Two loudspeakers are placed 3.00 m apart, as shown in Fig. 12-37. They emit 474-Hz sounds, in ph

step 1 Okay, so this is a very interesting problem. Here we'll be using physics and a little bit of geo

Two loudspeakers are placed 3.00 m apart, as shown in Fig....

Key Concepts

Coherent Sources

Coherent sources emit waves of constant phase difference and the same frequency, which is necessary for a stable and predictable interference pattern. In the context of the problem, the loudspeakers operate as coherent sources, and altering their phase relationship (in phase or out of phase) significantly affects the resulting interference pattern observed at the microphone.

Wavelength and Frequency Relationship

The relationship between wavelength and frequency, established by the wave speed (v = f?), is fundamental in interference problems. Knowing the frequency and the speed of sound allows the calculation of the wavelength, which in turn is used to determine the spacing of interference maxima and minima based on the conditions of constructive and destructive interferences.

Phase Difference

Phase difference refers to the offset in the phase of two waves when they meet at a point. For interference patterns, the relative phase at the sources or introduced by changes in the path length determines whether the interference is constructive or destructive. Control of phase differences is essential in setting up desired interference conditions, as seen in the question when altering the connectivity of the speakers.

Constructive and Destructive Interference

Constructive interference occurs when two waves meet in phase, meaning their crest and trough align, leading to an increase in amplitude and, consequently, intensity. Destructive interference happens when the waves meet out of phase, such that the crest of one coincides with the trough of another, reducing or canceling the amplitude. These principles are fundamental in predicting the locations of interference maxima and minima.

Interference of Waves

This concept refers to the superposition of waves from different sources, where the resulting wave amplitude and intensity depend on the phase relationship and path difference between the waves. In acoustics, interference plays a crucial role in determining the positions of loudness maxima and minima when waves from coherent sources overlap.

Path Difference

The path difference is the difference in the distances traveled by waves from their sources to a given point. It is a critical parameter in determining the phase relationship of the interfering waves. Specific values of path difference (in relation to the wavelength) will result in either constructive or destructive interference, which define the intensity pattern observed.

Transcript

00:01 Okay, so this is a very interesting problem.

00:03 Here we'll be using physics and a little bit of geometry to solve this problem.

00:09 So first of all, we have two mics, and then we have a microphone over here.

00:17 And sorry, we have two speakers and one microphone over here.

00:22 And what is given is they are three meters apart.

00:28 And then from midpoint to the microphone it's 3 .2 meters and then we have the distance between them now so it says that the microphone they're in phase and they emit 474 heart sound so yeah so they're in face and they're emitting 4 74 hertz of sound now now the thing is we should write that as loudspeaker or not mic so let's say that's speaker one and then this is speaker two so if they're in phase if we look at their wave pattern we see to be in in to be in phase they're maxima and minima should be matching with each other and that's what we're seeing we see a node over here which is same as here and then they have the same maxima at the same side now the question is how far must the microphone be moved to right to find the first intensity minimum now what we're doing is when moving this microphone now if we want to find the first intensity minimum now if we want to find the first intensity minimums at a minimum we see that it needs to go half of the wavelength because i mean if the microphone is around let's say this region they have this maximum value and then it needs to go around this point where they don't have the maximum so this is the node and then yeah that's the minimum so it's like half of the wave link so we do that so yeah so what's happening here is we have this intensity maxima around this region and then they are the nodes are matching over here which is the minima so that is the gap is half of the wavelength and we'll try and if we look at the geometry so what we did here is our microphone is now here which was initially there here which which is the midpoint so so the relation that we're concerned here is now we have two different distances d1 and d2 and the difference between their distance should be matching with half of the wavelength so that's what we did here so d2 is clearly greater than d1 so at this moment we should not bother by the numbers let's look at the geometry and try to solve for it so we're trying to solve for this x we are trying to find how far the microphone moved so yeah let's try to solve it using the letters and then we'll plug in the numbers right so yeah as we said d2 minus d1 is half of the wavelength now what's d2 and what's d1 let's try to find out so so the distance from this microphone to the midpoint of the speakers is l and the distance between the the mics is d.

04:24 So what's happening is we the so the so d2 is basically will be using pythagorean theorem and to get that we need this distance and this much distance so we know that this whole thing is d and this is the midpoint so that's half of d so let's see this right here is half of d and then we have this extra x over here so if we add that part so that's x so this distance right here this is half of d plus x that's here and this distance is l so that's l so what pythagorean theorem tells us is the square of this term should be equal to the square of this term plus this term right let me write that down so let's say you have a triangle and the x the the the distance between this point is a the distance between this and that is b and this is c so the theorem tells us that c square should be equal to and of course a and b must be perpendicular to each other so c square must be equal to a squared plus b squared and in other words c is square root of a squared plus b squared so that's what is done here so we found the distance and from this point to that point and from this point to this point, and we applied pythagorean theorem to replace d2.

06:31 Similarly, the distance from here to here is nothing but half of d minus this x extra term.

06:39 And that's what we got here.

06:40 And then again, we're applying pythagorean theorem to get d1.

06:44 So all we did was we replaced our d2 and d1.

06:47 And then we need to, so what we did was we need to square it.

06:52 We need to get rid of this square root term and to do so we are taking this term on the right on this side and that's that's how we are getting rid of this negative sign so it becomes positive when it's on the right and then we're taking square on both sides and then we end up having this whole big expression now what we did here is a formula that we use very commonly in maths so if you have a square plus b squared we write that as so if we have a plus b whole squared we write that as a squared plus two a b there's b squared and that's what we did here when we took the square on both sides this term is our a this term is our b so this term is b this term is a so so we're taking a squared plus b squared and so that's our a squared this guy is our b squared and in between we have two times a times b so yeah and once we did that we can get rid of this l square on both sides and then we have this expression now we're using another common formula in this problem which is a plus b whole squared minus a minus b whole square gives us two time uh sorry gives us four times a b right um so that's what's happening here so we have now we have a and we have b we have a and we have b we have a and we have b so we have we have a and we have b so we we're taking this guy on the left and that makes this expression and then we're replacing this whole big expression by simply writing four times this first time and second time.

09:16 So that's what's happening here...

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