3. Let\' s consider a very rough model of the vocal tract as a tube of length L

Question

3. Let' s consider a very rough model of the vocal tract as a tube of length L 0.15 m with one open end (mouth) and one closed end (vocal cords). The fundamental wavelength (first harmonic) of the vocal tract is 1-4L (For simplicity, ignore the fact used in the lab that L should be replaced by the "corrected length" L + 0.4D.) a) Compute the fundamental wavelength of the vocal tract, and the corresponding fundamental frequency. For the latter, you need the speed of sound in air at room temperature. (You can use your answer to the previous problem if you want.) b) Now, suppose the person inhales helium. The frequencies emitted by the sound source (the vibrating vocal cords) do not change when a person inhales helium, since the vocal cords vibrate the same as before. Instead, since the harmonic frequencies of the vocal tract depend on the speed of sound in the medium, they change when we replace air with helium. Compute the fundamental frequencies of the vocal tract when the medium of the sound waves is helium instead of air. (The fundamental wavelength is the same as for air. What changes is the fundamental frequency.) c) Was the fundamental frequency of the vocal tract when the medium is helium higher or lower than when the medium is air? What effect would inhaling helium has on the person' s voice (higher or lower pitch)?

Explanation / Answer

3. given length of volcal tract, L = 0.15 m

one open end

one closed end

fundamental wavelength, lambda1 = 4L

a. hence

lambda1 = 4*0.15 = 0.6 m

now c= 343 m/s

hence

f1 = c/lambda1 = 571.666666666 Hz

b. when helium is used,

c = 972 m/s

hence

f1 = c/lambda1 = 1620 Hz

c. hence the frequency of the person after inhaling helium increases, so inhaling helium would raise a person pitch

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