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| ### Statement |
| ### Statement |
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| $3.5.18.$ [Insert the problem statement] |
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| ### Solution |
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| ### Problem |
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| $3.5.18.$ 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 |
| $3.5.18.$ 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 |
| $$x = \frac{v_0}{\omega} \exp(-\gamma t) \sin \omega t,$$ |
| $$x = \frac{v_0}{\omega} \exp(-\gamma t) \sin \omega t,$$ |
| 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. |
| 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 | | ### Solution |
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| #### Connection to problem 3.5.12: | | #### Connection to problem 3.5.12: |
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| Problem 3.5.12 considers two oscillators with identical ratios $k/m = \omega_0^2$ and $b/m = 2\gamma$. | | Problem 3.5.12 considers two oscillators with identical ratios $k/m = \omega_0^2$ and $b/m = 2\gamma$. |
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| 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 $\gamma$. | | 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 $\gamma$. |
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| 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. | | 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. |
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| #### Connection to problem 3.5.15: | | #### Connection to problem 3.5.15: |
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| Let's find the velocity of the oscillator $v(t)$ by taking the time derivative of the coordinate $x(t)$: | | Let's find the velocity of the oscillator $v(t)$ by taking the time derivative of the coordinate $x(t)$: |
| $$ v(t) = x'(t) = \frac{v_0}{\omega} \left( -\gamma \exp(-\gamma t) \sin \omega t + \omega \exp(-\gamma t) \cos \omega t \right) $$ | | $$ v(t) = x'(t) = \frac{v_0}{\omega} \left( -\gamma \exp(-\gamma t) \sin \omega t + \omega \exp(-\gamma t) \cos \omega t \right) $$ |
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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, $\cos \omega t = \pm 1$. | | 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: | | Substituting these values into the velocity expression, we obtain the magnitude of the velocity at the moments of passing through equilibrium: |
| $$ |v(t)| = \frac{v_0}{\omega} \left| 0 + \omega \exp(-\gamma t) (\pm 1) \right| = v_0 \exp(-\gamma t) $$ | | $$ |v(t)| = \frac{v_0}{\omega} \left| 0 + \omega \exp(-\gamma t) (\pm 1) \right| = v_0 \exp(-\gamma t) $$ |
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| 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: | | 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: |
| $$ |v(T)| = v_0 \exp(-\gamma T) = \frac{v_0}{n} $$ | | $$ |v(T)| = v_0 \exp(-\gamma T) = \frac{v_0}{n} $$ |
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| From this, we obtain the decay factor over one period $T$: | | From this, we obtain the decay factor over one period $T$: |
| $$ \exp(-\gamma T) = \frac{1}{n} $$ | | $$ \exp(-\gamma T) = \frac{1}{n} $$ |
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| Let's determine the velocity after a time $2T$: | | Let's determine the velocity after a time $2T$: |
| $$ |v(2T)| = v_0 \exp(-2\gamma T) = v_0 (\exp(-\gamma T))^2 = v_0 \left(\frac{1}{n}\right)^2 = \frac{v_0}{n^2} $$ | | $$ |v(2T)| = v_0 \exp(-2\gamma T) = v_0 (\exp(-\gamma T))^2 = v_0 \left(\frac{1}{n}\right)^2 = \frac{v_0}{n^2} $$ |
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| Similarly, after a time $3T$: | | Similarly, after a time $3T$: |
| $$ |v(3T)| = v_0 \exp(-3\gamma T) = v_0 (\exp(-\gamma T))^3 = v_0 \left(\frac{1}{n}\right)^3 = \frac{v_0}{n^3} $$ | | $$ |v(3T)| = v_0 \exp(-3\gamma T) = v_0 (\exp(-\gamma T))^3 = v_0 \left(\frac{1}{n}\right)^3 = \frac{v_0}{n^3} $$ |
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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. | | 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. |
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