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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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| 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> |
| $3.3.3$ | | $3.3.3$ |
| A weight, oscillating freely on a spring, has moved from a distance of 0.5 cm from its equilibrium position to the largest one, equal to 1 cm, for a time of 0.01 s. What is the period of its oscillations? | | A weight, oscillating freely on a spring, has moved from a distance of 0.5 cm from its equilibrium position to the largest one, equal to 1 cm, for a time of 0.01 s. What is the period of its oscillations? |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So, | | For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So, |
| $$x(t) = A~\sin{~\omega t}$$ | | $$x(t) = A~\sin{~\omega t}$$ |
| where $A$ is the amplitude such that $A = 1~{\rm{cm}}$, because it is the maximum distance. Let's suppose that $x_0 = 0.5~{\rm{cm}}$ is achieved at $t=t_1$ and $A$ is achieved at $t = t_2$. Then, | | where $A$ is the amplitude such that $A = 1~{\rm{cm}}$, because it is the maximum distance. Let's suppose that $x_0 = 0.5~{\rm{cm}}$ is achieved at $t=t_1$ and $A$ is achieved at $t = t_2$. Then, |
| $$x_0 = A~\sin{~\omega t_1} \;(1)$$ | | $$x_0 = A~\sin{~\omega t_1} \;(1)$$ |
| $$A = A~\sin{~\omega t_2}$$ or | | $$A = A~\sin{~\omega t_2}$$ or |
| $$\sin{~\omega t_2} = 1$$ | | $$\sin{~\omega t_2} = 1$$ |
| Hence, $\omega t_2 = \frac{\pi}{2} + 2k\pi$ with $k\in\mathbb{Z}$, for $k=0$, $t_2 = \frac{\pi}{2\omega}$. As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01~{\rm{s}}$, | | Hence, $\omega t_2 = \frac{\pi}{2} + 2k\pi$ with $k\in\mathbb{Z}$, for $k=0$, $t_2 = \frac{\pi}{2\omega}$. As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01~{\rm{s}}$, |
| $$t_1 = \frac{T}{4} - \Delta t \;(2)$$ | | $$t_1 = \frac{T}{4} - \Delta t \;(2)$$ |
| Putting (2) into (1) and separating $T$, it is obtained | | Putting (2) into (1) and separating $T$, it is obtained |
| $$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin{\frac{x_0}{A}}}$$ | | $$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin{\frac{x_0}{A}}}$$ |
| </p> | | </p> |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$T = 0.06 {\rm{s}}$$ | | $$T = 0.06 {\rm{s}}$$ |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
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