Question

Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.6 - m - long hollow tube in the crest had connections to the nose and throat, leading some investigators to hypothesize that the tube was a resonant chamber for vocalization.

If you model the tube as an open-closed system, what are the first three resonant frequencies? Assume a speed of sound of 350 m/s.

Answer 1

in a open closed pipe

at open end always a antinode is formed

at closed end a node is formed

the distance between the node and antinode = wavelength / 4

for first resonant frequency L = wavelength / 4

wavelength = 4L

frequency fo = v/wavelength = v/4L = 350/(4*1.6) = 54.7 Hz

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for second resonant

there will be antinode - node - antinode - node = 3*wavelength/4

L = 3*wavelength/4

wavelength = 4L/3

frequeny = v/wavelength = 3v/4L = 3fo = 164.1 Hz

for third resonant

there will be antinode - node - antinode - node-antinode - node = 5*wavelength/4

L = 5*wavelength/4

wavelength = 4L/5

frequeny = v/wavelength = 5v/4L = 5fo = 273.5 Hz