Chapter 15.1, Problem 4PP

To determine

To calculate:The frequency of sound perceived by a submarine from another submarine.

Answer to Problem 4PP

The frequency of sound perceived by thedetector submarine from another submarine is $3.52\text{ MHz}$ .

Explanation of Solution

Given information:

Speed of sound in water ( $v$ ) = $1482\text{ m/s}$

Velocity of first submarine (source) ( $v_s$ ) = $9.20\text{ m/s}$

Velocity of second submarine (detector) ( $v_d$ ) = $0\text{ m/s}$ ( at rest)

Frequency of wave ( $f_s$ ) = $3.50\text{ MHz}$

The first submarine (source) is moving towards the detector submarine which is at rest.

Formula used:

Doppler Effect equation:

  $f_d=f_s\left(\frac{v-v_d}{v-v_s}\right)$

Where $f_d$ represents frequency perceived by the detector, $f_s$ represents wave’s frequency, $v$ represents speed of sound in air, $v_s$ represents velocity of source and $v_d$ represents velocity of detector.

Calculations:

As per problem,

To calculate frequency perceived by detector, $f_d$ , use formula

  $f_d=f_s\left(\frac{v-v_d}{v-v_s}\right)$

Substitute the values:

  $\begin{array}{l}v=1482\text{ m/s}\\ v_s=9.20\text{ m/s}\\ v_d=0\text{ m/s}\\ f_s=3.50\text{ MHz}\end{array}$

  $\begin{array}{l}{f}_d=3.50\text{ MHz}\left(\frac{1482\text{ m/s - }0\text{ m/s}}{1482\text{ m/s - 9}\text{.20 m/s}}\right)\\ f_d=3.50\text{ MHz}\left(\frac{1482\text{ m/s }}{1472.8\text{ m/s }}\right)\\ f_d=3.50\text{ MHz}(1.006\text{)}\\ f_d=3.52\text{ MHz}\end{array}$

Conclusion:

Thus, the frequency perceived by the detector submarine is $3.52\text{ MHz}$ .