$3.5.22.$ 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}<<1 $ and $\frac{2 \pi\lambda}{w}>>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.
$$ x(t)=A\cdot e^{-\lambda t /2}sin(wt), \lambda=\frac{\mu}{m}$$
In first let's consider $\frac{2 \pi\lambda}{w}<<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) $,
$$ Q_{-}=\int_{0}^{\frac{2 \pi}{w_{0}}}\lambda m v_{0}^{2}\cdot cos^{2}(w_{0}t)\,dt=\lambda m v_{0}^{2}\pi /w_{0}$$
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 $$ v_{0}=\frac{p}{m}\frac{1}{1-exp(-\frac{\pi\lambda}{w})}\approx\frac{p w_{0}}{\lambda m \pi}$$.
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}$.
#### Answer
$\frac{2 \pi\lambda}{w}<<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}>>1 $, $ v=\frac{p}{m}$.
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### Statement
### Statement
$3.5.22.$[Insert the problem 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}<<1 $ and $\frac{2 \pi\lambda}{w}>>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.
$$ x(t)=A\cdot e^{-\lambda t /2}sin(wt), \lambda=\frac{\mu}{m}$$
$$ x(t)=A\cdot e^{-\lambda t /2}sin(wt), \lambda=\frac{\mu}{m}$$
In first let's consider $\frac{2 \pi\lambda}{w}<<1 $. Drag is very weak.
In first let's consider $\frac{2 \pi\lambda}{w}<<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}<<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}<<1 $, $ v_{0}=\frac{p}{m}\frac{1}{1-exp(-\frac{\pi\lambda}{w})}\approx\frac{p w_{0}}{\lambda m \pi}$;
