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en/3.5.18.md
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| + | ### Statement | ||
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| + | $3.5.18.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Problem | ||
| + | |||
| + | $3.5.18.$ If at time $t = 0$ a damped oscillator is in its equilibrium position and its velocity is $v_0$, then its coordinate at time $t \neq 0$ is given by the formula | ||
| + | $$x = \frac{v_0}{\omega} \exp(-\gamma t) \sin \omega t,$$ | ||
| + | where $\omega = \sqrt{\omega_0^2 - \gamma^2}$, $\gamma < \omega_0 = \sqrt{k/m}$, and $k, m$, and $\gamma$ are the stiffness, mass, and damping coefficient of the oscillator, respectively. Show that the properties of the oscillator described in problems 3.5.12 and 3.5.15 do not contradict this statement. | ||
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| + | ### Solution | ||
| + | |||
| + | #### Connection to problem 3.5.12: | ||
| + | |||
| + | Problem 3.5.12 considers two oscillators with identical ratios $k/m = \omega_0^2$ and $b/m = 2\gamma$. | ||
| + | |||
| + | From the formula for the coordinate $x(t)$, it is evident that it depends only on the initial velocity $v_0$, the natural frequency $\omega_0$, and the damping coefficient $\gamma$. | ||
| + | |||
| + | Since the parameters $\omega_0$ and $\gamma$ of the two oscillators coincide, under identical initial conditions ($v_0$), the equation will yield identically equal dependencies $x(t)$. There is no contradiction. | ||
| + | |||
| + | #### Connection to problem 3.5.15: | ||
| + | |||
| + | Let's find the velocity of the oscillator $v(t)$ by taking the time derivative of the coordinate $x(t)$: | ||
| + | $$ v(t) = x'(t) = \frac{v_0}{\omega} \left( -\gamma \exp(-\gamma t) \sin \omega t + \omega \exp(-\gamma t) \cos \omega t \right) $$ | ||
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| + | The oscillator passes through the equilibrium position when $x(t) = 0$, which corresponds to the condition $\sin \omega t = 0$. At these moments, $\cos \omega t = \pm 1$. | ||
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| + | Substituting these values into the velocity expression, we obtain the magnitude of the velocity at the moments of passing through equilibrium: | ||
| + | $$ |v(t)| = \frac{v_0}{\omega} \left| 0 + \omega \exp(-\gamma t) (\pm 1) \right| = v_0 \exp(-\gamma t) $$ | ||
| + | |||
| + | According to the condition of problem 3.5.15, after a time $T$, the velocity decreases by a factor of $n$. We can write this as: | ||
| + | $$ |v(T)| = v_0 \exp(-\gamma T) = \frac{v_0}{n} $$ | ||
| + | |||
| + | From this, we obtain the decay factor over one period $T$: | ||
| + | $$ \exp(-\gamma T) = \frac{1}{n} $$ | ||
| + | |||
| + | Let's determine the velocity after a time $2T$: | ||
| + | $$ |v(2T)| = v_0 \exp(-2\gamma T) = v_0 (\exp(-\gamma T))^2 = v_0 \left(\frac{1}{n}\right)^2 = \frac{v_0}{n^2} $$ | ||
| + | |||
| + | Similarly, after a time $3T$: | ||
| + | $$ |v(3T)| = v_0 \exp(-3\gamma T) = v_0 (\exp(-\gamma T))^3 = v_0 \left(\frac{1}{n}\right)^3 = \frac{v_0}{n^3} $$ | ||
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| + | The obtained analytical expressions strictly show that the velocity at the equilibrium positions decreases in a geometric progression. There is no contradiction with the conclusions of problem 3.5.15. | ||
| + | |||
| + | ### Answer | ||
| + | |||
| + | See the solution. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.5.18.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | |||
| $3.5.18.$ If at time $t = 0$ a damped oscillator is in its equilibrium position and its velocity is $v_0$, then its coordinate at time $t \neq 0$ is given by the formula | |||
| $$x = \frac{v_0}{\omega} \exp(-\gamma t) \sin \omega t,$$ | |||
| where $\omega = \sqrt{\omega_0^2 - \gamma^2}$, $\gamma < \omega_0 = \sqrt{k/m}$, and $k, m$, and $\gamma$ are the stiffness, mass, and damping coefficient of the oscillator, respectively. Show that the properties of the oscillator described in problems 3.5.12 and 3.5.15 do not contradict this statement. | |||
| ### Solution | |||
| #### Connection to problem 3.5.12: | |||
| Problem 3.5.12 considers two oscillators with identical ratios $k/m = \omega_0^2$ and $b/m = 2\gamma$. | |||
| From the formula for the coordinate $x(t)$, it is evident that it depends only on the initial velocity $v_0$, the natural frequency $\omega_0$, and the damping coefficient $\gamma$. | |||
| Since the parameters $\omega_0$ and $\gamma$ of the two oscillators coincide, under identical initial conditions ($v_0$), the equation will yield identically equal dependencies $x(t)$. There is no contradiction. | |||
| #### Connection to problem 3.5.15: | |||
| Let's find the velocity of the oscillator $v(t)$ by taking the time derivative of the coordinate $x(t)$: | |||
| $$ v(t) = x'(t) = \frac{v_0}{\omega} \left( -\gamma \exp(-\gamma t) \sin \omega t + \omega \exp(-\gamma t) \cos \omega t \right) $$ | |||
| The oscillator passes through the equilibrium position when $x(t) = 0$, which corresponds to the condition $\sin \omega t = 0$. At these moments, $\cos \omega t = \pm 1$. | |||
| Substituting these values into the velocity expression, we obtain the magnitude of the velocity at the moments of passing through equilibrium: | |||
| $$ |v(t)| = \frac{v_0}{\omega} \left| 0 + \omega \exp(-\gamma t) (\pm 1) \right| = v_0 \exp(-\gamma t) $$ | |||
| According to the condition of problem 3.5.15, after a time $T$, the velocity decreases by a factor of $n$. We can write this as: | |||
| $$ |v(T)| = v_0 \exp(-\gamma T) = \frac{v_0}{n} $$ | |||
| From this, we obtain the decay factor over one period $T$: | |||
| $$ \exp(-\gamma T) = \frac{1}{n} $$ | |||
| Let's determine the velocity after a time $2T$: | |||
| $$ |v(2T)| = v_0 \exp(-2\gamma T) = v_0 (\exp(-\gamma T))^2 = v_0 \left(\frac{1}{n}\right)^2 = \frac{v_0}{n^2} $$ | |||
| Similarly, after a time $3T$: | |||
| $$ |v(3T)| = v_0 \exp(-3\gamma T) = v_0 (\exp(-\gamma T))^3 = v_0 \left(\frac{1}{n}\right)^3 = \frac{v_0}{n^3} $$ | |||
| The obtained analytical expressions strictly show that the velocity at the equilibrium positions decreases in a geometric progression. There is no contradiction with the conclusions of problem 3.5.15. | |||
| ### Answer | |||
| See the solution. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||