Question

Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.4 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? (Use 350 m/s for the speed of sound.)

Enter your answers using two significant figures in ascending order separated by commas.
f = _______Hz, ________Hz, ________Hz

Answer 1

Concepts and reason

The concepts required to solve the given problem are frequency, wavelength, and velocity.

Initially, derive the expressions for the first three resonant frequencies by using the relation between velocity, frequency, and wavelength. Finally, calculate the first three resonant frequencies by using the expressions of the frequencies of open closed pipe.

Fundamentals

The lowest frequency which is produced by the oscillation of an object is called as fundamental frequency.

The fundamental frequency of open closed pipe is,

$f = \frac{v}{{4L}}$

Here, f is the frequency of sound, v is the velocity of sound, and L is the length of the tube.

The velocity of the sound in air is,

$v = f\lambda$

Here, $\lambda$ is the wavelength.

For the first harmonic of the closed pipe, the length of the tube is same as the length of ¼ wavelength.

$\begin{array}{c}\\L = \frac{\lambda }{4}\\\\\lambda = 4L\\\end{array}$

The velocity of the sound in air is,

$v = {f_1}\lambda$

Here, ${f_1}$ is the first resonant frequency.

Substitute 4L for $\lambda$ .

$\begin{array}{c}\\v = {f_1}\left( {4L} \right)\\\\{f_1} = \frac{v}{{4L}}\\\end{array}$

For the third harmonic of the closed end pipe, the length of the tube is,

$\begin{array}{c}\\L = \frac{3}{4}\lambda \\\\\lambda = \frac{{4L}}{3}\\\end{array}$

Since 3/4 of a wavelength fit into the tube.

The velocity of the sound in air is,

$v = {f_2}\lambda$

Here, ${f_2}$ is the second resonant frequency.

Substitute $\frac{{4L}}{3}$ for $\lambda$ .

$\begin{array}{c}\\v = {f_2}\left( {\frac{{4L}}{3}} \right)\\\\{f_2} = \frac{{3v}}{{4L}}\\\end{array}$

For the Fifth harmonic of the closed end pipe, the length of the tube is,

$\begin{array}{c}\\L = \frac{5}{4}\lambda \\\\\lambda = \frac{{4L}}{5}\\\end{array}$

Since 5/4 of a wavelength fit into the tube.

The velocity of the sound in air is,

$v = {f_3}\lambda$

Here, ${f_3}$ is the third resonant frequency.

Substitute $\frac{{4L}}{5}$ for $\lambda$ .

$\begin{array}{c}\\v = {f_3}\left( {\frac{{4L}}{5}} \right)\\\\{f_3} = \frac{{5v}}{{4L}}\\\end{array}$

The first resonant frequency is,

${f_1} = \frac{v}{{4L}}$

Substitute 350 m/s for v and 1.4 m for L.

$\begin{array}{c}\\{f_1} = \frac{{350\;{\rm{m/s}}}}{{4\left( {1.4\;{\rm{m}}} \right)}}\\\\ = 62.5\;{\rm{Hz}}\\\end{array}$

The second resonant frequency is,

${f_2} = \frac{{3v}}{{4L}}$

Substitute 350 m/s for v and 1.4 m for L.

$\begin{array}{c}\\{f_2} = \frac{{3\left( {350\;{\rm{m/s}}} \right)}}{{4\left( {1.4\;{\rm{m}}} \right)}}\\\\ = 187.5\;{\rm{Hz}}\\\end{array}$

The third resonant frequency is,

${f_3} = \frac{{5v}}{{4L}}$

Substitute 350 m/s for v and 1.4 m for L.

$\begin{array}{c}\\{f_3} = \frac{{5\left( {350\;{\rm{m/s}}} \right)}}{{4\left( {1.4\;{\rm{m}}} \right)}}\\\\ = 312.5\;{\rm{Hz}}\\\end{array}$

Ans:

The first three resonant frequencies are 62 Hz, 190 Hz, and 310 Hz.