physics problem?

[Britney T]

assume 345 m/s

the average adult ear canal is about 25 mm long. If we model the ear canal as a closed tube (with the closed end at the ear drum), what is the fundamental frequency of the ear drum, as well as the next three existing harmonics?

[johnmc2]

Hi

The canal's harmonics will occur when you have a displacement node at the drum and an antinode at the open end.

So the longest wavelength that will produce this is when the length of the canal is 1/4 of a wavelength.
That gives the wavelength as 4*0.025=0.10m
The speed of sound is about 330m/s so as f=v/?, f=330/0.10=3300Hz

Because it's a closed tube, we won't get the second, fourth, sixth, or any even-numbered harmonics. That is because that would mean a node at the open end, which ain't possible.

So the next one will be the third harmonic (f=3300*3=9900Hz); followed by the fifth at 16,500Hz; and the seventh at 23,100Hz.

Hope this helps.