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The human outer ear contains a more-or-less cylindrical cavity called the auditory canal that behaves like a resonant tube to aid in the hearing process. One end terminates at the eardrum (tympanic membrane), while the other opens to the outside. (See the figure (Figure 1) .) Typically, this canal is approximately 2.21cm long.
part A
At what frequencies would it resonate in its first two harmonics?
Express your answers separated by a comma.
Part B
What are the corresponding sound wavelengths in part A?
Express your answers separated by a comma.
To find the frequencies at which the auditory canal would resonate in its first two harmonics, we can use the formula for the resonant frequencies of a closed cylindrical tube:
f = (n * v) / (2 * L)
Where:
- f is the frequency of resonance
- n is the harmonic number (1 for the first harmonic, 2 for the second harmonic, and so on)
- v is the speed of sound in air (approximately 343 m/s at room temperature)
- L is the length of the auditory canal
Part A:
For the first harmonic (n = 1):
f1 = (1 * 343 m/s) / (2 * 0.0221 m)
= 7815.83 Hz
For the second harmonic (n = 2):
f2 = (2 * 343 m/s) / (2 * 0.0221 m)
= 15631.66 Hz
So, the frequencies at which the auditory canal would resonate in its first two harmonics are 7815.83 Hz and 15631.66 Hz.
Part B:
To find the corresponding sound wavelengths, we can use the formula for the wavelength of a sound wave:
λ = v / f
Where:
- λ is the wavelength of the sound wave
- v is the speed of sound in air
- f is the frequency of the sound wave
For the first harmonic (n = 1) at 7815.83 Hz:
λ1 = 343 m/s / 7815.83 Hz
= 0.0439 m
For the second harmonic (n = 2) at 15631.66 Hz:
λ2 = 343 m/s / 15631.66 Hz
= 0.022 m
So, the corresponding sound wavelengths in part A are 0.0439 m and 0.022 m.