Explanation of a spring using “Artistotle's law of motion”, $\vec{F}=m\vec{v}$

So I was watching Susskind's Classical Mechanics lectures and I didn't understand something in the second lecture.

He was telling about Aristotle's Law of motion which is $$\vec F = m\vec v.$$

He applied this law on a spring-block system and got this function of position.

$$x(t)=x(0)\cdot e^{-\tfrac{k}{m}t}.$$

He then said that this law is not valid under classical mechanics as it does not help us to predict the past because after some time the particle would just be standing at the origin and predicting where it started from is not possible with finitely accurate measurements.

This is what I did not understand. There are many such laws like for example a damped harmonic motion which consist of an exponential decay. Are they also not valid?

• But $x(t)=x(0)\cdot e^{-\tfrac{k}{m}t}$ isn't damped harmonic motion! It's different from the equation of an overdamped harmonic oscillator. – user191954 Jan 13 at 15:28
• @Chair I meant the damped harmonic motion as an example. This equation is a simple harmonic motion equation derived from Aristotle's Law(which is wrong just like this equation). – harshit54 Jan 13 at 16:51
• @ZeroTheHero I didn't understand. Where exactly is $e^{+kt/m}$ not included? – harshit54 Jan 26 at 16:02
• @ZeroTheHero Wouldn't there be a $-k/m$ and not a $+k/m$ in the solution? It is the solution of the differential equation $-kx=m\dfrac{dx}{dt}$ – harshit54 Jan 26 at 16:10
• Nevermind I get it now... – ZeroTheHero Jan 26 at 16:19

The equation $$F = m v$$ is not time-symetrical, i.e. reversing the sign of $$t$$ gives a different equation because of the appearence of velocity. The equation $$F = m a$$ is second order in $$t$$, so it doesn't change under time inversion $$t \Rightarrow -\, t$$, but only if the left member ($$F$$) doesn't include terms with dependance on velocity (like friction). If there's a friction, then $$F$$ contains a non-symetrical term under time-inversion. Friction implies a loss of information.
If you have a strict exponential damping: $$x(t) = x_0 \, e^{- \lambda t}$$, then measuring position and velocity at time $$t_1$$ gives this system of equations: \begin{align} x_1 &= x_0 \, e^{- \lambda t_1}, \tag{1} \\[12pt] v_1 &= -\, \lambda \, x_1. \tag{2} \end{align} You then knows $$x_1$$ and $$v_1$$ (so you know $$\lambda$$ from (2)), but you don't know time $$t_1$$ and want to retrodict the initial position $$x_0$$ (at time $$t_0 = 0$$). Equation (1) gives you only one equation for two unknows. You need more information to retrodict the past. Measuring position and velocity again at time $$t_2$$ gives you two new equations: \begin{align} x_2 &= x_0 \, e^{- \lambda t_2}, \tag{3} \\[12pt] v_2 &= -\, \lambda \, x_2. \tag{4} \end{align} Equation (4) is useless. Combining (1) and (3) gives this, where $$\Delta t = t_2 - t_1$$ is known: $$$$\tag{5} x_2 = x_0 \, e^{- \lambda (t_1 \,+\, \Delta t)} = x_1 \, e^{- \lambda \, \Delta t},$$$$ so there's nothing new here, and you can't find $$x_0$$!
• But if I don't have the time $t_1$, I will not be able to predict the original state of even a valid system. – harshit54 Jan 13 at 15:35
• Because the damped harmonic oscillator doesn't have just an exponential. There's also a trigonometric function in which $x_0$ and $t_1$ (or $t_0$ or $t_2$) will appear, so you'll be able to solve for $x_0$. – Cham Jan 13 at 15:44