Two loudspeakers emit sound waves along the x-axis. The sound has...

Question

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 60 cm.
a. What is the wavelength of the sound?

Accepted Answer

Hey, everyone. So this problem is dealing with the phase difference between two waves. Let's see what it's asking us. A teacher uses two synchronous sound emitters placed on a graduated ruler and a sound detector to show her students the interference phenomenon. The sound detector is placed in a fixed position to the right of the two emitters on the graduated ruler. When the distance between the two emitters is 15 centimeters, the sound detector measures the maximum intensity. The teacher enhances the distance separating the two emitters until zero intensity is measured. The distance separating the two m is now 45 centimeters. We were asked to calculate the wavelength of the sound wave produced by the emitters. Our multiple choice answers are a 30 centimeters b 45 centimeters c 60 centimeters or d 90 centimeters. So the first step to this problem is recalling that our phase difference delta five is given by the equation 25 multiplied by delta R divided by LAMBDA where delta R is the distance between the sound sources and LAMBDA is our wavelength. Now, we can also recall that when we have our maximum intensity is when our delta five is equal to M multiplied by two pi our minimum intensity or zero intensity. In this case is going to be when delta five is equal to M plus one half multiply by two pi. And so we have two separate equations, two separate scenarios, one with the maximum intensity, one with the minimum intensity, but both have the same wavelength that same lambda. And so we can use our simultaneous equations to solve for lambda. So in the first case for our maximum intensity, we have M multiplied by two pi is equal to two pi delt R one divided by Lambda. And so those two pies cancel and we are left with M equals delta R one divide. And by Atlanta for our minimum intensity, we have the quantity M plus one half multiplied by two high is equal to two I delta R two divided by Lambda. And so delta R 11 and delta R two are known values. Those were given to us in the problem, we are going to the substitute for, we're gonna solve for delta R two. And then use the first equation to substitute. So as we simplify this again, a two prize cancel. And we have delta R two is equal to M lambda plus Lambda over two. And that's just multiplying both sides by lambda. M is that delta R one divided by Lambda. So put that in here. So delta R two equals delta R one divided by Lambda multiplied by lambda plus Lambda over two. Um those lambda cancel in that first term. And we're left with delta R two equals delta R one plus Lambda over two, 42, multiplied by the quantity delta R one, sorry, delta R two minus delta R one people slammed up. And so now we can plug in those known values. So the distance or delta R two was 45 centimeters when we had that minimum intensity or zero intensity. And we can rewrite that just using standard uh units of 0.45 m and R 15 centimeters. So that would be 0. equal slam down. And so our wavelength is going to be 0. m or centimeters. And that is all there is to this one, the correct answer is answer choice C All right, we'll see you in the next video.

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 60 cm. a. What is the wavelength of the sound?

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