The auditory canal, leading to the eardrum, is a closed pipe 2.7 cm long. Find the approximate value (ignoring end correction) of the lowest resonance frequency.
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The lowest frequency (also called the fundamental frequency or first harmonic):
f1 = u/(4L) = 340/(4*.027) = 3148 Hz
f1 = u/(4L) = 340/(4*.027) = 3148 Hz
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The shortest wavelength that will resonate in a closed end tube is equal to 4 times the length of the tube. Here lambda = 4(2.7 cm) = 10.8 cm = 0.108 m. Assuming the speed of sound is 340 m/s, we have: f = v / lambda = 340/0.108 = 3148 Hz.
NOTE: This is the highest resonance frequency, the lowest cannot be calculated.
NOTE: This is the highest resonance frequency, the lowest cannot be calculated.
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What an odd way to phrase things ( I'm not very biological).
Anyway: frequency = speed / wavelength.
remember to math the unit scale; probably have to express cm as meters, for example.
And this of coarse would be the speed of sound.
For resonance just divide in succession by two. One real world factor being that beyond three factors (resonants) the volume will drop to negligible.
Anyway: frequency = speed / wavelength.
remember to math the unit scale; probably have to express cm as meters, for example.
And this of coarse would be the speed of sound.
For resonance just divide in succession by two. One real world factor being that beyond three factors (resonants) the volume will drop to negligible.