Two loudspeakers in a 20°C room emit 686 Hz sound waves along the x-axis. b. If the speakers are out of phase, what is the smallest distance between the speakers for which the interference of the sound waves is maximum constructive?

Video transcript

Hey, everyone. This problem is dealing with sound waves and frequency. Let's see what it's asking us. Two sound emitters having high radiant phase difference are placed in a 25 degree Celsius hall on a horizontal graduated ruler. The frequency of the two emitters is 485 Hertz. A microphone probe is placed in a fixed position to the right of the two emitter of the two emitters on the graduated ruler, calculate the minimum distance that should separate the two sound emitters in order to measure with that microphone probe, a constructive interference. Our multiple choice answers here are a centimeters b 36 centimeters C centimeters or D 71 centimeters. So the key to this problem is recalling that a constructive interference means the maximum interference. And that happens when our phase difference or delta five is equal to two M pi. And our delta phi equation for that um change and phase difference is given by two pi multiplied by delta R divided by lambda plus delta five. Now the minimum distance tells us that M is equal to one M as an integer. And so for the minimum that's just going to be too high. And so we set our delta five equal to two pi, so we'll have two pi equals two pi delta R divided by lambda plus that initial phase difference was given to us. The problem as will subtract pie from both sides. And then we'll have and multiply both sides by lambda. So we'll have pie, lambda equals two pie. Delta R and delta R is that distant, That distance that we're solving for. So now we'll cancel the pies and solve for delta R. And that is going to be lambda divided by two. But we weren't given lambda in the problem. We were given frequency. So we can recall the relationship between lambda or wavelength and frequency is lambda equals V divided by F. We were also told the um the temperature in the hall 25 degrees Celsius. And so V, the speed of sound and air is dependent on temperature. So that's a value that you would look up um based on that temperature. So from there, we can plug V divided by F into our in four lambda. And then R delta R equation becomes V divided by two F. And so that speed of sound at 25 degrees Celsius is 346 m per second. That's divided by two, multiplied by the frequency given to us. In the problem. 485 Hertz, you plug that in and we get 0. 56 m. And we look at our answer choices, they are in centimeters. So that equals 36 centimeters and that aligns with answer choice B so that's all we have for this one. We'll see you in the next video.