Analysis

The difference equations for the oscillating wire can be derived by dividing it into N segments (10 or 20 should be sufficient) each of which can be considered rigid. This would be the case if the wire consisted of a series of rods connected together.

Let $x_{n}$ be the displacement of the bottom of the $n$th segment. Let $\theta_{n}$ be the angle it makes with the vertical. Let $T_{n}$ be the tension in the $n$th segment. Assume that the mass of the wire is located at the joints of the segments, $\Delta m$ at each joint. $\Delta m = \rho \delta z$ where $\delta z$ is the length of each segment, and $\rho$ is the mass per unit length of the wire. The equation of motion for the mass at the bottom of the nth segment is,

$$\Delta m \frac{\d^{2}x_{n}}{\d t^{2}} = T_{n+1} \sin ( \theta_{n+1} ) - T_{n} \sin ( \theta_{n} ) \quad .$$ (3.33)

Assume small oscillations so that $\sin ( \theta_{n} ) = (x_{n} - x_{n-1})/\delta z$.

$$\Delta m \frac{\d^{2}x_{n}}{\d t^{2}} = \frac{1}{\delta z... ... - ( T_{n+1} + T_{n} ) x_{n} + T_{n} x_{n-1} \right] \quad .$$ (3.34)

The mass associated with the end of the wire will be only $\Delta m / 2$ since there is no contribution from the (N+1)th segment. Consequently, the equation of motion for this point is,

$$( M + {\mathchoice{{\textstyle{\frac12}}}{{\textstyle{\frac... ...( \theta_{N} ) = -T_{N} ( x_{N} - x_{N-1} )/ \delta z \quad ,$$ (3.35)

where $M$ is any mass carried by the crane. In addition the displacement of the top of the wire is zero, so that $x_0 = 0$ in the equation for $n=1$.

The modes of oscillation are calculated by seeking solutions of the form $x_{n}(t) = y_{n} \exp ( i \omega t )$. Substituting this into the equations of motion gives,

$$- \Delta m\omega^{2} y_{n} = \left( T_{n+1} y_{n+1} - ( T_{n} + T_{n+1} ) y_{n} + T_{n} y_{n-1} \right) / \delta z$$ (3.36)

$$- ( M + {\mathchoice{{\textstyle{\frac12}}}{{\textstyle{\frac12}}}{{\scriptstyle{1/2}}}{{\scriptscriptstyle{1/2}}}}\Delta m )\omega^{2} y_{N} = T_{N} ( y_{N-1} - y_{N} ) / \delta z \quad .$$ (3.37)

The specification of the equations is completed by noting that, from the equilibrium conditions,

$$T_{N} = ( M + {\mathchoice{{\textstyle{\frac12}}}{{\textstyl... ...lta m ) g \quad , \qquad T_{n} = T_{n+1} + g \Delta m \quad ,$$ (3.38)

where $g$ is the acceleration due to gravity.

The equations can be organised in the form,

$$\bss{A}\bi{y} = - \bss{M}\omega^{2} \bi{y} \quad ,$$ (3.39)

where $\bi{y}$ is the column vector of displacements, $( y_{1}, y_{2}, \ldots ,y_{N} )$, $\bss{A}$ is a symmetric tridiagonal matrix and $\bss{M}$ is a diagonal matrix. The problem becomes one of finding the eigenvalues, $-\omega^{2}$, and eigenvectors, $\bi{y}$, of a generalised eigenvalue problem. The eigenvectors show how the wire distorts when oscillating in each mode and the eigenvalues give the corresponding oscillation frequencies. Low frequency modes are more important than high frequency modes to the crane manufacturer.

The problem can be solved most easily by using a LaPack routine which finds the eigenvalues and eigenvectors directly. However, before doing so it is necessary to eliminate the matrix $\bss{M}$ using the same method as discussed in problem 6.

You should investigate both the computational aspects, such as the dependence of the results on $\delta z$, as well as the physical ones, such as the dependence of the behaviour on the mass, $M$. Do your results make physical sense? You might even compare them with a simple experiment involving a weight on the end of a string.