System of ODEs - string-mass problem

[Benny]

Hi, can someone please help me do the following question?

Q. A light elastic string of length 3l is stretched between two fixed points a distance of 3L apart (3L > 3l), and two particles, each of mass m, are attached to the string, one at each of the two points of trisection, The system is set in motion so that the masses undergo small transverse oscillations.

If x an y are the displacements of the masses at time t, show that the equations of motion of the masses is

$$\left[ {\begin{array}{*{20}c} {\mathop x\limits^ \bullet } \\ {\mathop y\limits^ \bullet } \\ \end{array}} \right] = n^2 \left[ {\begin{array}{*{20}c} { - 2} & 1 \\ 1 & { - 2} \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} x \\ y \\ \end{array}} \right]$$

where $$n^2 = \frac{T}{{mL}}$$ and T is the tension in the string.

I don't know whether x and y are 'absolute' positions or just displacements from some position. Previous questions of this type that I've attempted ask for x and y relative to one end or relative to some other points. The fact that the question doesn't specify this is a bit confusing.

But I'll assume that it is meant that x and y are positions of the two particles from the left hand end.

The question refers to a string and not a spring but I think that since the string is elastic, I can use Hookes Law. The two masses lie along the same horizontal line. I'll denote the position of the left most mass by x and the position of the other mass by y (so x < y).

For the left most particle I used Newton's second law to obtain the following:

$$\begin{array}{l} m\mathop x\limits^{ \bullet \bullet } = - T - k\left( {x - l} \right) + k\left( {\left( {y - x} \right) - l} \right) \\ m\mathop y\limits^{ \bullet \bullet } = - k\left( {\left( {y - x} \right) - l} \right) + k\left( {\left( {3L - y} \right) - l} \right) + T \\ \end{array}$$

where T = k(L-l)..not sure about the expression for T.

The above simplifies to

$$\begin{array}{l} m\mathop x\limits^{ \bullet \bullet } = - T - 2kx + ky \\ m\mathop y\limits^{ \bullet \bullet } = kx - 2ky + 3kL + T \\ \end{array}$$

which doesn't correspond to the given answer. I was having trouble resolving forces. Most of the questions I've done are spring mass systems where the spring constant is given. In this case only the tension is given and I introduced a spring constant myself. I'm at a loss as to how to tackle this question so can someone please help? Thanks.

 

[HallsofIvy]

Hi, can someone please help me do the following question?

Q. A light elastic string of length 3l is stretched between two fixed points a distance of 3L apart (3L > 3l), and two particles, each of mass m, are attached to the string, one at each of the two points of trisection, The system is set in motion so that the masses undergo small transverse oscillations.

If x an y are the displacements of the masses at time t, show that the equations of motion of the masses is

$$\left[ {\begin{array}{*{20}c} {\mathop x\limits^ \bullet } \\ {\mathop y\limits^ \bullet } \\ \end{array}} \right] = n^2 \left[ {\begin{array}{*{20}c} { - 2} & 1 \\ 1 & { - 2} \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} x \\ y \\ \end{array}} \right]$$

where $$n^2 = \frac{T}{{mL}}$$ and T is the tension in the string.

I don't know whether x and y are 'absolute' positions or just displacements from some position.

I don't know what you mean by 'absolute' positions! Clearly x and y are displacements from some position; it doesn't matter from what position, it might be simplest to assume from the end of the string that the first mass is closest to.

Previous questions of this type that I've attempted ask for x and y relative to one end or relative to some other points. The fact that the question doesn't specify this is a bit confusing.

But I'll assume that it is meant that x and y are positions of the two particles from the left hand end.

Good!

The question refers to a string and not a spring but I think that since the string is elastic, I can use Hookes Law. The two masses lie along the same horizontal line. I'll denote the position of the left most mass by x and the position of the other mass by y (so x < y).

For the left most particle I used Newton's second law to obtain the following:

$$\begin{array}{l} m\mathop x\limits^{ \bullet \bullet } = - T - k\left( {x - l} \right) + k\left( {\left( {y - x} \right) - l} \right) \\ m\mathop y\limits^{ \bullet \bullet } = - k\left( {\left( {y - x} \right) - l} \right) + k\left( {\left( {3L - y} \right) - l} \right) + T \\ \end{array}$$

where T = k(L-l)..not sure about the expression for T.
No. T is not an external force. You can use T to find k. When the string, of natural length 3l, is stretched a distance 3L, the string is pulling with force T: T= k(3L- 3l) so k= T/(3(L- l)). If the masses are 3l
Now your equations are:
$$m\frac{d^2x}{dt^2}= -\frac{T}{3(L- l)}(x- l)+ \frac{T}{3(L- l)}(y- x- l)$$
and similarly for y.
The above simplifies to

$$\begin{array}{l} m\mathop x\limits^{ \bullet \bullet } = - T - 2kx + ky \\ m\mathop y\limits^{ \bullet \bullet } = kx - 2ky + 3kL + T \\ \end{array}$$

which doesn't correspond to the given answer. I was having trouble resolving forces. Most of the questions I've done are spring mass systems where the spring constant is given. In this case only the tension is given and I introduced a spring constant myself. I'm at a loss as to how to tackle this question so can someone please help? Thanks.

 

[Benny]

Thanks for the help but using k = T/(3L-3l) and then setting n^2 = T/(mL) doesn't appear to reduce the equations into the required form. For example, we should have x'' = -2n^2x + n^2y if the substitution n^2 = T/(mL) is used. I don't really see what I need to do. I thought about trying a change of variables but that is evidently something that the question suggests should not be needed.