3
$\begingroup$

I have a mass-spring system, which is as follows:

                                     enter image description here

I have derived the equations which are: $$ \begin{align} M_1 \frac{d^2x_1}{dt^2} &= M_1g + k_2(x_2 - x_1 - L_2) - k_1(x_1-L_1)\\ % M_2 \frac{d^2x_2}{dt^2} &= M_2g - k_2(x_2 - x_1 - L_2) \end{align} $$

However they ask me to find an expression for $x_2$ as a function of $x_1$. I cannot think of a way of doing this. It seems that I am missing something, how can I come up with such an expression?

EDIT:

They only said that $x_1(0) = 1$ and $x_2(0) = 2$, all parameters are equal to 1, and the initial momentum is 0, so I assume that the system is released from rest.

$\endgroup$

closed as off-topic by user36790, Sebastian Riese, Kyle Kanos, ACuriousMind, JamalS Oct 7 '15 at 8:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Community, Sebastian Riese, Kyle Kanos, ACuriousMind, JamalS
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ A system like this will have two "normal modes": in one mode, $m_1$ and $m_2$ are moving in the same direction; in the other mode, when one moves up the other moves down. For arbitrary starting conditions, the motion will be a linear combination of the two. Without knowing what the initial conditions are it is not possible to do what you ask. Was the system "released from rest" or some such? $\endgroup$ – Floris Oct 5 '15 at 23:32
  • $\begingroup$ Ups I have edited, they only said that $x_1(0) = 1$ and $x_2(0) = 2$, all parameters are equal to 1, and the initial momentum is 0, so I assume that the system is released from rest. $\endgroup$ – dpalma Oct 5 '15 at 23:42
  • $\begingroup$ Are the springs in their relaxed state at the moment the masses are released? See if you can find the frequency of the two normal modes of this system. See for example this set of notes. Expressing the motion in terms of simple (harmonic) basis functions should allow you to solve this. $\endgroup$ – Floris Oct 5 '15 at 23:44
  • 1
    $\begingroup$ There's a standard "trick" to solving coupled linear differential equations like this. Assume that the solutions are of the form x1=x1oExp(iwt) and x2=x2oExp(iwt). The two differential equations can then be reduced to a matrix equation involving the vector {x1o,x2o}. $\endgroup$ – Samuel Weir Oct 6 '15 at 0:29
  • $\begingroup$ So, should I apply superposition and find an expression? However, I don't see how solving the actual equation could help me in finding the required expression. Another important info: It is allowed using some numerical tools for solving this (like octave). However I need the expression, tried using Laplace transform but it was useless... $\endgroup$ – dpalma Oct 6 '15 at 1:30
0
$\begingroup$

Following the general approach given in this link, we can write down the equations of motion and solve for the normal modes.

I am using $x_1$ and $x_2$ as the displacement from equilibrium since it just removes a few $m\cdot g$, $-L_1$ and $-L_2$ terms but otherwise doesn't change the result in any fundamental way. You can then adapt this approach to solve your exact problem.

The equations of motion become:

$$m_1 \ddot x_1 = -k_1 x_1 + k_2(x_2 - x_1)\tag1$$ $$m_2 \ddot x_2 = -k_2(x_2 - x_1)\tag2$$

If we assume that there is a solution, it will be of the form:

$$x_1 = a_1 \cos(\omega t)\tag3$$ $$x_2 = a_2 \cos(\omega t)\tag4$$

where $a_1$ and $a_2$ might be complex (this would allow for arbitrary phase difference between the motion of the two masses) then we can find the relationship between $\omega$, $a_1$ and $a_2$:

$$- m_1 \omega^2 a_1 \cos\omega t = -k_1 a_1 \cos\omega t + k_2(a_2 - a_1)\cos\omega t\\ -m_2 \omega^2a_2 \cos\omega t = -k_2(a_2-a_1)\cos\omega t$$

Dividing out $\cos\omega t$ and rearranging, we get two equations for $a_1$ and $a_2$:

$$\begin{align}\left(-m_1\omega^2 +k_1+k_2\right) a_1 -k_2 a_2 &= 0\tag5\\ k_2 a_1 + \left(m_2 \omega^2 - k_2\right)a_2&= 0 \tag6 \end{align}$$

Since the right hand side is zero for these equations, the only non-trivial solution will be when the determinant on the left is zero, or

$$\left(-m_1\omega^2 +k_1+k_2\right) \left(m_2 \omega^2 - k_2\right)+k_2^2 = 0$$

Putting $\omega^2 = \Omega$, we can solve:

$$-m_1 m_2 \Omega^2 + (m_1 k_2 + m_2 (k_1+k_2)) \Omega - k_1 k_2 = 0$$

This leaves us with a messy expression for $\Omega = \omega^2$.

And then it gets interesting.

From the "assumed solution" (3) and (4) it immediately follows that

$$\frac{x_1}{x_2} = \frac{a_1}{a_2}$$

And we can find the ratio of amplitudes from either equation (5) or (6) by substituting our solution for $\omega^2$:

$$\frac{a_1}{a_2} = \frac{k_2}{-m_1\omega^2 +k_1+k_2}$$

I will leave it up to you to check my math, finish the solution. You might want to make sure that the solution makes sense for a simple case (like - what should happen when $m_1=0$ and $k_1 = k_2$?)

$\endgroup$
  • $\begingroup$ That makes sense, one last question. That is the solution of the homogeneus differential eq? to add the effects of gravity I have to include the constant term, right? $\endgroup$ – dpalma Oct 6 '15 at 21:58
  • $\begingroup$ Yes you would have to include a constant term - but that would just produce an offset to the equilibrium position. By defining $x_1$ as the displacement from equilibrium I circumvent the need to include $g$. At any rate I just wanted to give you a start... not do all the work for you. Note that the ratio will be different depending on the value of $\omega$ - which depends on which mode you are looking at... $\endgroup$ – Floris Oct 6 '15 at 22:02
  • $\begingroup$ I did not understand that about g. If you define $x_1$ as the displacement from equilibrium we would get rid of L, having: $m_1\ddot{x_1} = m_1g + k_2(x_2 - x_1) - k_1x_1$ mg is there, being the forcing function. Anyway, I get an idea for solving my problem, thanks for your patience. Answer accepted. $\endgroup$ – dpalma Oct 6 '15 at 22:18

Not the answer you're looking for? Browse other questions tagged or ask your own question.