Новое решение
en/3.5.3.md
+72 −0
| @@ -0,0 +1,72 @@ | |||
| + | ### Statement | ||
| + | |||
| + | $3.5.3.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Problem | ||
| + | |||
| + | $3.5.3^*.$ Under the conditions of problem 3.5.1, the pendulum had a velocity $v_0$ and a coordinate $x_0$ at the initial moment ($t=0$). What will be the amplitude of oscillations after $n$ impacts if the first one occurred at the initial moment? Plot the phase portrait. | ||
| + | |||
| + | *(Note: in the answer section of the problem book, the transferred momentum is denoted as $p$, and not $p_0$ as in the condition of problem 3.5.1. For consistency with the author's answer, the notation $p$ is used below).* | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | From problem 3.5.2, it is known that simple harmonic motion on the phase plane with coordinates $(x, \frac{p_{\text{imp}}}{m\omega})$ corresponds to motion along a circle. The radius of this circle is numerically equal to the oscillation amplitude $A$. | ||
| + | By the Pythagorean theorem, the amplitude for any state is expressed as: | ||
| + | $$ A = \sqrt{x^2 + \left(\frac{p_{\text{imp}}}{m\omega}\right)^2} $$ | ||
| + | |||
| + | It is important to note that at the moment of a brief impact, the coordinate $x$ of the pendulum does not have time to change, but its momentum increases abruptly by the value of the transferred momentum $p$. | ||
| + | |||
| + | Let us consider two cases for the impact frequency. | ||
| + | |||
| + | #### Case 1: Impacts follow each other after a period $T_0$ | ||
| + | |||
| + | Over a time equal to the period of free oscillations $T_0 = 2\pi/\omega$, the representative point on the phase portrait makes a full revolution (a rotation by $360^\circ$) and returns to the same coordinates. | ||
| + | |||
| + | 1. At the initial moment ($t=0$), the initial momentum is $mv_0$. Immediately after the first impact, the momentum becomes: | ||
| + | $$ p_1 = mv_0 + p $$ | ||
| + | 2. By the time of the second impact (after a time $T_0$), the pendulum completes a full oscillation and returns to the point with coordinate $x_0$, and its momentum is again equal to $p_1$. The second impact instantly adds another momentum $p$: | ||
| + | $$ p_2 = (mv_0 + p) + p = mv_0 + 2p $$ | ||
| + | 3. Since every period the system returns to its initial position $x_0$, the momenta simply accumulate algebraically. Immediately after the $n$-th impact, the total momentum at this point will be: | ||
| + | $$ p_n = mv_0 + np $$ | ||
| + | 4. Substituting the coordinate $x_0$ and the accumulated momentum $p_n$ into the amplitude formula, we obtain: | ||
| + | $$ A_n = \sqrt{x_0^2 + \left(\frac{mv_0 + np}{m\omega}\right)^2} = \sqrt{x_0^2 + \left(\frac{v_0}{\omega} + \frac{np}{m\omega}\right)^2} $$ | ||
| + | |||
| + | On the phase portrait, this motion looks like an unwinding spiral consisting of concentric circular arcs of increasing radius. | ||
| + | |||
| + |  | ||
| + | |||
| + | #### Case 2: Impacts follow each other after half a period $T_0/2$ | ||
| + | |||
| + | Over the time $T_0/2$, the point on the phase plane makes half a revolution (a rotation by $180^\circ$). In this process, both coordinates (both position and momentum) change their sign to the opposite. | ||
| + | |||
| + | 1. First impact ($n=1$, odd): Occurs at $t=0$. The momentum abruptly becomes $p_1 = mv_0 + p$. The amplitude after the first impact is: | ||
| + | $$ A_1 = \sqrt{x_0^2 + \left(\frac{v_0}{\omega} + \frac{p}{m\omega}\right)^2} $$ | ||
| + | 2. Free oscillations until the second impact: After the time $T_0/2$, both coordinates change sign. The position becomes $-x_0$, and the momentum becomes $-(mv_0 + p)$. | ||
| + | 3. Second impact ($n=2$, even): An impact is delivered, adding a momentum $+p$. We calculate the new momentum: | ||
| + | $$ p_2 = -(mv_0 + p) + p = -mv_0 $$ | ||
| + | Let us calculate the amplitude for this state (taking into account the coordinate $-x_0$): | ||
| + | $$ A_2 = \sqrt{(-x_0)^2 + \left(\frac{-mv_0}{m\omega}\right)^2} = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} $$ | ||
| + | The pendulum "sheds" the energy of the impact and transitions to an orbit of a smaller radius (the amplitude decreases). | ||
| + | 4. Free oscillations until the third impact: After another $T_0/2$ (a full period $T_0$ has passed in total), the coordinates change sign again: $-(-x_0) = x_0$ and $-(-mv_0) = mv_0$. The system has returned exactly to its initial state (which was before the first impact). | ||
| + | 5. Third impact ($n=3$, odd): Once again adds a momentum $p$ to the initial $mv_0$. The state becomes completely identical to the state after the first impact, therefore, $A_3 = A_1$. | ||
| + | |||
| + | Thus, for all odd $n$, the amplitude is equal to $A_1$, and for all even $n$, it is equal to $A_2$. On the phase portrait, the representative point infinitely jumps between two closed semicircles (outer and inner). | ||
| + | |||
| + |  | ||
| + | |||
| + | ### Answer | ||
| + | |||
| + | If the impacts follow each other at time intervals $T_0$, then the amplitude is: | ||
| + | $$ A_n = \sqrt{\left[\frac{v_0}{\omega} + \frac{np}{m\omega}\right]^2 + x_0^2} $$ | ||
| + | If at intervals $T_0/2$, then the amplitude is: | ||
| + | $$ A_n = \sqrt{\left[\frac{v_0}{\omega} + \frac{p}{m\omega}\right]^2 + x_0^2} \quad \text{for odd } n, $$ | ||
| + | $$ A_n = \sqrt{\frac{v_0^2}{\omega^2} + x_0^2} \quad \text{for even } n, $$ | ||
| + | where $\omega = 2\pi/T_0$. | ||
| + | |||
| + |  | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,72 @@ | |||
| ### Statement | |||
| $3.5.3.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | |||
| $3.5.3^*.$ Under the conditions of problem 3.5.1, the pendulum had a velocity $v_0$ and a coordinate $x_0$ at the initial moment ($t=0$). What will be the amplitude of oscillations after $n$ impacts if the first one occurred at the initial moment? Plot the phase portrait. | |||
| *(Note: in the answer section of the problem book, the transferred momentum is denoted as $p$, and not $p_0$ as in the condition of problem 3.5.1. For consistency with the author's answer, the notation $p$ is used below).* | |||
| ### Solution | |||
| From problem 3.5.2, it is known that simple harmonic motion on the phase plane with coordinates $(x, \frac{p_{\text{imp}}}{m\omega})$ corresponds to motion along a circle. The radius of this circle is numerically equal to the oscillation amplitude $A$. | |||
| By the Pythagorean theorem, the amplitude for any state is expressed as: | |||
| $$ A = \sqrt{x^2 + \left(\frac{p_{\text{imp}}}{m\omega}\right)^2} $$ | |||
| It is important to note that at the moment of a brief impact, the coordinate $x$ of the pendulum does not have time to change, but its momentum increases abruptly by the value of the transferred momentum $p$. | |||
| Let us consider two cases for the impact frequency. | |||
| #### Case 1: Impacts follow each other after a period $T_0$ | |||
| Over a time equal to the period of free oscillations $T_0 = 2\pi/\omega$, the representative point on the phase portrait makes a full revolution (a rotation by $360^\circ$) and returns to the same coordinates. | |||
| 1. At the initial moment ($t=0$), the initial momentum is $mv_0$. Immediately after the first impact, the momentum becomes: | |||
| $$ p_1 = mv_0 + p $$ | |||
| 2. By the time of the second impact (after a time $T_0$), the pendulum completes a full oscillation and returns to the point with coordinate $x_0$, and its momentum is again equal to $p_1$. The second impact instantly adds another momentum $p$: | |||
| $$ p_2 = (mv_0 + p) + p = mv_0 + 2p $$ | |||
| 3. Since every period the system returns to its initial position $x_0$, the momenta simply accumulate algebraically. Immediately after the $n$-th impact, the total momentum at this point will be: | |||
| $$ p_n = mv_0 + np $$ | |||
| 4. Substituting the coordinate $x_0$ and the accumulated momentum $p_n$ into the amplitude formula, we obtain: | |||
| $$ A_n = \sqrt{x_0^2 + \left(\frac{mv_0 + np}{m\omega}\right)^2} = \sqrt{x_0^2 + \left(\frac{v_0}{\omega} + \frac{np}{m\omega}\right)^2} $$ | |||
| On the phase portrait, this motion looks like an unwinding spiral consisting of concentric circular arcs of increasing radius. | |||
|  | |||
| #### Case 2: Impacts follow each other after half a period $T_0/2$ | |||
| Over the time $T_0/2$, the point on the phase plane makes half a revolution (a rotation by $180^\circ$). In this process, both coordinates (both position and momentum) change their sign to the opposite. | |||
| 1. First impact ($n=1$, odd): Occurs at $t=0$. The momentum abruptly becomes $p_1 = mv_0 + p$. The amplitude after the first impact is: | |||
| $$ A_1 = \sqrt{x_0^2 + \left(\frac{v_0}{\omega} + \frac{p}{m\omega}\right)^2} $$ | |||
| 2. Free oscillations until the second impact: After the time $T_0/2$, both coordinates change sign. The position becomes $-x_0$, and the momentum becomes $-(mv_0 + p)$. | |||
| 3. Second impact ($n=2$, even): An impact is delivered, adding a momentum $+p$. We calculate the new momentum: | |||
| $$ p_2 = -(mv_0 + p) + p = -mv_0 $$ | |||
| Let us calculate the amplitude for this state (taking into account the coordinate $-x_0$): | |||
| $$ A_2 = \sqrt{(-x_0)^2 + \left(\frac{-mv_0}{m\omega}\right)^2} = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} $$ | |||
| The pendulum "sheds" the energy of the impact and transitions to an orbit of a smaller radius (the amplitude decreases). | |||
| 4. Free oscillations until the third impact: After another $T_0/2$ (a full period $T_0$ has passed in total), the coordinates change sign again: $-(-x_0) = x_0$ and $-(-mv_0) = mv_0$. The system has returned exactly to its initial state (which was before the first impact). | |||
| 5. Third impact ($n=3$, odd): Once again adds a momentum $p$ to the initial $mv_0$. The state becomes completely identical to the state after the first impact, therefore, $A_3 = A_1$. | |||
| Thus, for all odd $n$, the amplitude is equal to $A_1$, and for all even $n$, it is equal to $A_2$. On the phase portrait, the representative point infinitely jumps between two closed semicircles (outer and inner). | |||
|  | |||
| ### Answer | |||
| If the impacts follow each other at time intervals $T_0$, then the amplitude is: | |||
| $$ A_n = \sqrt{\left[\frac{v_0}{\omega} + \frac{np}{m\omega}\right]^2 + x_0^2} $$ | |||
| If at intervals $T_0/2$, then the amplitude is: | |||
| $$ A_n = \sqrt{\left[\frac{v_0}{\omega} + \frac{p}{m\omega}\right]^2 + x_0^2} \quad \text{for odd } n, $$ | |||
| $$ A_n = \sqrt{\frac{v_0^2}{\omega^2} + x_0^2} \quad \text{for even } n, $$ | |||
| where $\omega = 2\pi/T_0$. | |||
|  | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||