: "path length difference".
If I have 2 speakers emitting 1250hz, unknown distance apart. And a person stands 12m directly in front of one of them. How would I calculate the distance between speakers so the person would hear no sound? I've been stuck on this problem for awhile now.
Redgie,I suspect that there is something wrong with the information given because that makes it impossible to solve the problem. Maybe the frequency is 1250 MHz? (1250 million Hz). In that case the wavelength would be 0.24 m (instead of 240,000 m). Then the xtra distance would be a half wavelength = 0.12 m. If the diagonal distance to the other speaker is h = sqrt (x^2 + 12^) then h - 12 = 0.12 and solving for x gives you 1.7 m
Thank you for the reply. The exact wording of the question is:
"Two loudspeakers (S&S) are a certain distance apart, and person P stands 12m in front of one of them on a line perpendicular to the baseline of the loudspeakers. If both loudspeakers emit identical 1250hz tones, calculate the distance between the two loudspeakers so that the person:
i. Hear no sound.
ii. Hears maximum sound. "
My mistake, we are talking about sound waves here, so we need to assume the velocity of sound (since it wasn't given). If we take the velocity of sound to be 340 m/sec then the wavelength will be 0.272 m. Half a wavelength will be 0.136 m. Thus for no sound we take xdis to be 0.136 m and for sound we take xdis to be 0.272 m. Use the method above and you'll get for no sound x = 1.81 m and for sound you'll get x = 2.57 m (h - 12) = 0.272 m while h = sqrt (x^2 + 12^)
Thank you very much. I've been scratching my head over this one. Ready for the exam now. Thanks again!
i cant wait to pass the exams
Why didnt I find this guy sooner than 2 hours before the final
All the best on your final.
@@MichelvanBiezen thank you, your videos are very helpful
@@MichelvanBiezen I think I did well, this specific video helped a lot!
Great, keep up the good work!
New subscriber ❤️
Welcome to the channel!
@@MichelvanBiezen Thank you. All the best. If you are interested visit my channel please
You have some interesting demonstrations.
@@MichelvanBiezen Thank you ❤️
This is confusing, where is the original question? what is the question asking? you go over this in your video however i would like to see the question written down :)
The video is just an illustration of wave interference. The concept is: determine the path length difference between the two waves. If the path length difference is an integer multiple of a wavelength, then the waves will experience constructive interference. If the path length difference is an integer multiple of a wavelength + (1/2) wavelength, then the waves will experience destructive interference.
Why not use Vsound= 331m/s
That is the speed of sound at 0 C. It doesn't matter what velocity you use.
thumbnail image is wrong
😊
🙂
how comes we never experience this ? you never find a spot in a room with two speaker where there's no music..
and also you never find a spot where music get louder.. it sounds the same everywhere in the room
you need a single frequency sound source to make this work. Music contains sound of many frequencies and different wave lengths.
@@MichelvanBiezen thank you
Velocity of sound is 332m/s
That depends on the temperature and the air pressure. At room temperature it is between 340 m/sec and 345 m/sec
Funny, you sound exactly like an American. But your name is completely Dutch. Parents or a further ancestor?
I was born and raised in Belgium. (good guess)
@@MichelvanBiezen Ah so close haha