Posted by Sammi on Thursday, March 24, 2011 at 10:22pm.
Since it one end is open and the other is shut, the ear canal is considered a closed pipe. We set L = 2.3cm = 0.023m. The frequency for a closed pipe is fn=n(v/4L)=v/wavelength where n=1,2,3,.. referring to the number of overtones/harmonics. L=(1/4)*wavelength so we solve for wavelength 0.023m=(1/4)*wavelength and get wavelength=0.092m. Since the audible frequency range is 20Hz-20,000Hz, we solve the equation fn=n(v/wavelength) using n=1,2,3,... until we reach 20,000Hz. Here, we f1=3728 Hz, f2=11185 Hz, f3=18642 Hz (f4 is higher than the audible range so we know to stop at n=3).
Oh and use the speed of sound of air (at 20 degrees C) to be v=343m/s.
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