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What frequencies (in Hz) will & 1.85tube producethe audible range (20 Hz 20,000 Hz) at 17.0*C for the following cases?the tube closcdendlowest frequency second ...

Question

What frequencies (in Hz) will & 1.85tube producethe audible range (20 Hz 20,000 Hz) at 17.0*C for the following cases?the tube closcdendlowest frequency second lowest frequency highest frequency (roundedthe nearest Hz)the tubeopenboth endslowest frequencysecond lowest frequencyhighest frcqucncy (roundednearcst Hz)

What frequencies (in Hz) will & 1.85 tube produce the audible range (20 Hz 20,000 Hz) at 17.0*C for the following cases? the tube closcd end lowest frequency second lowest frequency highest frequency (rounded the nearest Hz) the tube open both ends lowest frequency second lowest frequency highest frcqucncy (rounded nearcst Hz)



Answers

The lowest frequency in the audible range is $20 \mathrm{Hz}$. What are the lengths of (a) the shortest open-open tube and (b) the shortest open-closed tube needed to produce this frequency?

This'll question number 54 from Chapter 17 and this problem. We're given a 1.8 meter long tube and air temperature of 20 degrees Celsius. Were asked If I'm with the two will produce in the audible range if a The tomb is open at one end and B if it's open at both ends. So first thing. What is this audible range? Just as a reminder. The audible range of hearing frequency is 20 to 20,000 hertz. So that's the range will be looking for. And then I wanna hunt, wrote the equations that will be using. We need to find the speed of sound and air first, and then we'll be using the frequency equations. This is F equals envy. Over to l is when both ends are open and F equals and V over for l is when only one end is open. All right, so let's find this airspeed airspeed, uh, sound of speed and air, speed of speed, of sound and air. Excuse me 331 meters per second times the square root of 20 plus 2 73 over to 73. And that gives us a speed of 342 0.91 meters per second. All right, so now we have our speed. Let's go on to part a. So frequencies the two will produce in the audible range if it's open at one end, which is going to be n equals V over four of this equation. Okay, so first thing, we need to find the lowest range which we need to find out. If the low. If the resonance frequency would be setting this and equal toe, one is going to give us something in the lower range. So let's all for the resident's frequency one times 3 42.91 meters per second barn by four times 1.8 meters and that is going to give us a frequency 47.63 hurts. So that's the lowest frequency. That this that tube is going to produce of one end is open and that does fall in the range. So that's the lowest. And now we need to find the highest. So to do that, we can set our frequency as 20,000 the upper range of the audible range, and we can solve for N And what that will dio is that will be the exact, um, the exact number, Um, the exact end that would produce this frequency. And we can go down to the energy or value right below that to find what the actual frequency would be, Um, at that residence. So four times 1.8 meters per second. Excuse me, meters, not for a second meters. All right, so we most fly four times 1.8 over the other side. We divide by this speed of air, and that's going to give us an end equal to 419.93 OK, so that's our end that produces exactly 20,000 hurts. But to get the upper range, we're gonna lower this to the energy right below it, because that will give us a realistic frequency. So now we saw for that frequency, So and is gonna be 4 19 times divided by four times 1.8 meters. And that gives us a frequency of 19,955 hurts. And that is your range. Let's see your low waas 43 47.63 So there's a range if it's open at one end. So now we go to be we do the same process, but it's gonna be open at two ends, which was gonna be f equals and the over to l. All right, so it's all for the lowest frequency we could get and see if it's in the range. T o f equals one times 3 42 0.91 meters per second, divided by two times 1.8. And that gives us a frequency of hold on one second calculators of 95.25 It hurts. Okay, so that's gonna be our low. It does fall above 20 hertz, which is the audible range. And now we're gonna solve for the high the same way. So we plug in the upper range and we're gonna solve for end 3 40 2.91 and sulfur end. You get 209.96 and that is the exact number of, um, harmonics. Exact number. So now we go one energy blow that to find the realistic actual frequency that would fall 209 times well 3 42.91 meters per second over two times 1.8 meters, and that gives us a frequency of 19th i 1009 08 Hertz. So arrange here is 95 0.25 hertz to 19,000 908 turds, and that's the range if the tube is open at what?

Since the difference between conjugated harmonics is equal to the fundamental frequency. Therefore, F one will be 3 90 minus 3 25 hearts on this gifts F one Toby 65 foots. This means that the fault that the next harmonic after 1 95 hearts will be equal. Do let's go on that FN induction week. We'll do 1 95 forts less F one. It is 65 4th and back gifts the next harmonic often 1 95 votes to be off No. 60 hoods Now, since f N is a cold o n times F one. This means that Emily B. F an over F one and lugging these two values off f N N f one that we found in the first part. We get in to be equal to full now for part C So only old harmonics are present into bay. No. So the difference between consequence harmonics is twice the fundamental frequency. Unlike, ah, the first case where the difference is equal to the fundamental frequency and that's all we found. F one. So here the difference is twice the fundamental frequencies. Twice off. F one is hearts and this gives F one Toby one drink D. Hurwitz. Remember to take this to into account while doing a calculation. So now that we have f once of the mixed harmonic after 600 boards, then be given by 600. So let's call that again. That will be six under hearts plus twice off F one. Because that's the difference. I'm using this value off a fun. Over here we get f end to be 8 40 huts Now again, since if n is n times if one Andi, we should get an order here because oh, only old Harmon Exactly our present into Beano. That's the reason we took the frequency. The difference between the consecrated harmonics, Toby twice off F one. So this gifts and the FN over F one on dhe substituting half end on a phone from part C get and be killed seven

For this problem on the topic of waves were shown in the figure a small loudspeaker that is driven by oscillator, which has a frequency varied between 1000 and 2000 Hz. We don't have a cylindrical pipe D with two open ends and the length of 45.7 cm. And the speed of sound in the air filled pipe is 344 m/s. We want to know the number of frequencies that the sound from the loudspeaker will set up residence in the pipe. And we want to know the lowest and second lowest frequencies at which resonance occurs. Now if L. Is the pipeline and lambda the wave land, then we know lambda is equal to two L. Over N. Where N is an integer and that is an integer number of half wave lands fit into the length of the pipe. And if he is the speed of sound than the resonant frequencies are given by F. Is able to be over Lunda, which is envy over two out. And so The frequencies are equal to n. Times 344 meters per second, Divided by two times the length of the pipe, which is two times 0 457 m. Which gives the frequency F two, B 376 0.4 times in. Yeah. And so with this expression for the frequencies, if we want to find the resident frequencies that lie between 2000 and 2000 hertz, well first set F equal to 1000 hertz, Which, from the equation about returns in two B 2.66. And then we'll set f equal to 2000 Hz. And we solve for N. Again and this time we get in equal to five 0.32 This implies that in can be 34 and five. And so there are the appropriate for the appropriate values of N. We can see that there are three frequencies. Now we want to find the lowest and second lowest frequencies at which residents occurs well, using the values from above. The lowest frequency occurs when at which resonance occurs is when Is equal to three. And the equation above this means that this frequency F is three into 376.4 huts, Which is 1000, hurts. The second laws frequency occurs When N is equal to four and plugging this into the equation above. We get this resonance frequency F to be for into 300 and 76 0.4 hertz, Which gives the resonant frequency of 1000 506 Hz.

Okay. And open organ pipe has a length of five centimeter. Okay, in discussion there is a there is an open organ pipe. Okay. And its length is given. Okay, L it cost 25 centimeters so or we can say it is five multiplied. But I understood the part minus two m. So in part we have to find out the fundamental frequency of the vibration of the pipe. Okay, so the fundamental frequency of an open open organ pipe can be given as if it was to be divided by two. Okay, so velocity of the sound in the air that is 340 m per second divided by two, multiplied by L. So early is five multiplied. But I'm just to the bar minus 25 multiplied. But I'm just to the pound minus two when we saw it will be 3400 hertz. Okay. Or we can say it will be F equals to 3.4 kg hertz. So this will be the answer. A part of this question and now part B. Okay. And barbie, what will be the highest harmonic of such a tube? That is in the audible range? Inaudible ranges from 20 to 20,000 hearts. Okay, so 3400 hearts. That is the fundamental frequency. Okay, so for this frequency we can find out the required a harmon, highest harmonic. Okay, so required highest harmonic it will be okay, the highest frequency for audible range. That is 20,000 hearts divided by our fundamental frequency. That is 3400 hearts. For 3400 when we saw it will be 5.8. Okay and it will not reach six harmonic. Okay And just lower in teacher than 5.8. It will be five. So five will be the answer of party of this question that is required. I just harmony. Thank you.


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