Two organ pipes, open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2.00 cm. Find the beat frequency that they produce when playing together in their fundamentals.

Video transcript

Hey everyone. This problem is working with sound and beat frequency. So let's read the question and see what they're asking for. A set of hollow pipes are closed on one end by immersing them in water, leaving only one open end that's open to air. The lengths of the pipes above the water differ by 20 centimeters, determine the beat frequency for a pipe with a length of two m above the water. And this is a two part question. The first part is asking where the second pipe is longer. By this differing of 220 cm. And the second part of the question is the second pipe is shorter by 20 centers. So let's write down some some givens here. We know length. One of the pipe is two m in the first part of the question length too, Is that 20 m plus because it's longer. 20 cm, 20 cm can be rewritten as .2 m. So our second length is 2.2 m For the 2nd part of the equation. Sorry for the second part of the question, This L. two is shorter by .2 m. So we have two m -1 m and that's 1.8 m. We also know that the speed of sound and air is 343 m/s. We need to recall that constant and this uh they tell us to consider waves with fundamental frequency only and that tells us the fundamental frequency and is one. So what we need to do now is recall our beat frequency equation. So that's F. Beat is F one minus F. To the absolute value of that. Where the frequency recall that is given by N. V. Over four L. We already know that N. Is one. So now we're looking at the over four L. So F one minus up to can be rewritten as V over four. L one minus V over four L. Two. The velocity the velocities are the same. It's just the sound, the velocity of sound and air. So 403 143 m per second over four times two m minus 343 m per second Over four times 2.2 m equals 3. Hz. So we can look at our potential answers here and only one of them actually has the first correct answer for part one. But let's let's solve part two anyway and make sure that we're on the right track with how we're solving these equations. So we're gonna just take the same equation again, the only difference is going to be this L2 term, So V 1 -4. L one minus B. 2/4 L. Two, 343 m per second, times four times two m that's L. One stayed the same And then m/s Over four times 1. meters. That L two term is what changed between the first question and the second question is this of this problem, Plug all of that in and we get 4. hertz and looking back at our answer. That is the answer for part B or part two of this question. So answer B is correct. That's all we have for this video, folks. We'll see you in the next one.