# Modeling a car connected to a mass $m$ by an spring and a damper, moving with constant speed along a gently undulating road with profile $h(x)$

I was solving some questions from the book "Classical Mechanics - R. Douglas Gregory", when I faced this problem:

A car that is represented by a chassis which keeps contact with the road, connected to an upper mass $m$ by a spring and a damper. The car is moving with constant speed $c$ along a gently undulating road with profile $h(x)$, where $h'(x)$ is small. At time $t$ the upper mass has displacement $y(t)$ above it's equilibrium level. Show that, under suitable assumptions, $y$ satisfies a differential equation of the form:

$y'' +2Ky'+\Omega^2y = 2Kch'(ct)+ \Omega^2 h(ct)$ where $K$ and $\Omega$ are positive constants.

I could not reach this conclusion in my attempts, but reached a slightly different answer. So I checked the solution in the solutions book.

The solution is the following:

Since the undulation is small we suppose that the horizontal displacement of the car in $t$ is simply given by $x = ct$. The the extension $\Delta$ of the spring at the time $t$ is

$\Delta = y - h(ct) \implies \Delta' = y' - h'(ct)$

The equation of motion for the vertical oscillation of the car is therefore

$my'' = -\alpha\Delta -\beta\Delta'$

writing $\alpha = m\Omega^2$ and $\beta = 2mK$ you can reach the wanted differential equation.

The problem of my resolution was that instead of

$my'' = -\alpha\Delta -\beta\Delta'$

I used the equation

$m\Delta'' = -\alpha\Delta -\beta\Delta'$

And for me, my equation still makes more sense than the other. So I want to understand: Why I use $my''$, instead of the $m\Delta''$ that would make sense using the damped simple harmonic equation?

## 1 Answer

To make thing easier for me and possibly for you I have inverted the system as follows with $L$ as the length of the spring when in static equilibrium. There seems to be agreement as to the two forces acting on the mass as $-\alpha \Delta$ and $-\beta \dot \Delta$ where $\Delta (= y-h)$ is the extension of the spring.

The author has then used Newton's second law which resulted in the equation of motion for the mass $$m\ddot y = -\alpha \Delta - \beta \dot \Delta$$.I have put a set of axes labelled 1 which are fixed relative to the ground which are used to measure distance $y$.

You have wanted to use $\ddot \Delta$ as the acceleration of the mass.
The problem with that is that you are measuring the extension $\Delta$ relative to a set of axes, labelled 2, which are pinned to a point on the spring $B'$ which is $L$ away from the point of support $A'$.
The problem with this is that set of axes is accelerating relative to the ground with an acceleration of $\ddot h$ and so in that accelerating frame of reference you cannot use Newton's laws directly.

This means that $$m\ddot \Delta \ne -\alpha \Delta - \beta \dot \Delta$$

However all is not lost as you could introduce a pseudo force of $-m\ddot h$ and then use Newton's laws to get

$$m\ddot \Delta = -\alpha \Delta - \beta \dot \Delta - m \ddot h \Rightarrow m\ddot y = -\alpha \Delta - \beta \dot \Delta$$ because $\ddot \Delta = \ddot y - \ddot h$

• Thanks! I now get it why we need to use $m \ddot y$. But this raised another question. We use $-\beta \dot \Delta$ because of the damped spring. If the damping was caused, for example, by the air resistance. Then I should have used $-b\dot y$? – MrBr Apr 26 '18 at 23:05
• @MrBr in a car the suspension damping is done by something like a piston and cylinder containing oil and the damping depends on the rate of change of the extension of the spring ie the speed of the mass relative to the oil. – Farcher Apr 27 '18 at 4:19