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| + | <body style=""> | ||
| + | <header style="text-align:center;"> | ||
| + | <h2>Solutions of Savchenko Physics Textbook</h2> | ||
| + | <p class="author"> | ||
| + | Aliaksandr Melnichenka <br/> | ||
| + | October 2023 | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <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;"> | ||
| + | $$\tau = \frac{l_{1} }{V + u} = \frac{(V - u) \tau_{0} }{V + u}$$ | ||
| + | </p> | ||
| + | <p> | ||
| + | From here | ||
| + | </p> | ||
| + | <p style="text-align: center;"> | ||
| + | $$u = V \frac{ \tau_{0} - \tau }{ \tau_{0} + \tau }$$ | ||
| + | </p> | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$v = c(\tau _{0} − τ)/(\tau _{0} + τ)$$ | ||
| + | </p> | ||
| + | |||
| + | |||
| + | <footer class="row container"> | ||
| + | <br> | ||
| + | <p> | ||
| + | <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | ||
| + | </p> | ||
| + | <p> | ||
| + | <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small> | ||
| + | </p> | ||
| + | </footer> | ||
| + | </body> | ||
| + | |||
| + | </html> | ||
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| <!DOCTYPE html> | |||
| <html lang="en"> | |||
| <head> | |||
| <meta charset="utf-8"> | |||
| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |||
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| <meta name="keywords" content="solutions, savchenko, physics problems, olympiad physics, physics book"> | |||
| <meta name="description" content="A website with solutions to physics problems from Savchenko Textbook"> | |||
| <meta property="og:title" content="Savchenko Solutions"> | |||
| <meta property="og:image" content="img/logo.png"> | |||
| <meta property="og:description" content="A website with solutions to physics problems from Savchenko Textbook"> | |||
| <meta name="yandex-verification" content="6cfda41f74038368"> | |||
| <title>Savchenko Solutions</title> | |||
| <link rel="stylesheet" href="https://savchenko-physics.github.io/css/css-latex/style.css"> | |||
| <link rel="icon" href="https://savchenko-physics.github.io/img/logo.png" type="image/png"> | |||
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| MathJax.Hub.Config({ | |||
| extensions: ['tex2jax.js'], | |||
| jax: ['input/TeX', 'output/HTML-CSS'], | |||
| tex2jax: { | |||
| inlineMath: [['$', '$'], ['$', '$']], | |||
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| ignoreClass: 'html' | |||
| }, | |||
| showProcessingMessages: false, | |||
| messageStyle: 'none' | |||
| }); | |||
| </script> | |||
| </head> | |||
| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <h2>Solutions of Savchenko Physics Textbook</h2> | |||
| <p class="author"> | |||
| Aliaksandr Melnichenka <br/> | |||
| October 2023 | |||
| </p> | |||
| </header> | |||
| <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;"> | |||
| $$\tau = \frac{l_{1} }{V + u} = \frac{(V - u) \tau_{0} }{V + u}$$ | |||
| </p> | |||
| <p> | |||
| From here | |||
| </p> | |||
| <p style="text-align: center;"> | |||
| $$u = V \frac{ \tau_{0} - \tau }{ \tau_{0} + \tau }$$ | |||
| </p> | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$v = c(\tau _{0} − τ)/(\tau _{0} + τ)$$ | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
| </html> | |||