# Homework Help: Particles joined by light elastic string

1. Nov 6, 2009

Hi guys,

I was thinking of a problem the other day; consider a group of particles, connected to each other by light elastic strings, there is no restriction on which particles can be connected to which, then can one determine the position of a particle at time t.

So I set about trying to determine a way to model that system, and decided to start simple, consider the case of two particles joined by a single light elastic string, moving in one dimension and not under the influence of gravity, so my system would look like this:

http://weald10k.fileave.com/PF007_pic1_2.png​

where the particles are free to move in one dimension, x, with positive x to the right. I will also make the assumption that the string is initially taught such that the distance between P1 and P2 is greater than or equal to the natural length of the string.

Also of course at some point in the future the string will become slack, at which point the positions and velocities of the particles can be determined and the normal equations of motion will apply. When the string becomes taught again the system model can be reapplied with different initial conditions.

The concern of my question only relates to determining the position and velocity of the two particles at any time t before the string becomes slack.

Each particle P1 and P2 has properties:
Mass = mi
Position Vector = $\vec x_{i}(t)$
and we know the initial position and velocity of both $\vec x_{i,0}$ and $\vec v_{i,0}$​

and the String has properties:
modulus of elasticity = λ
Natural length = ℓ
where both are scalars

So resolving forces for P1 and P2:
$$m_1\frac{d^2\vec{x_1}}{dt^2} = \vec{T_1} \ \ \ \ \ \ \ \ \ \ \ \ m_2\frac{d^2\vec{x_2}}{dt^2} = \vec{T_2}$$

Now here is where I start to run in to trouble, by Hooke's law, the tension will be:

$$\vec{T} = -\frac{\lambda}{\ell}\vec{x_e}$$​

where $\vec x_e$ is the extension vector of the string. What I quickly came to realise is that it isn't as easy as it appeared to be. In order to determine the extension vector, I had to first ascertain the concept of extension when there is no "fixed end" of the string. I first considered the force acting on P1. consider the diagram below where $\vec x_e$ is the extension related to the tension force acting on P1:

http://weald10k.fileave.com/PF007_pic2_2.png​

So I realised, lets take $\vec x_{2}$ to be a "psudo fixed end", such that the extension vector will be pointing away from P2.

Initially I can up with some odd equation for the extension in terms of $\vec x_{1}$, $\vec x_{2}$ and ℓ, then I realised they contradicted each other, and realised its because ℓ is a scalar and in my mind I was treating it as a vector in some peculiar fashion.

What I realised is that the extension vector is dependant on the vector ℓ, by the diagram we see that,

$$\vec{x_e} = \vec{x_{2}x_{1}} + \vec{\ell}$$​

but we have yet to define the vector ℓ, and it turns out that the vector ℓ is not infact constant, and is dependant on weather P1 is to the left or right of P2. In the diagram above P1 is to the left, below

http://weald10k.fileave.com/PF007_pic3_2.png​

P1 is the the right, and as you can see the vector ℓ must be pointing in the opposite direction as before. What I then concluded is that ℓ must point in the direction opposite to $\vec{x_{2}x_{1}}$. This leads to the equation of the extension of the string for the force T1:

$$\vec{x_e} = \vec{x_{2}x_{1}} - \ell \frac{\vec{x_{2}x_{1}}}{|\vec{x_{2}x_{1}}|}$$​

which would mean that the initial force resolution equations become:

(i) $$\begin{array}{rcl} \displaystyle m_1\frac{d^2\vec{x_1}}{dt^2}&=& \displaystyle \vec{T_1} \\ &=& \displaystyle -\frac{\lambda}{\ell} \left( \vec{x_{2}x_{1}} - \ell \frac{\vec{x_{2}x_{1}}}{|\vec{x_{2}x_{1}}|} \right) \\ &=& \displaystyle -\frac{\lambda}{\ell} \left( (\vec{x_{1}} - \vec{x_{2}}) - \ell \frac{(\vec{x_{1}} - \vec{x_{2}})}{|\vec{x_{2}x_{1}}|} \right) \end{array}$$

(ii) $$\begin{array}{rcl} \displaystyle m_2\frac{d^2\vec{x_2}}{dt^2}&=& \displaystyle \vec{T_2} \\ &=& \displaystyle -\vec{T_1}\\ &=& \displaystyle \frac{\lambda}{\ell} \left( \vec{x_{2}x_{1}} - \ell \frac{\vec{x_{2}x_{1}}}{|\vec{x_{2}x_{1}}|} \right) \\ &=&\displaystyle -\frac{\lambda}{\ell} \left( \vec{x_{1}x_{2}} - \ell \frac{\vec{x_{1}x_{2}}}{|\vec{x_{1}x_{2}}|} \right) \\ &=&\displaystyle -\frac{\lambda}{\ell} \left( (\vec{x_{2}} - \vec{x_{1}}) - \ell \frac{(\vec{x_{2}} - \vec{x_{1}})}{|\vec{x_{2}x_{1}}|} \right)$$

From here I have no idea how to go any further really. Initially I thought that this problem may boil down to the acceleration of a particle in terms of the position vectors of the two particles, and perhaps through a Laplace transform it would be solvable as I would have a system of two equations in two variables. But the ℓ and having to use a modulus of a variable vector has thrown a spanner in the works.

I thought perhaps there were two ways of proceeding further. One was that T1 = -T2, therefore:

$$m_1\frac{d^2\vec{x_1}}{dt^2} = -m_2\frac{d^2\vec{x_2}}{dt^2}$$​

and through repeated integration w.r.t t knowing the initial conditions one can find an expression for x1 in x2, which can be used to eliminate x1 or x2 from the force resolution equations (i) and (ii), but that doesn't make them any simpler to solve, I'm still left with a form that is nearly as complicated as before.

I did think I could express the modulus of $\vec{x_{2}x_{1}}$ as a piece wise function using the heavyside function, but I almost feel that that makes the problem more complicated then if feels it should be.

Thank you so much for reading my "essay". What I am asking is if anyone could suggest if any of the ways I have thought of moving forward with this problem are viable, or any suggestions at all, perhaps I have missed something very obvious that I have not factored into my model that will make things easier. Thank you soo much guys :D.

Last edited: Nov 6, 2009
2. Nov 7, 2009

### tiny-tim

In the first diagram, shouldn't T1 and T2 be equal?

3. Nov 7, 2009