# Spring-mass function

Here is the question that I'm stuck on:

What is the smallest value of $b$ for which we get solutions that, when viewed in the position-velocity plane, lie along a straight line? Algebraically support your conclusion.

I'm given Newton's Law of motion equation which is $$m \frac{d^2x}{dt^2}+b \frac{dx}{dt} + kx = 0$$

where $x$ is the position of an object (a block, for example) that's attached to the end of the spring, $m$ is the mass of the object, $b$ is the firction parameter (damping coefficient) and $k$ is the spring constant. I'm not sure how to approach the question that I imposed above, but I do know that

$$\frac{dx}{dt} = y$$ where $y$ is the velocity. Also, I know that $$\frac{dy}{dt} = \frac{d^2x}{dt^2}$$

I could then re-write my second order linear differential equation as a system of first order linear differential equations, which is

$$\frac{dx}{dt} = y$$ $$\frac{dy}{dt} = -\frac{k}{m}x - \frac{b}{m}y$$

Even if the work above doesn't pertain to how to solve the problem, this is what I know so far.

• I speak only for myself on this: what are “solutions that, when viewed in the position-velocity plane, lie along a straight line”? – Chase Ryan Taylor Nov 21 '17 at 22:06
• This occurs when both $\frac{dx}{dt}$ and $\frac{dy}{dt} = 0$. – John Smith Nov 21 '17 at 22:08
• I guess straight line in that plane means $\frac{\partial x}{\partial t} = kx+m$ – mathreadler Nov 21 '17 at 22:11
• @mathreadler that seems more like it than what OP states. The definition provided by Leo will simply give us $x = y= 0$. – Piotr Benedysiuk Nov 21 '17 at 22:28
• Do you know how to relate the eigenvalues of the system to its phase portrait? – amd Nov 22 '17 at 0:07

Define the velocity $v = dx/dt$. If the solution in the position-velocity plane is a straight line, then we have $v \;=\; c\,x$ for some unknown, real constant $c$. We know it must have this form, as opposed to a more general form like $v \;=\; c\, x \;+\; a$, because $x \;=\; 0$, $v \;=\; 0$ is an equilibrium point of this damped system, so it must lie along the line.
Newton's law for this system is: $$m\frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x \;=\; m\frac{d v}{dt} + b v + k x \;=\; 0\, .$$ The fact that $v = c\,x$ means that $dv/dt = c\, dx/dt = c\,v$, so: $$m\,c\,v \;+\; b\,v + k\,x \;=\; 0\; \rightarrow\; v \;=\; \frac{-k}{mc \,+\, b}\, x$$ However, we already know that $v = c\, x$, which implies that $$c \;=\; \frac{-k}{mc \,+\, b} \; \rightarrow\; m\, c^2 \;+\; b\, c \;+\; k \;=\; 0$$ Solving this quadratic equation for $c$ yields $$c \;=\; \frac{-b \;\pm\; \sqrt{b^2 \;-\; 4\,m\,k}}{2 m}\, .$$ Since $m$, $k$, and $b$ are all positive quantities, the only way that $c$ can be real is if the quantity inside the square root is greater than or equal to $0$, i.e. if $b \ge \sqrt{4 m k}$. This answers the original question.
Such a condition on $b$, leading to a straight line in the $v-x$ plane, is known as overdamping.