Question

The human ear canal is approximately 2.3 cm long. It is open to the outside and is closed at the other end by the eardrum.

Estimate the frequencies (in the audible range) of the standing waves in the ear canal. Express your answers using two significant figures. If there is more than one answer, enter them in ascending order separated by commas.

Answer 1

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).

Answer 2

Oh and use the speed of sound of air (at 20 degrees C) to be v=343m/s.