A sperm whale (Fig. 17-44a) vocalizes by producing a series of clicks. Actually, the whale makes only a single sound near the front of its head to start the series. Part of that sound then emerges from the head into the water to become the first click of the series. The rest of the sound travels backward through the spermaceti sac (a body of fat), reflects from the frontal sac (an air layer), and then travels forward through the spermaceti sac. When it reaches the distal sac (another air layer) at the front of the head, some of the sound escapes into the water to form the second click, and the rest is sent back through the spermaceti sac (and ends up forming later clicks).

Figure 17-44bshows a strip-chart recording of a series of clicks. A unit time interval of$\mathbf{1}\mathbf{.0}\mathbf{\text{\hspace{0.17em}ms}}$is indicated on the chart. Assuming that the speed of sound in the spermaceti sac is$\mathbf{1372}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$, find the length of the spermaceti sac. From such a calculation, marine scientists estimate the length of a whale from its click series.

Step by step solution

01

The given data

1. Speed of sound is$1372\text{\hspace{0.17em}m}/\text{s}$
2. Unit time interval is$1\text{\hspace{0.17em}ms}$

02

Understanding the concept of distance

Here we have to use the basic formula for speed in terms of distance and time to find the length of the spermaceti sac.

Formula:

The speed of the wave,

$\mathbf{v}\mathbf{=}\frac{\mathbf{d}}{\mathbf{t}}$ …(i)

03

Calculation of spermaceti sac

The total distance traveled by the sound is 2L,which is equal to twice the length of spermaceti sac.This distance is traveled in $1\mathrm{ms}$.Using this information in the formula of equation (i), we get the length value as:

So

$\begin{array}{c}\mathrm{v}=\frac{2\mathrm{L}}{\mathrm{t}}\\ 1372\text{\hspace{0.17em}m}/\text{s}=\frac{2\mathrm{L}}{3\times {10}^{-3}\text{\hspace{0.17em}s}}\\ \mathrm{L}=2.1\text{\hspace{0.17em}m}\end{array}$

Hence, the length of spermaceti sac is,$2.1\text{\hspace{0.17em}m}$ .

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