Sound Waves Legacy Problem #23 Guided Solution
Problem*
An organ pipe which acts as a open-end resonator has a length of 83 cm. Its fundamental frequency is 210 Hz.
- Determine the speed of sound waves in the air column of the pipe.
- Determine the frequency of the second, third and fourth harmonics of the organ pipe.
Audio Guided Solution
Here is a question about an organ pipe which is acting as an open-end resonance column. As an open-end resonance column, it will have antinodes at both of the open ends. If it vibrates in its fundamental frequency pattern, then between the ends will be a single node right in the center, in such a manner that there is one half of a wavelength within the length of the open-end air column. Now the length of this open-end air column is 83 centimeters, and that's the length of it, not the wavelength. That 83 centimeters is equal to a half of a wavelength. We're also given that the fundamental frequency for this pattern is 210 Hertz. Or I ask two questions. Find the V, or the speed of sound waves within the air column, and find the frequency of the second, third, and fourth harmonics of the organ pipe. Find V, and find F2, F3, and F4. So to do part A of this problem, I know that to find the speed, I need to know the frequency and the wavelength. And as mentioned, the 83 centimeters is the length of the tube, and it's equal to one half of a wavelength. So if I say 83 equal 0.5 lambda, and then multiply both sides by 2, I'll end up getting lambda equal 166 centimeters, or 1.66 meters. Now I know frequency, and I know wavelength for the first harmonic. And I can say V equal F times lambda, or F times wavelength, 210 times 1.66 meters, gives me 348.6 meters per second, or rounding to two significant digits, 350 meters per second. Part B demands that we find the second, third, and fourth harmonic frequencies for this same air column. To do that, what I need to do is recognize that the frequency of the nth harmonic is equal to n times the frequency of the first harmonic, where n can be any whole number integer, 2 for the second, 3 for the third, and 4 for the fourth. So the second harmonic frequency is twice the 210 Hertz, or 20 Hertz. The third harmonic frequency is three times the 210 Hertz, or 630 Hertz. And the fourth harmonic frequency is four times 210 Hertz, or 840 Hertz.
Solution
- 350 m/s (rounded from 349 m/s)
- 420 Hz, 630 Hz, 840 Hz
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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