CHAPTER XI. ENERGY OF VIBRATIONS. 119. AN ordinary pendulum affords a very good example of the transformation of energy. When we draw it aside from its lowest position we do work against gravity, the amount of this work being equal to the weight of the pendulum multiplied by the height of the new position of its centre of gravity above its lowest position. If we now release the pendulum it falls back, and in the fall gravity does as much work upon it as was previously done against gravity. This amount of work may be called the energy of vibration of the pendulum, and it is continually undergoing transformation. In the two extreme positions it is all in the shape of 'potential energy,' or as it may be better called 'statical energy,' because its amount is computed without any reference to the laws of motion. In the lowest position it is all in the form of 'kinetic energy' or energy of motion, the amount of which is measured by multiplying each element of the mass by half the square of its velocity and adding these products for the whole pendulum. In intermediate positions the energy is partly in the one form and partly in I the other. The statical energy in any position is computed by multiplying the weight of the pendulum by the height of its centre of gravity above the lowest position of the centre of gravity, and is continually diminishing as the pendulum descends, while the kinetic energy undergoes a corresponding increase. In the ascent a converse change takes place, and the total amount (computed by taking the sum of the two energies) is the same in all positions. 120. Let the pendulum be a 'simple pendulum,' consisting of a mass m suspended by a string of length / without weight or mass, and let a be the angle which the string in the extreme positions makes with the vertical. Then the statical energy in the extreme positions is mg l vers a. The velocity which the mass m will acquire in descending through the vertical distance / vers a to its lowest position is given by the formula v2=2g l vers a. Hence the value of 1⁄2 m v2, or the energy of motion, in the lowest position, is 1⁄2 mv2=m gl vers a, which is the same as the statical energy in the extreme positions. In any intermediate position, let ◊ be the inclination of the string to the vertical. Then the statical energy is mgl vers 0, and the kinetic energy is mo2, where v2 = 2 g l (vers a—vers 6), which, added to the statical energy, gives for the total energy mgl vers a, as before. 121. When the angle a is small, the total energy is α α mgl vers a=2mgl sin2 2 which is proportional to the square of the amplitude, la being the amplitude. Also, the effective force on the mass m in any position—that is, the component force along the tangent-is mg sin 0, which increases from zero in the lowest position to mg sin a in the extreme positions. The distance through which this effective force works during the movement from an extreme position to the lowest is la, and if we multiply this by half the maximum force, that is, by mg sin a, which is the same as mg a, we obtain m gl a2, which is identical with the expression above obtained for the whole work done or whole energy. This rule always holds for vibrations in which the force called out follows Hooke's law, that is to say, the mean working force (which if multiplied by the whole displacement gives the work done) is just half the maximum force. This can be proved as follows. Let o (Fig. 46) be the position of equilibrium, a thẻ extreme position, B and C two points equally distant from o and a respectively. Then if μ. OA be the expression for the force at A, the forces at в and c will be μ. O B р. ов and μ. oc respectively. Let denote the length of an element of the path at B, and of an equal element at c. Then the works done in moving through these elements are μ. O B. σ and μ. oc. 5, the sum of which is μ. (OB+OC) σ = μ..0 A. 6, which is the same as the sum that would have been obtained if the working force had the constant value p. OA; for the work in each of the two elements would then have been ↓ μ. O A. σ. Since the line o A can be exhaustively divided into such pairs of elements, the whole work will be the same as if the working force had the constant value μ. o A, and will therefore be μ. O A2, which is proportional to the square of the amplitude О А. 122. Using the notation of the integral calculus, denoting the amplitude by a and any smaller displacement by x, the force in the position x is μx, the work done in moving over the element dx is μx dx, and the whole work is 123. The proof can also be exhibited graphically. Draw OA equal to the amplitude, and oA making with o A an angle whose tangent is p. Then the ordinate A A' represents the force at A, and the ordinate B B' the force at any other point в; the strip B B' represents the work done in moving over the base of the strip at B, and the sum of all such strips is the triangle o A A', which is half the rectangle under oa and a a'; hence the whole work is half the work done by a force A A' working through a distance O A. 124. When a string executes simple harmonic stationary vibration in one plane, as described in § 64, the energy of its vibration is most easily computed by considering the velocities of its different points at the moment when the string is straight, and when the energy is therefore entirely in the kinetic form. The velocity at the point x (the origin being midway between two nodes) is (with the notation of § 64) |