Question

The crest of a Parasaurolophus dinosaur skull contains a nasal passage in the shape of a long, bent tube open at both ends. The dinosaur may have used the passage to produce sound by setting up the fundamental mode in it.

(a) If the nasal passage in a certain Parasaurolophus fossil is$\mathbf{2}\mathbf{.0}\mathbf{}\mathbf{\text{m}}$long, what frequency would have been produced?

(b) If that dinosaur could be recreated (as in Jurassic Park), would a person with a hearing range of$\mathbf{60}\mathbf{}\mathbf{\text{Hz}}\mathbf{}\mathbf{\text{\hspace{0.17em}to\hspace{0.17em}}}\mathbf{}\mathbf{20}\mathbf{}\mathbf{\text{kHz}}$be able to hear that fundamental mode and, if so, would the sound be high or low frequency? Fossil skulls that contain shorter nasal passages are thought to be those of the female Parasaurolophus.

(c) Would that make the female’s fundamental frequency higher or lower than the male’s?

Short answer

1. The frequency that would have been produced is$86\text{Hz}$
2. Yes, the person would be able to hear the frequency but it would be a low frequency wave.
3. The female’s frequency would be higher to that of the male.

Step-by-step solution

Identification of given data

1. Length of the fossil$\mathrm{L}=2\text{\hspace{0.17em}m}$
2. Velocity of sound,$\mathrm{v}=343\text{\hspace{0.17em}m/s}$
3. Person with hearing range, $60\text{Hzto\hspace{0.17em}}20\text{kHz}$

Significance of frequency

The number of waves passing a fixed location in a unit of time is referred to as frequency in physics.

Assuming the nasal cavity is a pipe open at both ends; we can find the fundamental frequency and answer the remaining questions.

Formula:

The frequency of body with n number of oscillations, $\mathrm{f}=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{L}}$ …(i)

(a) Determining the frequency

Using equation (i), we can write the formula for the resonant frequency of the pipe, which is open at both ends. Hence, the frequency that would have been produced is given as:

$\begin{array}{c}\mathrm{f}=1\times \frac{343\text{\hspace{0.17em}m/s}}{2\times 2\text{\hspace{0.17em}m}}\text{(}\because \text{n=1forfundamentalfrequency)}\\ =85.75\text{\hspace{0.17em}Hz}\\ \approx 86\text{Hz}\end{array}$

Hence, the value of frequency is$86\text{Hz}$

(b) Checking whether a person of hearing range 60Hz to 20kHz would hear the frequency

From the above part (a), the frequency is found to be $86\text{Hz}$.

Hence, the sound would be perceptible as the sound frequency, which is$60\text{Hz,}$ is within the hearing range of the person, but it is still a very small frequency. So, the sound of the frequency will be low.

(c) Checking whether the fundamental frequency of female be high or low than that of male

The pitch of the sound waves depends on the frequency. Higher the frequency, higher would be the pitch. Since the length of the nasal passage is smaller for the female Parasaurolophus, the frequency would be higher by the above formula. Therefore, the pitch would also be higher.