Page:Encyclopædia Britannica, Ninth Edition, v. 15.djvu/759

Rh MECHANICS 727 . endu- Still keeping to easy examples, suppose the cord of the ordi- im with nary simple pendulum to be extensible, according to Hooke s law. xten- Let&quot; A be its length at time t. Then the tension is E(A - l)/l, ible and the work it can do in contracting is the integral of this with ord. regard to A from I to A., i.e., Hence we have Thus Lagrange s equations become dt in -m 2 + E(A - 1)11 mgcos6 = ; equations which could be obtained immediately from the application of the second law, with the help of the kinematical expressions for acceleration perpendicular to, and along, the radius-vector of a plane curve ( 47). Instead of the complex pendulum treated in 177, we will now take the case of two masses attached at different points to an elastic string, or light helical spring, and consider their vertical vibrations. Let a, b be the unstretched lengths of the parts of the string, M and TO the masses. Then if, rj be the vertical displacements at time t, we have V- V ~ &quot; a b so that Lagrange s equations are -2 mfl r = 0. m)0=0, The equilibrium positions are found by supposing the accelera tions to vanish, so that, if we suppose and r) to be measured from them, the terms in g will disappear. Hence the solution is of exactly the same nature as that already given for an apparently different problem ( 178). We may mention that equations practically the same as these are obtained when we consider the motions of a watch and its balance- wheel, the watch being supported in a horizontal position by means of a wire, and oscillating in its own plane by the torsion-elasticity of the wire. The reader of 242 below will have no difficulty in obtaining this result. It suggests a practical method of &quot; setting &quot; a watch to true time, without turning the hands forward or backward, and without letting it run down. The following is a simple, but very instructive, example of the transference of energy (back and forward) between two parts of a system. Two bar-magnets of equal mass, length, and strength (fig. 58), are supported horizontally by pairs of parallel strings, so that when at rest they are in one line. One of them is slightly displaced in the direction of its length, find the subsequent motion of both. 1 If we can, by any process, find two fundamental states of motion which, once established, will be permanent, any other possible motion Fig. 58. of the system will be a superposition of these two. The amplitudes and phases in the components may have any values, so long as the whole disturbance is small. This follows from the fact that the system has two degrees of freedom only, since we are concerned only with motions in the plane of the figure, (A) Now one obviously possible motion is a simple harmonic vibration of the whole, without change of distance between the magnets. The period of this vibration is obviously the same as that of either magnet if the other were removed. (B) Another obviously possible motion is that in which the magnets are, at every instant, equally and oppositely 1 We suppose the bars to be so long, in comparison with the dis tance between them, that we need take account only of the action ol their poles which are turned towards one another. deflected. The period of this oscillation will be less or greater than that of the former according as the poles attract or repel one another. Now the initial state of motion proposed evidently con- is ts of the superposition of (A) and (B) in such a way but a definite velocity of one of them only. This cor responds to simultaneous zero of displacement, with equal velocities, for each of (A) and (B). There is therefore at that instant no displacement of either mass ; and one is at rest while the other is moving with double the assigned velocity. If 2ir/n, 2-n-jn be the periods of the two motions, it is obvious that after the time v(n - n } the magnets will have interchanged their states so that the arrangement will present exactly the same appearance as at first, if looked at from the other side. Let a be the distance between the ends of the bars when all four strings are vertical. Then, if 0, &amp;lt;f&amp;gt; be at time t the inclinations of the pairs of strings to the vertical, a becomes where I is the common length of the strings. The expression for the potential energy due to magnetism is of the form /x/D, where p is positive if like poles be turned to one another. Hence Forming the equations as usual, and omitting powers of and &amp;lt;f&amp;gt; above the first, we have - (d&amp;gt; - i a ^ from which the results already given may be deduced. Finally, let us take the case of Atwood s machine ( 173) when Atwood s the masses are equal, and one of them is vibrating through small machine. arcs. Let r, 6 be the polar coordinates of the vibrating mass ; then, One mass neglecting powers of 6 higher than the second, we have the general- vibrating ized equations Put gr for r, and 0/2 for 6, and we get Transform to rectangular coordinates in the plane of motion, x being vertically downwards ; then x = y 2 jx 2, y=-1y]x. This shows that the vertical acceleration of the vibrating particle is very small, but constantly downward. Hence the energy of the vibratory motion is steadily converted into energy of translation of the masses. When both the equal masses vibrate through small arcs, it is found that the mass whose angular range is the greater has down ward acceleration with diminishing angular range. Hence it would appear that, if the string be long enough, the entire motion would be periodic. 216. Before leaving this subject we may form, from the com plete value of 8A given in last section, the generalized equations corresponding to those of Hamilton s &quot; varying action,&quot; as given in 204. We have at once M But, by the value of T, we have &c. =&c.
 * hat there is, at starting, no displacement of either mass,