This result shews that the length of the required pendulum will be equal to a. Daniel Bernoulli: De Motibus oscillatoriis corporum humido insidentium; Comment. Acad. Petrop. 1739, p. 106. 2. A body, which is symmetrical with regard to a vertical section, executes small oscillations in a fluid in such a manner that the motions of all the particles of the body take place parallel to the vertical section: to determine the nature of the vertical and angular oscillations of the body, the centre of gravity of the plane of floatation, when the body is at rest, not lying in the vertical line through the centres of gravity of the body and the fluid displaced. Let AB (fig. 63) be the projection, on the plane of symmetry, of the section of the body which, when the body is in equilibrium, constitutes its plane of floatation: let C be the centre of gravity of the plane AB, ab the projection of the plane of floatation at any time t. Draw a'b' parallel to ab, through C. Let G be the centre of gravity of the body, H that point of the body which, when the body is at rest, coincides with the centre of gravity of the fluid displaced. Let x be the vertical distance of the centre of gravity of the body below its place of rest at the time t, the angular deviation from the position of equilibrium. Then the equations of motion are M being the mass of the body, M' that of the fluid displaced, viz. a Db, y the horizontal distance of the centre of gravity of the fluid displaced from G at the time t, parallel to ba, Mk2 the moment of inertia of the body about a line through G at right angles to the section of symmetry. We must express M' and M'y approximately in terms of x and 0. Produce HG to h: the line HGh will of course be at right angles to AB: draw Ck vertically. Let K = the area of the section AB, a = Ch, a LCGh, p = the density of the fluid. = = Then the distance of G below the surface Ck+ CG sin GCb' = Ck + CG cos (a – 0) = = Ck+ CG (cos a + sin a), nearly, Ck+CG cos a + a0: but, when the body is floating at rest, the depth of G is equal to CG cos a: hence, x being the difference between these two depths, x = Ck+ a0. Again, λ denoting a small area at a point P in the section a' Cb', z the length of a perpendicular upon it, cut off by the section ACB, and x, the distance of λ from a line through C, at right angles to the plane of symmetry, we have M' = mass of ADB – pΣ (λz) + mass of aa'b'b, Σ denoting summation for the two wedges A Ca', BCb', in the former of which z is positive, in the latter negative; or M' = M-p0Σ (xx) + pK. Ck, nearly, but, C being the centre of gravity of the area ACB, and therefore approximately of the area a' Cb', Σ (λxx,) = 0 ; hence Again, the action of the fluid a Db on the body is equivalent to that of aa'b'b together with that of a' Db'. Now the moment of a a' b'b about G = gpK. Ck . CG cos GCb', nearly, = gpK. (x – a0).a, = nearly, and tends to twist the body in the direction of the arrows in the figure. Also, the moment of the fluid a' Db' about G, if HG = b, and c = the projection of CG on Cb', is equal to Mgb0-gp.λz (x, + c), in the direction of the arrows, the Σ being supposed to extend to the two wedges ACa', BCb', in the former of which, z and x' are positive, in the latter, negative; or the moment of the fluid a' Db' is equal to and therefore, Σ(x) being nearly zero, since the centre of gravity of the area ACB and therefore aproximately of the area a' Cb' is at C, we see that the moment of a' Db' = g@ { Mb – pΣ (λx,2)} = g0 (Mb – pI), where I denotes the moment of inertia of the area a'b', or, which is approximately the same, of the area ACB about the line through C which cuts the section of symmetry at right angles. = Thus My.ggpK (x - a0). a + g✪ (Mb – pI). We have therefore, from (3) and (2), d2x gpK dt2 M do pKag which are two linear differential equations which may easily be integrated. Assume, for the sake of simplicity, Let μ, μ, be the two values of μ deducible from this equation: then, A1, A2, B1, B2, ε1, ε, being arbitrary constants, x = A, sin (μ,t + ε) + A2 sin (μ2t + ε2), 0 = ß, sin (μ1t + ɛ) + ß2 sin (μ2t + ɛ2), 2 by which equations the motion is completely determined. It must be observed that COR. Suppose that the centre of gravity of the plane of floatation, when the body is at rest, lies in the vertical line through G and H. Then, putting a = 0, the two differential equations degenerate into the period of the vertical and angular oscillations being respectively M\ π (Mk2)3 g1 (pl – Mb)}' The amplitudes A and B of the oscillations are independent of each other, and their periods are different. Such, it may be observed, is not the case under the more general form of the problem, where x and both involve two terms of periods the amplitudes A1, ß1, and Α,, B2, being not independent of each other. π με π and " με 3. A paraboloid of revolution, with its axis vertical and vertex downwards, is oscillating in a fluid; having given the period of its small oscillations, to determine the depth of its immersion when it is in a position of rest. If denote the volume of the fluid displaced by the paraboloid in a position of rest, K the area of its plane of floatation, and x the distance of the centre of gravity of the paraboloid below its place of rest at the end of any time t; then, observing that M is equal to pV, we have Let a represent the length of the axis immersed, when the body is at rest: then, the volume of a paraboloid of revolution being equal to one-half of its circumscribing cylinder, |