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Lagrange’s Equations for Conservative Systems

In a conservative system, the generalised forces can be derived from a potential function V. Potential functions can represent the effects of:

• Ideal springs (energy storage) – Energy storage =

• Gravity – energy storage =

When dealing with such systems, a so-called ‘Lagrangian’ L is defined as follows:

(2)

The appropriate form of Lagrange’s equations for conservative systems is given by:

(3)

Application to a 2-degree-of-freedom system

This is a conservative system, and so we can use form (3) of the equation above. The springs have negligible mass, and there is no energy loss in system. We define generalised co-ordinates x (displacement from equilibrium – this eliminates any gravity forces) and , the angular displacement of the pulley whose moment of inertia through its axis is I.

Kinetic and power and mention Bezout Theorem for other complex function

]]>• Poles

In the Laurent expansion

f(z) = an (z – z0)n,

If an = 0 for n < -m < 0 and a-m 0, we say that z0 is a pole of order m. For instance, if m = 1, that is a-1/(z-z0) is the first non-vanishing term, we have a pole of order one, often called a simple pole.

If, on the other hand, the summation continues to n = – , the z0 is a pole of infinite order and is called an essential singularity. One point of fundamental difference between a pole of finite order and an essential singularity is that a pole of order m can be removed by multiplying f(z) by (z-z0)m. This obviously cannot be done for an essential singularity.

The behavior of f(z) as z is defined in terms of the behavior of f(1/t) as t 0. Consider the function

As z , we replace the z by 1/t to obtain

Clearly, from the definition, sin z has an essential singularity at . This result could be expected from the following analysis.

When x = 0, sin z = sin iy = = i sinh y,

which approaches exponentially as y . Thus, the absolute value of sin z is not bounded.

• Branch points

There is another sort of singularity. Consider

f(z) = z a,

in which a is not an integer, As z moves around the unit circle from e 0i to e 2i,

f(z) e 2ai e 0i.

We have a branch point at the origin and another at infinity. The points e 0i and e 2i in the z -plane coincide but they lead to different values of f(z).

The problem is resolved by constructing a cut line joining both branch points so that f(z) will be uniquely specified for a given point in the z-plane.

Note that a function with a branch point and a required cut line will not be continuous across the cut line. In general, there will be a phase difference on opposite sides of this cut line. Hence line integrals on opposite sides of this cut line will not generally cancel each other.

The contour line used to convert a multiply connected region into a simply connected region (Section 1.3) is completely different. The function is continuous across this contour line, and no phase difference exists.

Example:

Consider

f(z) = (z2 – 1) 1/2 = (z + 1)1/2(z – 1)1/2

The first factor on the right hand side(RHS), (z+1)1/2, has a branch point at z = -1. The second factor has one at z = 1. At infinity f(z) has a simple pole. The cut line has to connect both branch points. To check on the possibility of taking the line segment jointing z = +1 and -1 as a cut line, let us follow the phases of these two factors as we move along the contour shown in Fig2.1.

Fig.2.1

For convenience, let z + 1 =r e i and z – 1 = e i. Then the phase of f(z) is ()/2. At point 1, = 0; 1 to 2, , but – unchanged; then stays constant until 6; 6 to 7, = . increases by 2 as we move from 3 to 5. The phase of f(z) is tabulated in the final column of table2.1.

Table2.1 Phase Angle

(1). The phase at points 5 and 6 is not the same as the phase at 2 and 3. This behavior can be expected at a branch point cut line.

(2). The phase at 7 exceeds that at 1 by 2 and f(z) = (z -1)1/2 is therefore single-valued for the contour shown, encircling both branch points.

If we take the x-axis -1 ≤ x ≤ 1 as a cut line, f(z) is uniquely specified. Alternatively, the positive x-

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