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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#1.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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? | | $1.1.7.$ 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> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| <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. | | 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> |
| <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 $t + \tau_{0}$. | | 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> |
| <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. | | 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> |
| <p> | | <p> |
| Indeed, let $AC = u \tau_{0}$, where $u$ is the submergence velocity of the boat. | | Indeed, let $AC = u \tau_{0}$, where $u$ is the submergence velocity of the boat. |
| </p> | | </p> |
| <p> | | <p> |
| The leading edge of the signal will move to point $B$ in time $\tau_{0}$, having travelled a distance $l_{0}$. | | The leading edge of the signal will move to point $B$ in time $\tau_{0}$, having travelled a distance $l_{0}$. |
| </p> | | </p> |
| <p> | | <p> |
| Hence, the signal length is equal to | | Hence, the signal length is equal to |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$l_{1} = BC = ( V - u ) \tau_{0}$$ | | $$l_{1} = BC = ( V - u ) \tau_{0}$$ |
| </p> | | </p> |
| <p> | | <p> |
| This signal reaches the bottom, discharges from it without changing its extent, and travels towards the boat. | | This signal reaches the bottom, discharges from it without changing its extent, and travels towards the boat. |
| </p> | | </p> |
| <p> | | <p> |
| At some point in time, the leading edge of the signal will align with the boat. | | At some point in time, the leading edge of the signal will align with the boat. |
| </p> | | </p> |
| <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 | | 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> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$\tau = \frac{l_{1} }{V + u} = \frac{(V - u) \tau_{0} }{V + u}$$ | | $$\tau = \frac{l_{1} }{V + u} = \frac{(V - u) \tau_{0} }{V + u}$$ |
| </p> | | </p> |
| <p> | | <p> |
