91Ó°ÊÓ

1. Length of the fossil$\mathrm{L}=2\text{″¾}$
2. Velocity of sound,$\mathrm{v}=343\text{″¾/s}$
3. Person with hearing range, $60\text{±á³ú³Ù´Ç }20\text{kHz}$

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, $\mathrm{f}=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{L}}$ …(¾±)

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:

$\begin{array}{c}\mathrm{f}=1×\frac{343\text{″¾/s}}{2×2\text{″¾}}\text{(}âˆ\mu \text{n=1forfundamentalfrequency)}\\ =85.75\text{ H³ú}\\ ≈86\text{Hz}\end{array}$

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

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.

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.