75 lines
4.1 KiB
Markdown
75 lines
4.1 KiB
Markdown
---
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title: 11. Motion in one dimension
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---
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The motion of a system having one degree of freedom is said to take place in one dimension. The most general form of the Lagrangian of such a system in fixed external conditions is
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```load
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1/11.1
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```
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where $a(q)$ is some function of the generalised co-ordinate $q$. In particular,
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if $q$ is a Cartesian co-ordinate ($x$, say) then
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```load
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1/11.2
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```
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The equations of motion corresponding to these Lagrangians can be integrated in a general form. It is not even necessary to write down the equation of motion; we can start from the first integral of this equation, which gives the law of conservation of energy. For the Lagrangian `1/11.2` (e.g.) we have $\mfrac{1}{2}m\dot{x}^2+U(x)=E$. This is a first-order differential equation, and can be inte- grated immediately. Since $\dd{x}/\dd{t} = \sqrt{2[E - U(x)]/m}$, it follows that
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```load
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1/11.3
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```
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The two arbitrary constants in the solution of the equations of motion are here represented by the total energy E and the constant of integration.
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Since the kinetic energy is essentially positive, the total energy always exceeds the potential energy, i.e. the motion can take place only in those regions of space where $U(x) \lt E$. For example, let the function U(x) be of the form shown in 'TODO Fig. 6 (p. 26)'. If we draw in the figure a horizontal line corresponding to a given value of the total energy, we immediately find the possible regions of motion. In the example of Fig. 6, the motion can occur only in the range AB or in the range to the right of C.
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The points at which the potential energy equals the total energy,
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```load
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1/11.4
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```
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give the limits of the motion. They are *turning points*, since the velocity there is zero. If the region of the motion is bounded by two such points, then the motion takes place in a finite region of space, and is said to be *finite*. If the region of the motion is limited on only one side, or on neither, then the
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motion is *infinite* and the particle goes to infinity.
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A finite motion in one dimension is oscillatory, the particle moving repeatedly back and forth between two points (in Fig. 6, in the potential well AB between the points X1 and x2). The period T of the oscillations, i.e. the time during which the particle passes from X1 to X2 and back, is twice the time
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from X1 to X2 (because of the reversibility property, `1/5`) or, by `1/11.3`),
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```load
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1/11.5
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```
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where $x_1$ and $x_2$ are roots of equation `1/11.4` for the given value of $E$. This formula gives the period of the motion as a function of the total energy of the particle.
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<!-- PROBLEMS -->
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<!-- PROBLEM 1. Determine the period of oscillations of a simple pendulum (a particle of mass -->
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<!-- m suspended by a string of length l in a gravitational field) as a function of the amplitude of -->
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<!-- the oscillations. -->
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<!-- SOLUTION. The energy of the pendulum is E = 1ml2j2-mgl cos = -mgl cos to, where -->
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<!-- o is the angle between the string and the vertical, and to the maximum value of . Calculating -->
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<!-- the period as the time required to go from = 0 to = Do, multiplied by four, we find -->
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<!-- -cos -->
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<!-- po) -->
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<!-- The substitution sin $ = sin 10/sin 100 converts this to T = /(l/g)K(sin 100), where -->
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<!-- 1/75 -->
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<!-- - -->
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<!-- is the complete elliptic integral of the first kind. For sin 100 22 100 < 1 (small oscillations), -->
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<!-- an expansion of the function K gives -->
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<!-- T = -->
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<!-- §12 -->
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<!-- Determination of the potential energy -->
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<!-- 27 -->
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<!-- The first term corresponds to the familiar formula. -->
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<!-- PROBLEM 2. Determine the period of oscillation, as a function of the energy, when a -->
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<!-- particle of mass m moves in fields for which the potential energy is -->
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<!-- (a) U = Alx -->
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<!-- (b) U = Uo/cosh2ax, -U0 0, (c) U = Uotan2ax. -->
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<!-- SOLUTION. (a): -->
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<!-- T = -->
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<!-- By the substitution yn = u the integral is reduced to a beta function, which can be expressed -->
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<!-- in terms of gamma functions: -->
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<!-- The dependence of T on E is in accordance with the law of mechanical similarity (10.2), -->
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<!-- (10.3). -->
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<!-- (b) T = (7/a)V(2m/E). -->
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<!-- (c) T =(t/a)v[2m/(E+U0)] -->
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