Правка разделов «Statement», «Solution», «Answer»
en/3.5.22.md
+8 −21
| @@ -1,7 +1,7 @@ | |||
| ### Statement | |||
| $3.5.22.$ Each time the oscillator passes through the equilibrium position in the same direction, an additional impulse | |||
| − | $p$ is imparted to it by a kick in the direction of velocity. What will be the motion of the oscillator, and what steady-state maximum speed will be established? The characteristics of the oscillator are known. Consider two limiting cases: $ \frac{2 \pi \lambda}{w} | ||
| + | $p$ is imparted to it by a kick in the direction of velocity. What will be the motion of the oscillator, and what steady-state maximum speed will be established? The characteristics of the oscillator are known. Consider two limiting cases: $ \frac{2 \pi \lambda}{w} \ll 1 $ and $ \frac{2 \pi \lambda}{w} \gg 1. $ | ||
| ### Solution | |||
| One of the ways to solve this task is by considering conservation of energy. | |||
| Drag force takes energy out of the system, adding the momentum once in a period should compensate the losses. | |||
| To calculate the energy losses: | |||
| $$ Q_{-}=\int_{0}^{T} \mu v(x)\,dx = \int_{a}^{b} \mu\cdot v(t)^{2}\,dt $$ | |||
| We take that: | |||
| @@ -16,40 +16,27 @@Solution | |||
| $$ \frac{d v }{d t}+w_{0}^{2}x+\frac{\mu}{m}v=0 $$ | |||
| And with $x(0)=0:$ | |||
| − | $$ x(t)=A\cdot e^{-\lambda t /2}sin(wt), \lambda=\frac{\mu}{m} $$ | ||
| + | $$ x(t)=A\cdot e^{-\lambda t /2}\sin(wt), \lambda=\frac{\mu}{m} $$ | ||
| − | In first let's consider $ \frac{2 \pi \lambda}{w} | ||
| + | In first let's consider $ \frac{2 \pi \lambda}{w} \ll 1 $. Drag is very weak. | ||
| That means $ w\approx w_{0} $ and also means that we can neglect the exponent while calculating integral, because it will lead only to additional terms of 2nd (and greater, thus very small) power of $\lambda$. (one can check by Taylor series). | |||
| − | SO, $ v(t)=v_{0}cos(w_{0}t) $, | ||
| − | $$ Q_{-}=\int_{0}^{\frac{2 \pi}{w_{0}}} \lambda m v_{0}^{2} \cdot cos^{2}(w_{0}t)\,dt=\lambda m v_{0}^{2} \pi /w_{0} $$ | ||
| + | SO, $ v(t)=v_{0}\cos(w_{0}t) $, | ||
| + | $$ Q_{-}=\int_{0}^{\frac{2 \pi}{w_{0}}} \lambda m v_{0}^{2} \cdot \cos^{2}(w_{0}t)\,dt=\lambda m v_{0}^{2} \pi /w_{0} $$ | ||
| Where $ v_{0} $ is speed right after impulse boost. | |||
| $$ Q_{+}=\frac{m}{2}(v_{0}^{2}-(v_{0}-\frac{p}{m})^{2})\approx pv_{0}, $$since $p$ is small, because of weak drag. | |||
| Eventually we have $ v_{0}=\frac{p w_{0}}{\lambda m \pi } $. | |||
| − | Also, since it is decaying harmonic oscillator, we could take $ v_{0}\cdot e^{-\frac{1}{2}\frac{2 \pi \lambda }{w}}=v_{0}-\frac{p}{m} $. And instantly get $$ v_{0}=\frac{p}{m}\frac{1}{1-exp(-\frac{\pi \lambda}{w})}\approx \frac{p w_{0}}{\lambda m \pi } $$. | ||
| + | Also, since it is decaying harmonic oscillator, we could take $ v_{0}\cdot e^{-\frac{1}{2}\frac{2 \pi \lambda }{w}}=v_{0}-\frac{p}{m} $. And instantly get $$ v_{0}=\frac{p}{m}\frac{1}{1-\exp(-\frac{\pi \lambda}{w})}\approx \frac{p w_{0}}{\lambda m \pi } $$. | ||
| Note that we can't use it for the other case, since the solutions $ x(t) $ are equal only in complex form, and in real they're two different types of motion. | |||
| One has decaying sinusoid, and another have just multiplication of decaying real exponents: $x(t)=e^{-\lambda/2t} \cdot A e^{-\sqrt{\lambda^2/4 - ω_0^2}t} $. So the solution from Savchenko book is not correct. | |||
| Physically, solution shows that body will eventually reach $ x=0 $ with zero velocity. So, $ v=\frac{p}{m} $. | |||
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| #### Answer | |||
| − | $ \frac{2 \pi \lambda}{w} | ||
| + | $ \frac{2 \pi \lambda}{w} \ll 1 $, $ v_{0}=\frac{p}{m}\frac{1}{1-\exp(-\frac{\pi \lambda}{w})}\approx \frac{p w_{0}}{\lambda m \pi } $; | ||
| − | $ \frac{2 \pi \lambda}{w} | ||
| + | $ \frac{2 \pi \lambda}{w} \gg 1 $, $ v=\frac{p}{m} $. | ||
| @@ -1,7 +1,7 @@ | |||
| ### Statement | ### Statement | ||
| $3.5.22.$ Each time the oscillator passes through the equilibrium position in the same direction, an additional impulse | $3.5.22.$ Each time the oscillator passes through the equilibrium position in the same direction, an additional impulse | ||
| $p$ is imparted to it by a kick in the direction of velocity. What will be the motion of the oscillator, and what steady-state maximum speed will be established? The characteristics of the oscillator are known. Consider two limiting cases: $ \frac{2 \pi \lambda}{w} |
$p$ is imparted to it by a kick in the direction of velocity. What will be the motion of the oscillator, and what steady-state maximum speed will be established? The characteristics of the oscillator are known. Consider two limiting cases: $ \frac{2 \pi \lambda}{w} \ll 1 $ and $ \frac{2 \pi \lambda}{w} \gg 1. $ | ||
| ### Solution | ### Solution | ||
| One of the ways to solve this task is by considering conservation of energy. | One of the ways to solve this task is by considering conservation of energy. | ||
| Drag force takes energy out of the system, adding the momentum once in a period should compensate the losses. | Drag force takes energy out of the system, adding the momentum once in a period should compensate the losses. | ||
| To calculate the energy losses: | To calculate the energy losses: | ||
| $$ Q_{-}=\int_{0}^{T} \mu v(x)\,dx = \int_{a}^{b} \mu\cdot v(t)^{2}\,dt $$ | $$ Q_{-}=\int_{0}^{T} \mu v(x)\,dx = \int_{a}^{b} \mu\cdot v(t)^{2}\,dt $$ | ||
| We take that: | We take that: | ||
| @@ -16,40 +16,27 @@Solution | |||
| $$ \frac{d v }{d t}+w_{0}^{2}x+\frac{\mu}{m}v=0 $$ | $$ \frac{d v }{d t}+w_{0}^{2}x+\frac{\mu}{m}v=0 $$ | ||
| And with $x(0)=0:$ | And with $x(0)=0:$ | ||
| $$ x(t)=A\cdot e^{-\lambda t /2}sin(wt), \lambda=\frac{\mu}{m} $$ | $$ x(t)=A\cdot e^{-\lambda t /2}\sin(wt), \lambda=\frac{\mu}{m} $$ | ||
| In first let's consider $ \frac{2 \pi \lambda}{w} |
In first let's consider $ \frac{2 \pi \lambda}{w} \ll 1 $. Drag is very weak. | ||
| That means $ w\approx w_{0} $ and also means that we can neglect the exponent while calculating integral, because it will lead only to additional terms of 2nd (and greater, thus very small) power of $\lambda$. (one can check by Taylor series). | That means $ w\approx w_{0} $ and also means that we can neglect the exponent while calculating integral, because it will lead only to additional terms of 2nd (and greater, thus very small) power of $\lambda$. (one can check by Taylor series). | ||
| SO, $ v(t)=v_{0}cos(w_{0}t) $, | SO, $ v(t)=v_{0}\cos(w_{0}t) $, | ||
| $$ Q_{-}=\int_{0}^{\frac{2 \pi}{w_{0}}} \lambda m v_{0}^{2} \cdot cos^{2}(w_{0}t)\,dt=\lambda m v_{0}^{2} \pi /w_{0} $$ | $$ Q_{-}=\int_{0}^{\frac{2 \pi}{w_{0}}} \lambda m v_{0}^{2} \cdot \cos^{2}(w_{0}t)\,dt=\lambda m v_{0}^{2} \pi /w_{0} $$ | ||
| Where $ v_{0} $ is speed right after impulse boost. | Where $ v_{0} $ is speed right after impulse boost. | ||
| $$ Q_{+}=\frac{m}{2}(v_{0}^{2}-(v_{0}-\frac{p}{m})^{2})\approx pv_{0}, $$since $p$ is small, because of weak drag. | $$ Q_{+}=\frac{m}{2}(v_{0}^{2}-(v_{0}-\frac{p}{m})^{2})\approx pv_{0}, $$since $p$ is small, because of weak drag. | ||
| Eventually we have $ v_{0}=\frac{p w_{0}}{\lambda m \pi } $. | Eventually we have $ v_{0}=\frac{p w_{0}}{\lambda m \pi } $. | ||
| Also, since it is decaying harmonic oscillator, we could take $ v_{0}\cdot e^{-\frac{1}{2}\frac{2 \pi \lambda }{w}}=v_{0}-\frac{p}{m} $. And instantly get $$ v_{0}=\frac{p}{m}\frac{1}{1-exp(-\frac{\pi \lambda}{w})}\approx \frac{p w_{0}}{\lambda m \pi } $$. | Also, since it is decaying harmonic oscillator, we could take $ v_{0}\cdot e^{-\frac{1}{2}\frac{2 \pi \lambda }{w}}=v_{0}-\frac{p}{m} $. And instantly get $$ v_{0}=\frac{p}{m}\frac{1}{1-\exp(-\frac{\pi \lambda}{w})}\approx \frac{p w_{0}}{\lambda m \pi } $$. | ||
| Note that we can't use it for the other case, since the solutions $ x(t) $ are equal only in complex form, and in real they're two different types of motion. | Note that we can't use it for the other case, since the solutions $ x(t) $ are equal only in complex form, and in real they're two different types of motion. | ||
| One has decaying sinusoid, and another have just multiplication of decaying real exponents: $x(t)=e^{-\lambda/2t} \cdot A e^{-\sqrt{\lambda^2/4 - ω_0^2}t} $. So the solution from Savchenko book is not correct. | One has decaying sinusoid, and another have just multiplication of decaying real exponents: $x(t)=e^{-\lambda/2t} \cdot A e^{-\sqrt{\lambda^2/4 - ω_0^2}t} $. So the solution from Savchenko book is not correct. | ||
| Physically, solution shows that body will eventually reach $ x=0 $ with zero velocity. So, $ v=\frac{p}{m} $. | Physically, solution shows that body will eventually reach $ x=0 $ with zero velocity. So, $ v=\frac{p}{m} $. | ||
| #### Answer | #### Answer | ||
| $ \frac{2 \pi \lambda}{w} |
$ \frac{2 \pi \lambda}{w} \ll 1 $, $ v_{0}=\frac{p}{m}\frac{1}{1-\exp(-\frac{\pi \lambda}{w})}\approx \frac{p w_{0}}{\lambda m \pi } $; | ||
| $ \frac{2 \pi \lambda}{w} |
$ \frac{2 \pi \lambda}{w} \gg 1 $, $ v=\frac{p}{m} $. | ||