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Two speakers emitting sounds with the same frequency are both equal distance from listener; (1 point) and produce constructive interference. The distance from the l...

Question

Two speakers emitting sounds with the same frequency are both equal distance from listener; (1 point) and produce constructive interference. The distance from the listener of one of the speakers is doubled while the other speaker remains in the same place. Which can now be said about the ound at the listener? The speed of sound is 340 m sec.Perfect destructive interference occuIS. The frequencies of the sounds double Perfect constructive interference OCCUIS_ The wavelengths of the sounds doubleB

Two speakers emitting sounds with the same frequency are both equal distance from listener; (1 point) and produce constructive interference. The distance from the listener of one of the speakers is doubled while the other speaker remains in the same place. Which can now be said about the ound at the listener? The speed of sound is 340 m sec. Perfect destructive interference occuIS. The frequencies of the sounds double Perfect constructive interference OCCUIS_ The wavelengths of the sounds double Beat frequencies are caused by (1 point) neither constructive nor destructive interference destructive interference only constructive interference Only both constructive and destructive interference



Answers

Two identical loudspeakers are located at points $A$ and $B, 2.00 \mathrm{~m}$ apart. The loudspeakers are driven by the same amplifier and produce sound waves with a frequency of $784 \mathrm{~Hz}$ Take the speed of sound in air to be $344 \mathrm{~m} / \mathrm{s}$. A small microphone is moved out from point $B$ along a line perpendicular to the line connecting $A$ and $B$ (line $B C$ in Fig. $P 16.65$ ) (a) At what distances from $B$ will there be destructive interference?
(b) At what distances from $B$ will there be constructive interference?
(c) If the frequency is made low enough, there will be no positions along the line $B C$ at which destructive interference occurs. How low must the frequency be for this to be the case?

Two identical loudspeakers. Air located at a different 20.2 m away. There's too little off speaker driven by the same empty flash and play fire. The frequency of sound produced by the loudspeaker is 7 84. Hearts feed off. Sounding air is 3 44 m per second and the distance between the two say distance between the two. Loudspeaker is 2 m, so the wavelength will be Lambda physical too. Or 3 44 upon 7 84. That is their 0.43 it it Mito Mm. Yeah, this is pick a one B hair speaker A distances to meter find. See, this is find peace. Distance from Speaker. We took peace X here I e saw f p is equal toe loot over a squared plus x square. So a P minus BP is BP is float over X square plus x square minus thanks. So here, from the condition off destructive interference route over s square plus X square minus X is equal. Tow em plus one upon to Linda. So a square plus X square is equal to x plus M plus half Linda squaring on both side s square plus X square is equal to X square. That's twice I am blessed Half one Lambda X plus and plus half square Linda Square. Yeah, so no ex physical too s square upon tow. Am less Have Who's Linda minus M Pless Half hole square, Lambda Square one Toy and Bless half Linda So Mexico Toe Square E boy am less half lim down minus hand Let's half lambda the front Do This is a question one. So for Mexico toe sort of X equal to 9.1 m and Chemical tau zero em is equal to one X equal to 2.71 m where I am a culture to x equal to 1.27 meter chemical toe three x equal 2.534 m I am called to forest X equal to 0.26 Mito now off for the second plan For constructive indifference You do it over a squared plus X squared minus X tickle toe M Linda. The route over square plus X square is equal to X plus m lambda squaring The both side s square plus X square is equal to X square plus two Isom Lambert X plus m squared member square so twice, and lambda X equal toe square minus and square Linda, square the distances from bes Mexico Little s Square Oh, yeah, Twice. Um, Lambda minus m lambda off this equation toe now for chemical toe one X is for pine tree for meter. I am equal to two x is 1.84 m mm ical toe three x 0.861 m on chemical toe four Exists there a point to 62 m. Thank you. Now, if there is no destructive interference along the line busy than substituting X equal to zero and m equal to zero in the question one that is ex ical Toe s square upon twice I am blessed Half lambda minus AM plus half lambda have fun to X is zero square one lambda minus lambda pond for so square Linda is limbed up on four. So Lambda Square is a world forest square. The four Lambda is yeah to s. Now let us calculate d low frequency if iss new upon lambda. No, upon lamb dyes to isis and newest 3 44 m for a second. Upon to and to Mr That is 86. Hard step for low frequencies 86 hearts

This is the problem based on phenomena off interference here it is giving two speakers as fun and esto mhm separated apart by the disclaims off 2 m. So distance between as one and as to is 2 m on their producing that sound off frequency 18 hearts and the speed of sound is given 3 40 m per second. Oh, there is a point directly. 4 m in front off one of the speakers. So there is a white B on B s bunnies for Britain. Hm e yeah. P point is fight Come to me, people, I mean, is point off constructive or destructive or any other Hey, super position that we have to find by using the pie Tabaro store. Um, we have toe file their distance ps two. So that will be forest squared, plus to his square. So it is to be 4.47 to Britain. Hence the father difference with me b s to my must be Essman that is poor quite four seven to minus four. So quite 472 weeks violet after Wales will beat speed upon frequency. So 3 40 upon 1800 that is 18000.18 89 meted for destructive interference. No party friends to me am plus half into equivalent. Yeah, So from here part differences. Quite 44 Sorry. Point 472 M plus half 1.1889 hence M plus half fiscal toe. Quite +472 apart. 0.1889 Mhm eso It is to me 2.5 minus five. That is two hands constructive. Enter. Sorry. Destructive interference. Take place. Yeah, that son.

Hi there. So for this problem we have two identical loud speakers that are located appoints A And point being and they are separated by the distance of two m and these two large speakers are Driving by the same amplifier. So you produced some waves we know that plus the speed of sound waves are 344 m/s And the frequency is 784 hairs. No, as my my microphone is move out from point B just parallel to death um to a perpendicular line connecting the points A. M. B. So it has chosen this figure. So the first question is at what distance from B will there be destructive interference? So part A. Is at one point of adds there will be destructive interference. First, the first thing that we can do is to determine the wavelength because the Waveland is defined as the velocity over the frequency. We are given those two values. So plugging this into the calculator, we obtain that the wave length is equal to 0.939 m. So that's the wavelength of this week. Some waves. So to solve this for part and we are going to say that the separation of the speakers is age as I put in here. So if we consider that separation, the condition for destructive interference is given by the following. Since we know this distance in here, we won um we want the path difference. So between this speaker, that is an exposition and this speaker. But this speaker is we can calculate the distance from the speaker to this point in red by taking the hypothesis and in this case it will be a change square plus at the square the square root of thought. So taking the path difference, we will obtain that the destructive interference of course when the path difference minus X, which is the position of the loudspeaker B is equal to beat at times the wave land. Now, in this case for destructive interference, B is an odd multiple of one half. So we will have that Byetta can have values of one half, three over to um fi over two and so on. No, we need to solve for ads because that's what we want. A convenient way into a convenient way to do this is by adding adds to both sides and elevate that to the square. So I'm going to do that. If we add if we add in the left side of this equation plus adds, We will have we eliminate the S. one. Since I said as I said that we can take we can take the square of that. We will change this. And for the other side we will have peter this plus adds and we take the square root of this. So taking the square root of that. We will be the following. We will have peter's square land squared Plus two times. Beater λ Ads plus at the square. And we can simplify this order because we can counsel this with this one and we can now solve for um ads. It is very straightforward that we are going to obtain for adds that this is equal to edge square the distance between the two speakers over two times. Pita lum, dum minus pita over two times lambda. Now we are given all of this information. The thing that we need to do is to is tar um trying for values of beat huh? And to see what we obtain so well forbid to 1/2 we will obtain that the X value is going to be well for this first part I'm gonna substitute all of the values but for the others I'm just gonna suppose. You understand? So for Beatles one half we put one half For the wavelength we know that the wasteland is zero yeah 0.439. If I'm correct. Yes. And this times minus Pita which is one half over to This again 0.439 m. When we plot this into the calculator we obtained that for Peeta equals to one half. We will obtain a value of 9.01 m. So this is the first one when vita is equal to one half. But we can also have that Vita has a value of 3/2. So when we substitute that value for Dita and now in here is just changing this this for three and this for three. So we will obtain that. The adds value for this case is going to be Um 2.71 m. So for this case we can keep going With Vita equals to fight over two and well for that will obtain the value of ads of um destructive interference is equal to 0.53 m. And you can keep going with that weekend, keep going with better equals two, 7/2. And Vita equals tonight over to but you get okay where to. So those are the available values were also going to have for 7/2 and for beaten night over to and those values are for 7/2, we are going to obtain 0.0 20 six m. And for B to 9/2 we will obtain a value off. Um it's over 9.2. So those are all the allowable values for um S so that we will obtain destructive interference. Now for part B. Of this problem we are asked them at what distance from being will there be constructive interference? So the past uh the steps to do this this part again we need to find the values for ads but in this case for constructive inter friends so um we need to apply the same, we need to use the same equation but in this case the values of great the values of pete huh? Need to be uh integral values. So we need to try for vita equals 2123 and so on. Yeah. And when we do that again we're gonna take the same expression as before this expression right here and I'm gonna put that I can hear that as it's equal to Age over 2 2 times. Pita London minus beat over to London So we need to put all yourself beat huh? Um, so we will have for the first one with bitter equals to want we will obtain that adds is equal to 4.34 m for beach and equals two will obtain that ed's has a value of yeah 1.84 m and we can keep going with bitter equal stream. Yeah, That will give us a value of 0.86 m for beta equals for we will have that adds has a value of Um 0.26 Natures and and this is it for part B. Now for parsi of this problem we are asked if the frequency is made low enough there will be no positions allow the line B and C at which destructive interference of course. How long must frequency be for this to be the case. So yeah, if the problem says age is one half of the one half of the wave land. We will obtain destructive inter prints added speakers be but um so that will obtain destructive. Came to her friends uh B but if age is less than half of the wave land, the path difference can never be as large as one half of the wave land. So and this is also came from the from the expression of that we obtain with X. Equals to zero and B. To equals to one half. So the minimum frequency that we can have from this is the velocity over two times a judge so that we will obtain that this is 344 m per second over four m. And this will give us 86 hers. Yeah. So this is it for this problem. Thank you.

This question were given that there is a speaker a and then is one mostly could be and did there at two meter distance apart. And there is a queue point here, which is that distance one meter from the speaker. And there is a point b here which is have distance X. So the remaining distances two minus x. So let's say to relent is and because it was two meter, so we will use this as a variable here for the moment. So for Pat, eh? We have frequency equals two, two zero sis hunch. We can find out the land off this wave so the land becomes mellow city over frequency, which is equality for four meter second, which is sound wave velocity in air over two zero six hours, which is it caused to one point six seven meters. So that's the prevalent off the sound which is produced by the speakers. The questions is about the destructive interference and constitution different. So for destructive interference, we should have just doing defendants. We should have the butt defense should be, But difference should be toe end plus one over lambda by two. Just the road multiple off the Lambda where and can be zero one two Don't, don't don't. And for constructive interference we should have d apart Different speedy should be And Lambda where and can be zero one two Don't don't So let's find out the past defense first So but difference should be close to We have got violent elf and this is X So this is l minus X No, my part defense becomes, Let's say this is Delta So l minus X miners. This becomes L minus two X or ex becomes l minus Delta and or two for the safety of interference. We already found her there for any question Jiro and equals to Jiro so D. But difference should be equals to land a by two. So my del tile should be lambda by two. So using this situation, you find out that Dan Tile equals point eight three five meter. So this gives us X equals two point five eight meter and for and equals two minus one. We can also find out the DealTime Jessica's appoint a three five meter, which gives us X equals to one point for two meter. So these are two values which makes the difference into the destructive interference. So let's go to the park. Be here. This was part eh? To Mr by beena for constructive interference, we can have any cause to one or any question minus one. So for dental also, we can have any question Jiro just the delta and will be cause, which is a trivial Justin. They should be caused to sit the same point, which is not possible. So then tell it calls to lander, which is of course, to one point six seven meter which gives us X equals two point one centimeter and for any cause to minus one, we have del del equals two minus one point six centimeter or X equals to one point three two for a speaker A and B. So these this access out of the just one point in time that this means that this is the point one point two. And for this we have point once and point one centimeter. That means that this is the point. Frank wants him. So these are two points. This this one in the second point. Similarly, the same for the destructive interference. We have two points. This one and this one and the question. See, the sense is that why no, the usual? Because we to use at home. I know what they produce is some kind of just the independence of the cast of independence. There isn't but police easy and say well, because the speakers, which we have at home, are very big, are very big and men. It produces some kind of the They're here, so it produces a wave in much, much broader way or having the Lambda may be the same. But the sound wave within at every point is all almost same, so we don't have any kind of destructive interests for constructive interference. That's it.


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