Steven W. answered • 07/27/16
Tutor
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(4,319)
Physics Ph.D., college instructor (calc- and algebra-based)
Hi Piper!
It looks like they mean you to approximate the jugs as tubes or air open at one end. Since hearing ranges are usually thought of in terms of frequencies, let's use the expression for the harmonic frequencies generated by an air tube open at one end:
f = (nv)/(4L)
where
L = length of the tube
v = speed of sound (in air, in this case)
n = index number of the harmonic frequencies; in this case, it takes on odd positive integer values: 1, 3, 5... and so on.
Most tubes resonate principally in their first, or fundamental, frequency (n=1), which is also the lowest frequency that tube can produce. So let's concentrate on the n=1 case, the fundamental frequency:
f1 = v/(4L)
Assuming the speed of sound is relatively fixed under given conditions (assuming a value of v = 340 m/s is often a reasonable approximation for standard conditions), the fundamental frequency is inversely proportional to the length of the tube. Since the longest limit is the one being probed, meaning the largest value for L, we must be interested in the lowest value for the frequency. Since the question posits a case where the lowest frequency produced is at the limit of human hearing, we are interested in f1 being lowest frequency the human ear can generally hear.
I took a quick look online and found a typical low-frequency threshold for human hearing is f = 20 Hz.
With f1 = 20 Hz and v = 340 m/s, you can calculate the length L for a jug whose fundamental (lowest) frequency would reach the low-frequency threshold for human hearing.
If you would like to compare a number, just let me know; and also if you have any other questions or want to talk more about this.
Steven W.
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07/27/16
Steven W.
tutor
I apologize. I just had a longer post lost, so I was making sure I could comment here.
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07/27/16
Steven W.
tutor
The reason that I looked at the lowest fundamental frequency of human hearing, given the question's desire to look at the largest container that would push the bounds of human hearing, is because fundamental frequency and length of the container are inversely related. So the largest value length for the container would be matched with the smallest value for human hearing frequency (the lowest frequency).
This inverse relationship between f1 and L can be seen in the equation f1 = v/(4L). Say, for example, that the speed of sound, v, were 120 m/s (a number I just made up), and L was 10 m. Then f1 = 120/(4*10) = 120/40 = 3 Hz. If the tube were lengthened to 30 m, then f1 = 120/(4*30) = 120/120 = 1 Hz. The longer tube has a smaller value for fundamental frequency. So the longest tube would have the lowest value for fundamental frequency. In this case, the longest tube is set to match the lowest frequency of human hearing (~20 Hz).
This relationship can be heard in a "slide whistle." If you are not familiar with a slide whistle, you can find demonstrations on YouTube and elsewhere that show this effect. By pulling on a plunger, you change the pitch of the whistle (pitch is the perception of frequency; higher pitch = higher frequency and lower pitch - lower frequency). If you pull the plunger out, you lengthen the tube and the pitch (frequency) goes down. If you push the plunger back in, you shorten the tube, and the pitch (frequency) goes back up higher.
With this in mind, I solved for the length of the tube above using f1 = 20 Hz and the speed of sound, v = 340 m/s. My equation then looked like this:
20 = 340/(4L), which I then solved for L.
If having L in the denominator throws you off for solving for it, try inverting both sides and solving for L with (4L/340) =(1/20).
Let me know what you get from that. I got a number quite a bit smaller than 68 (mine was under 10). Just let me know if you want to talk about any of these steps more.
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07/27/16
Piper M.
4.25?
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07/27/16
Steven W.
tutor
Yes, that is what I got, too. So the pnly remaining question is whether 4.25 m is a reasonable jug size. Just let me know if you have any mote questions. Good luck!
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07/27/16
Steven W.
tutor
Only remaining question. Any more questions. Stupid iphone keypad.
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07/27/16
Piper M.
Haha I know mine does that too! Thank you so much.
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07/27/16
Piper M.
07/27/16