Savchenko Solutions

Solutions of Savchenko Physics Textbook

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.1.7.$ A submarine diving vertically and uniformly emits sound pulses of duration $\tau_0$. The duration of the pulse reflected from the bottom is $\tau$. The speed of sound in water is $c$. At what speed does the submarine dive?
    </p>

    <h3>Solution</h3>
    <p>
                      <p>
          Let's consider what the change of signal duration is related to. For this purpose, it is convenient to pass from the signal duration to its length. 
        </p>
          <p>
          Let the signal emitted by the sonar on the boat have the form shown in the figure, i.e. its intensity jumps from zero to some value at time $t$, remains constant during time $\tau_{0}$ and again jumps to zero at time $t + \tau_{0}$. 
        </p>
          <p>
          If the boat is stationary, the extent of the radiated signal (the distance between the leading and trailing edges) is $l_{0} = V \tau_{0}$. If the boat is sinking, the extent of the signal becomes smaller. 
        </p>
          <p>
          Indeed, let $AC = u \tau_{0}$, where $u$ is the submergence velocity of the boat. 
        </p>
          <p>
          The leading edge of the signal will move to point $B$ in time $\tau_{0}$, having travelled a distance $l_{0}$. 
        </p>
          <p>
          Hence, the signal length is equal to 
        </p>
        <p style="text-align: center;">
          $$l_{1} = BC = ( V - u ) \tau_{0}$$
        </p>
          <p>
          This signal reaches the bottom, discharges from it without changing its extent, and travels towards the boat.
        </p>
          <p>
          At some point in time, the leading edge of the signal will align with the boat.  
        </p>
          <p>
          The rear front of the signal continues to move with velocity $V + u$ relative to the boat, therefore, the duration of the signal received on the boat is equal to
        </p>
        <p style="text-align: center;">

$$\tau = \frac{l_{1} }{V + u} = \frac{(V - u) \tau_{0} }{V + u}$$

From here

$$u = V \frac{ \tau_{0} - \tau }{ \tau_{0} + \tau }$$

    <h4>Answer</h4>
    <p>
        $$v = c(\tau _{0} − τ)/(\tau _{0} + τ)$$
    </p>


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