Oscillations and Waves > Forced and damped oscillations > Problem 3.5.22

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} \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:
$$ \frac{d v }{d t}+w_{0}^{2}x+\frac{\mu}{m}v=0 $$

And with $x(0)=0:$
$$ x(t)=A\cdot e^{-\lambda t /2}\sin(wt), \lambda=\frac{\mu}{m} $$

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) $,
$$ Q_{-}=\int_{0}^{\frac{2 \pi}{w_{0}}} \lambda m v_{0}^{2} \cdot \cos^{2}(w_{0}t)\, dt=\lambda m v_{0}^{2} \pi /w_{0} $$
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 for $x$ shows that body will eventually reach $ x=0 $ with zero velocity. So, $ v=\frac{p}{m} $.

Answer

$ \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} \gg 1 $, $ v=\frac{p}{m} $.