A 280 Hz sound wave is directed into one end of the trombone slid...

Question

A 280 Hz sound wave is directed into one end of the trombone slide seen in FIGURE P17.55. A microphone is placed at the other end to record the intensity of sound waves that are transmitted through the tube. The straight sides of the slide are 80 cm in length and 10 cm apart with a semicircular bend at the end. For what slide extensions s will the microphone detect a maximum of sound intensity?

Accepted Answer

Welcome back, everyone. We are making observations about the following open end U shaped pipe. We are told that a beam of frequency of Hertz is entering the top of the pipe and comes out the bottom where there is a special type of microphone. Now, we are told that both the top and bottom bars of our pipe start out at 65 centimeters or 650.65 m and we are told that they can extend up to an additional 0.65 m apiece. Now, we are tasked with finding what is going to be the proper amount I or the extension length that will allow us to detect a sound level that is the highest level of sound. So how are we going to do this here? Well, for open tube conditions to create standing waves, we can say that the frequency is equal to N times V or the speed of sound at room temperature divided by two times the length of our pipes with N being numbers that come from the set of positive integers, 123, etcetera. Now we know that the speed of sound at room temperature is 343 m per second. And we're given our frequency, but we still need to solve for our L variable. So let's go ahead and do that. We need to calculate the entire length of our tube including any potential extensions here. So what is this going to be? Well, our length is going to be two times 0.65, which is the length of each of our initial straightaways plus two I because each straightaway can be extended by up 2. m plus the length of our curved part of our tube, which is just going to be one half times the formula for the circumference, which is two pi times the radius A K a hour separation of 15 centimeters or 150.15 m divided by two. What this gives us for our length is 1.536 plus two. I now we are going to sub in this formula into our formula for frequency and then we will solve for I. So let's go ahead and do that. Let me go ahead and change colors here, quick to do that. All right. So we have that our frequency given is 340 Hertz. This is equal to N times 3 divided by two times our length of 1. plus two. I, what I'm gonna do is I'm gonna multiply both sides by our length term, which is 1.536 plus two, I divided by 3 40. Let me do this on the right hand side as well, 1.536 plus two, I divided by 3 40. On the left hand side, this cancels out the 3 40 on the right hand side, this cancels out the length term. What this gives us on the right hand side here is we get 43 N divided by 3 times two, which really gives us 0.5044 10. So we have 1.536 plus two. I is equal to 20.5044 N and subtract 1.536 from both sides of our equation and divide both sides of our equation by two. And what we finally get is N is equal to actually my apologies. I is equal to 0.5044, N minus 1.53 6/2. Now remember our limitation on our extension interval is between zero and 00.65 m. So let's plug in different values of N and see what I that we get here. I've already calculated these out beforehand. So let me write some out here. We have for N equals three, negative 0.114 or N equals four. We have 0.24 and this is in meters by the way. So we'll have to convert back into centimeters for N equals five, we have 0.49 and for N equals six, we have 0.74. Now, you'll see for N equals four and N equals five, it is less than 50.65 and greater than zero. So the, those are our two possible values for I, we'll convert those back into centimeters by multiplying 100 giving us a final answer of 24 49 centimeters for our possible extensions which correspond to our final answer. Choice of B Thank you all so much for watching. Hope this for your help. We will see you all in the next one.

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