Sound Waves Legacy Problem #27 Guided Solution
Problem*
A musical recorder acts as an open-end air column, with a vibrational antinode located at the hole near the mouthpiece (known as the windway hole) and a vibrational antinode located at the nearest open tone hole. Blowing gently on the mouthpiece will force the air column to vibrate at its fundamental frequency. Assuming a speed of sound of 345 m/s, what length of air would be required to cause the recorder to sound out at 1050 Hz?
Audio Guided Solution
Sometime in your life you've likely have blown on a recorder to produce a sound. It's just a small little toy that kids often use. You blow gently on one end and it begins to produce a rather high-pitched sound. You can adjust the frequency of the sound it produces by covering some of the holes within that little recorder. And the idea is it acts as an open-end air column which stretches from the windway hole to the first open hole from the windway hole. Now the windway hole is a little slot near the mouthpiece of the recorder that has an antinodal position at that point. In other words, air is vibrating vigorously in and out at that point. And the first opened hole from the windway hole is the end of that open-end air column. At that end, air is also vibrating back and forth vigorously to form an antinode. Here we're told that the fundamental frequency of a recorder is 1050 Hz and the speed is 345 meters per second. We want to know what length of the air column from windway hole to first open hole would produce such a fundamental frequency. To do so, we have to think in terms of how things are related. And what we know right away is that by the wave equation, the V, F, and a lambda are related. So if I know V and F as I do, I could calculate the wavelength. The other thing that I know about standing wave patterns within air columns is that the wavelength is related to the length according to the standing wave pattern. And for an open-end air column in the fundamental frequency, within the length of the tube, there's a half of a wavelength. So if I know wavelength, I can calculate length. So my strategy is calculate wavelength using wave equation. And then second step, find the length using the standing wave pattern. So finding the wavelength demands that you take the V equal F times lambda equation and rearrange it for lambda. Lambda is equal to V over F, where lambda's wavelength, V, is the speed, 345, and F is the frequency, 1050 Hz. Doing so will get you a wavelength of 0.3286 meters. Now, second step, find the length. Within the length of the tube, there's a half of a wavelength. So L equals one-half times lambda. Lambda is 0.3286 meters, and so L is 0.1643 meters, or 16.43 centimeters. I can round that to three significant digits, where it's 0.164 meters, or 16.4 centimeters.
Solution
0.164 m or 16.4 cm
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities and record in an organized manner, often times they can be recorded on the diagram itself. Equate given values to the symbols used to represent the corresponding quantity (e.g., \(\descriptive{v}{v,velocity} = 345\unit{\meter\per\second}\), \(\descriptive{λ}{λ,wavelength} = 1.28 \unit{m}\), \(\descriptive{f}{f,frequency} = \colorbox{gray}{Unknown}\)).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read About It!
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