4
$\begingroup$

I'm working through a Differential Equations book and came across a question that puzzles me:

"Consider a spring-mass system where there are two masses connected in series by springs, with each spring having a different constant ($k_1$ and $k_2$)" (set up like this):

|<><>($m_1$)<><>($m_2$)

Where $m_1$ and $m_2$ are the masses, and $k_1$ and $k_2$ would be the spring constants. The position of each mass is given by $x_1$ and $x_2$.

It asks to show that (using $F=ma$) the system can be described by this pair of DEs: \begin{align} m_1\ddot{x}_1 &= -(k_1+k_2)x_1 + k_2x_2\\ m_2\ddot{x}_2 &= k_2x_1 - k_2x_2 \end{align}

and then reduce these to a single equation for the position of mass 2 ($x_2$).

Any suggestions on how to get these equations? Been a while since my Physics days but I think it's primarily a mathematics question. Thanks in advance

$\endgroup$
4
  • $\begingroup$ For the first mass, the acceleration is $\ddot{x_1}$ and the force exerted on the mass is just the sum of the forces from the springs. Similarly for the second mass. $\endgroup$
    – copper.hat
    Nov 4, 2014 at 17:44
  • $\begingroup$ When you write 'reduce these to a single equation...', do you mean reduce them to a single ODE? $\endgroup$
    – copper.hat
    Nov 5, 2014 at 6:18
  • $\begingroup$ Yes - I'm asked to reduce it to a single ODE for the position of Mass 2 (X2), and then am given values for k1, k2, m1, m2 and initial conditions and am then asked to solve for the position. $\endgroup$ Nov 6, 2014 at 0:06
  • $\begingroup$ I don't see how you can obtain a single ODE in $x_2$. $\endgroup$
    – copper.hat
    Nov 6, 2014 at 0:48

2 Answers 2

2
$\begingroup$

To get to your first to equation, we need to examine the free body diagram for each mass. For mass one, when we displace it by $x_1$ (to the right), $k_1$ acts in the opposite direction. As this is occurring, $m_1$ is pushing into spring $k_2$ which causes $k_2$ to push back against $m_1$. Since $k_2$ is pushing against $m_1$, it is also pushing against $m_2$ in the opposite direction (to the right) by $x_2$. That is, $$ m_1\ddot{x}_1 = -k_1x_1 - k_2x_1 + k_2x_2 = -(k_1 + k_2)x_1 + k_2x_2 $$ Now when we displace $m_2$ by $x_2$ (to the right), we have $k_2$ pulling back and then pulling (to the right) $m_1$ by $x_1$ so $$ m_2\ddot{x}_2 = -k_2x_2 + k_2x_1 $$


For the next part, do you not have initial conditions or is there a forcing $f(t)$? If not, we can only go to as far as I have gotten.

What I would do next is take the Laplace Transform of both DE (under assumptions of zero IC). $$ F(s) = \int_0^{\infty}f(t)e^{-st}dt $$ The Laplace transform of both are \begin{align} m_1s^2X_1(s) &= -(k_1 + k_2)X_1(s) + k_2X_2(s)\tag{1}\\ m_2s^2X_2(s) &= -k_2X_2(s) + k_2X_1(s)\tag{2} \end{align}


If we have IC or a forcing function $f(t)$ we could continue in such a manner. We can solve equation (1) for $X_1(s) = \frac{k_2X_2(s)}{m_1s^2 + k_1 + k_2}$ which can then plug into equation (2). $$ m_2s^2X(s) = -k_2X(s) + \frac{k_2^2}{m_1s^2 + k_1 + k_2}X(s)\tag{3} $$ Now, we need IC or $f(t)$.


With zero initial conditions and no driving force, we can solve equation (3) as is; that is, $$ X_2(s)\bigg[m_2s^2 + k_2 - \frac{k_2^2}{m_1s^2 + k_1 + k_2}\bigg] = 0\tag{4} $$ Then we have $X_2(s) = 0\Rightarrow\mathcal{L}^{-1}(X_2(s)) = \mathcal{L}^{-1}(0)$ which is $$ x_2(t) = 0 $$

$\endgroup$
6
  • $\begingroup$ Thanks for your help, I'm still a little confused here. I do get initial conditions in the next section, maybe after giving these it'll become a little more clear to me. They're K1 = 3, K2 = 2, M1 = 1, M2 = 1, x1(0) = 0, x1'(0)= 0, x2(0) = 0, and x2'(0) = 0 $\endgroup$ Nov 6, 2014 at 0:14
  • $\begingroup$ Thanks mate, appreciate it! $\endgroup$ Nov 6, 2014 at 0:36
  • $\begingroup$ @RyanMorgan with zero initial conditions and no forcing function, the system never oscillates and the solution is zero. $\endgroup$
    – dustin
    Nov 6, 2014 at 16:37
  • $\begingroup$ So then in solving for the position of mass 2 (x2) it would simply be 0? $\endgroup$ Nov 6, 2014 at 18:59
  • $\begingroup$ @RyanMorgan $x(t)$ is displacement of masses where when the masses havent been displaced $x_1(0) = x_2(0) = 0$, hit with a velocity $\dot{x}_1(0) = \dot{x}_2(0) = 0$, or there is no forcing function $f(t)$, the system never oscillates. Therefore, the position of $x_2$ with respect to time is always zero since we regard that point as zero. However, this question is strange because the way it is worded seems like you should be finding something useful but you can't. $\endgroup$
    – dustin
    Nov 6, 2014 at 19:14
0
$\begingroup$

Assume that $\xi_i$ denotes the position of mass $m_i$ with respect to the wall where the sequence of springs is attached. Thus, the elongation of the first spring is $\Delta \ell_1 = \xi_1 - L_1$, where $L_1$ is the length of the spring at rest. Similarly, $\Delta \ell_2 = \xi_2 - \xi_1 - L_2$. Thus, the forces $F_i = -k_i \Delta \ell_i$ at the end of each spring are \begin{aligned} F_1 &= -k_1 (\xi_1 - L_1),\\ F_2 &= -k_2 (\xi_2 - \xi_1 - L_2) . \end{aligned} Using Newton's laws, we find \begin{aligned} m_1 \ddot \xi_1 &= F_1 - F_2 = -k_1 (\xi_1 - L_1) + k_2 (\xi_2 - \xi_1 - L_2),\\ m_2 \ddot \xi_2 &= F_2 = - k_2 (\xi_2 - \xi_1 - L_2) . \end{aligned} At equilibrium $\ddot \xi_i = 0$, we find the equilibrium positions $\bar\xi_1=L_1$ and $\bar\xi_2=L_1+L_2$, i.e. \begin{aligned} 0 &= -k_1 (\bar \xi_1 - L_1) + k_2 (\bar \xi_2 - \bar \xi_1 - L_2),\\ 0 &= - k_2 (\bar \xi_2 - \bar \xi_1 - L_2) . \end{aligned} Subtraction from the above differential system and introduction of $x_i = \xi_i - \bar \xi_i$ yields the desired system.

Alternatively, one could start by introducing $x_i$, the displacement of mass $m_i$ from its equilibrium position. Thus, the elongation of the first spring is $\Delta \ell_1 = x_1$, and the elongation of the second spring is $\Delta \ell_2 = x_2-x_1$. Application of Newton's laws with $F_i=-k_i\Delta \ell_i$ yields directly the system \begin{aligned} m_1 \ddot x_1 &= -(k_1+k_2) x_1 + k_2 x_2 ,\\ m_2 \ddot x_2 &= - k_2 (x_2 - x_1) . \end{aligned} Note also that $M\ddot X + K X = 0$ with $$ M=\begin{bmatrix} m_1 & 0 \\ 0& m_2 \end{bmatrix},\qquad K=\begin{bmatrix} k_1+k_2 & -k_2\\ -k_2 & k_2 \end{bmatrix} $$ and $X = (x_1,x_2)^\top$.

Unless additional assumptions are made, it might not be an easy task to decouple these equations. In fact, if the motion is assumed periodic in time, say $X \propto \text{e}^{\text i \omega t}$, then $(K-\omega^2 M) X = 0$. Diagonalization leads to the eigenvalues $\omega_i^2$ (related to the resonance frequencies $\omega_i/(2\pi)$ in Hz) and to the eigenvectors $X_i$ (normal modes). However it appears that $X = (0,1)^\top$ does not satisfy the above algebraic equation for any $\omega$, and therefore the component $x_2$ cannot be decoupled from the system.

One could assume that $m_1 \ll m_2$ so that the first mass is at equilibrium. This way, we get $x_1 = \frac{k_2}{k_1+k_2} x_2$ and $$ m_2 \ddot x_2 = -K_\text{eq} x_2, \qquad K_\text{eq} = \frac{k_1k_2}{k_1+k_2}. $$ Here, we have introduced the spring of stiffness $K_\text{eq}$ which is equivalent to the sequence of springs with stiffness constants $k_1$, $k_2$. One observes that the relationship $$ \frac{1}{K_\text{eq}} = \frac{1}{k_1} + \frac{1}{k_2} $$ is satisfied.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .