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I'm struggling to understand the reasoning in question $b)$. Basically, I had to write Newton's Second Law and do a change of variables to put the equation of motion of a mass $m$ in a specific form, which will give the driven harmonic oscillator time function once solved.

My Physics assistant was unable to answer my question/problem. It's about the change of variables itself, what it depicts/means, that I've done wrong because of my understanding of the problem and concepts (I think).


Here is the wording of the problem :

"A car, viewed as a point of mass $m$, is moving on a bumpy road with an horizontal and constant velocity $\vec v_x$. The mass is linked up to a spring device of spring constant $k$ and of length at rest $l_0$. At the end of the device, there is a wheel without mass and of negligible radius that follows the road (see the diagram here https://i.imgur.com/4RthHYD.png ).

We suppose that the device which keeps the spring straight doesn't affect the movement of the mass. The values of the parameters of this problem are such that the wheel doesn't take off the ground and the car never hits the wheel. The road has a sinusoidal shape of height $H$ and length $L$."

a) Express the vertical position of the wheel $h(t)$ as a function of the time.

I've already done that one. With a vertical axis $y$ pointing up, whose origin is set at $H/2$, we get : $$h(t) = \frac {H}{2} sin (\frac{2 \pi v_x}{L} t) = \frac {H}{2} sin(\omega t)$$

b) Using $h(t)$, find the equation of motion of the car in the vertical direction $y$.

Indication : Put the equation of motion in the form : $\ddot u + \omega_0^2 u = \alpha_0 sin(\omega t)$. A change of variables $y \rightarrow u$ may be needed.


Firstly, I've written down Newton's Lex Secunda :

$$-mg -k(y -h(t) - l_0) = m\ddot y$$where $(y -h(t) - l_0)$ is the extension/compression of the spring in function of the position $y$ of the mass from the $y$-axis. Indeed, the position of the mass minus the position of the wheel minus the length of the spring at rest gives us the extension/compression of the spring.

Now, my try was to immediately do the change of variables. I defined an axis $u$ pointing up, whose origin is set at the equilibrium position and then I expressed the extension/compression of the spring like this : $\Delta u = u - d$, where $d$ is the compression of the spring when the system is at the equilibrium position (as if it were a simple harmonic system with a mass) such that $-mg + kd = 0$.

Therefore I have the equality of the extension/compression :

$$y -h(t) - l_0 = u-d \Rightarrow y = h(t) + l_0 + u - d$$

So I substituted $y$ the lex secunda and got :

$$- mg - k(u-d) = m \ddot y$$ But I know that $-mg + kd = 0$, so I finally got : $$\ddot u + \frac{k}{m} u = \frac{H}{2} \omega^2 sin (\omega t)$$ because $\ddot y = \ddot h(t) + \ddot u$. $$\ddot u + \omega_0^2 u = \frac{H}{2} \omega^2 sin (\omega t)$$

But the problem is that apparently, my change of variables is wrong. In the corrected version, they firstly developped the lex secunda equation and got :

$m \ddot y = -mg -k(y -h(t) - l_0) = -mg - ky + kl_0 + k \frac{H}{2}sin(\frac{2\pi v_x}{L}t)$

$\ddot y = -g - \frac {k}{m}y + \frac {k}{m}l_0 + \frac {k}{m} \frac{H}{2}sin(\frac{2\pi v_x}{L}t)$

$\ddot y = -g - \omega_0^2y + \omega_0^2l_0 + \omega_0^2 \frac{H}{2}sin(\omega t)$

$\ddot y + \omega_0^2 (\frac{g}{\omega_0^2} + y - l_0) = + \omega_0^2 \frac{H}{2}sin(\omega t)$, where $\frac{g}{\omega_0^2} = d$.

Here is what I don't understand :

In the corrected version, they set, without explanations, $u = (d + y - l_0)$ and $\ddot u = \ddot y$ and substituted them in the equation of motion. However, I don't understand how can $u$ express the position of the spring relative to an axis $u$, whose origin would be set at the equilibrium position (since we then want to find the driven harmonic oscillator function $u(t)$ which gives the position of the mass in respect to the equilibrium position of the system). I don't understand it because if $u$ is the position of the mass in respect to the equilibrium position, then $u - d = y -l_0$ is the extension of the spring, but we saw above that the extenion is $y - h(t) - l_0$. So it makes no sense to me.

In brief, my problem is that I don't understand how $u - d = (y - l_0)$ is the extension/compression of the spring.

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In the expression they give, it looks to me that $u$ is the displacement of the car relative to its equilibrium position, not the amount by which the spring is compressed. These are related, but you can't just work in the frame of reference of the road: if you did, you would have an accelerating frame of reference. That is usually a way to make the problem harder.

Once you recognize that you are looking at the mass relative to its equilibrium position, with an inertial frame of reference, then I think their approach makes sense. You chose the wrong $u$.

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  • $\begingroup$ You're right about "u". It's a mistake on my part (I will edit my post). Yet, if "u" is the displacement of the car relative to its equilibrium position, then "u - d = y - l0" is the extension of the spring. But we saw earlier that it is "y - h(t) - l0", so there is an inconsistency, am I wrong ? (By the way, what is the difference between an accelerating and a non-inertial frame of reference ?). And sorry if I didn't exactly get your point. $\endgroup$ – Desura Oct 13 '15 at 22:47
  • $\begingroup$ I mis spoke - I meant inertial, not non-inertial... As for the extension of the spring - draw a diagram for yourself. I think It must depend on the instantaneous height of the road (in other words the equilibrium is defined relative to a non-undulating road surface). $\endgroup$ – Floris Oct 14 '15 at 5:53

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