Chapter 20, Problem 69A

(a)

To determine

The distance of the submarine from the seamount.

Answer to Problem 69A

The distance of the submarine from the seamount is $1,380\text{ m}$ .

Explanation of Solution

Given:

The velocity of the submarine is $v_s=12.0\text{m}/\text{s}$ .

The frequency of the sonar ping is $f_{\text{s}}=1.50\times {10}^3\text{Hz}$ .

The time for echo $t_{\text{echo}}=1.800\text{ s}$ .

Formula used:

The expression for the distance is,

$d=vt$

Here, $v$ is the velocity and $t$ is the time.

Calculation:

Consider the velocity of sound in water as $v=1,533\text{m}/\text{s}$ .

The time taken to reach the seamount from the submarine is,

$\begin{array}{l}t=\frac{t_{\text{echo}}}{2}\\ t=\frac{1.800\text{ s}}{2}\\ t=0.900\text{ s}\end{array}$

The distance of the submarine from the seamount is,

$\begin{array}{l}d=vt\\ d=\left(1,533\text{m}/\text{s}\right)\left(0.900\text{ s}\right)\\ d=1,380\text{ m}\end{array}$

Conclusion:

Thus, the distance of the submarine from the seamount is $1,380\text{ m}$ .

(b)

To determine

The frequency of the sonar wave that strikes the seamount.

Answer to Problem 69A

The frequency of the sonar wave that strikes the seamount is $1,510\text{ Hz}$ .

Explanation of Solution

Given:

The velocity of the submarine is $v_s=12.0\text{m}/\text{s}$ .

The frequency of the sonar ping is $f_{\text{s}}=1.50\times {10}^3\text{Hz}$ .

Formula used:

The expression for the frequency of the sonar wave is,

$f_{\text{d}}=f_{\text{s}}\left(\frac{v-v_{\text{d}}}{v-v_{\text{s}}}\right)$

Here, $f_{\text{s}}$ is the frequency of the sonar ping, $v$ is the velocity, $v_{\text{d}}$ is the velocity of the sonar wave, and $v_{\text{s}}$ is the velocity of the submarine.

Calculation:

Consider the velocity of sound in water as $v=1,533\text{m}/\text{s}$ and the velocity of the sonar wave $v_d=0$ .

The frequency of the sonar wave that strikes the seamount is,

$\begin{array}{l}{f}_{\text{d}}=f_{\text{s}}\left(\frac{v-v_{\text{d}}}{v-v_{\text{s}}}\right)\\ f_{\text{d}}=\left(1.50\times {10}^3\text{Hz}\right)\left(\frac{\left(1,533\text{m}/\text{s}\right)-\left(0.0\text{m}/\text{s}\right)}{\left(1,533\text{m}/\text{s}\right)-\left(12\text{m}/\text{s}\right)}\right)\\ f_{\text{d}}=1,510\text{ Hz}\end{array}$

Conclusion:

Thus, the frequency of the sonar wave that strikes the seamount is $1,510\text{ Hz}$ .

(c)

To determine

The frequency of the echo.

Answer to Problem 69A

The frequency of the echo is $1,520\text{ Hz}$ .

Explanation of Solution

Given:

The velocity of the submarine is $v_s=12.0\text{m}/\text{s}$ .

The frequency of the sonar ping is $f_{\text{s}}=1.50\times {10}^3\text{Hz}$ .

Formula used:

The expression for the frequency is,

$f_{\text{d}}=f_{\text{s}}\left(\frac{v-v_{\text{d}}}{v-v_{\text{s}}}\right)$

Here, $f_{\text{s}}$ is the frequency of the sonar wave, $v$ is the velocity, $v_{\text{d}}$ is the velocity of the echo, and $v_{\text{s}}$ is the velocity of the sonar.

Calculation:

Consider the velocity of sound in water as $v=1,533\text{m}/\text{s}$ and the velocity of the echo $v_d=-12\text{m}/\text{s}$ .

The velocity of the sonar is $v_{\text{s}}=0$ .

The frequency of echois,

$\begin{array}{l}{f}_{\text{d}}=f_{\text{s}}\left(\frac{v-v_{\text{d}}}{v-v_{\text{s}}}\right)\\ f_{\text{d}}=\left(1,510\text{Hz}\right)\left(\frac{\left(1,533\text{m}/\text{s}\right)-\left(-12\text{m}/\text{s}\right)}{\left(1,533\text{m}/\text{s}\right)-\left(0\text{m}/\text{s}\right)}\right)\\ f_{\text{d}}=1,520\text{ Hz}\end{array}$

Conclusion:

Thus, the frequency of the echo is $1,520\text{ Hz}$ .