Answer to Question #163186 in Optics for Yolande

Question

 If a human ear canal  can be  thought of as resembling an organ pipe,  closed at one end  that resonates  at a fundamental frequency of 3000  Hertz. what is the length of the canal?

Answer 1

Let's assume that the normal body temperature is $37^{\circ}$ and find the speed of sound in the human ear canal:

$$v=\sqrt{\dfrac{\gamma RT}{M}},$$ $$v=\sqrt{\dfrac{1.4\cdot8.314\ \dfrac{J}{mol\cdot K}\cdot310.15\ K}{0.02896\ \dfrac{kg}{mol}}}=353\ \dfrac{m}{s}.$$

Then, we can find the wavelength of wave, from the wave speed equation:

$$v=f\lambda,$$ $$\lambda=\dfrac{v}{f}=\dfrac{353\ \dfrac{m}{s}}{3000\ Hz}=0.118\ m.$$

For the fundamental (first) harmonic in closed-end pipe we have the following length-wavelength relationship:

$$\lambda=4L,$$ $$L=\dfrac{\lambda}{4}=\dfrac{0.118\ m}{4}=0.0295\ m=2.95\ cm.$$