The Variants of Energy Integral Induced by the Vibrating String

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1 Applied Mathematical Sciences, Vol. 9, 15, no. 34, HIKARI td, The Variants of Energy Integral Induced by the Vibrating String Hwajoon Kim Kyungdong University School of IT Engineering Yangju 48-1, Gyeonggi, Korea Corresponding author Copyright c 15 Hwajoon Kim. This is an open access article distributed under the Creative Commons Attribution icense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract The equation which epresses the principle of energy conservation for the vibrating string can be derived from the homogeneous or inhomogeneous wave equation. In this article, we have checked the variants of energy equation induced by the vibrating string. Mathematics Subject Classification: 355 Keywords: energy equation, vibrating string 1 Introduction We have checked the representation of the solution of the energy equation by aplace transform in [1], and here, we have eplored the variants of energy integral induced by the vibrating uniform string stretched between fied end points whose eternal forces acted on it. It is a well-known fact that the energy integral E(t) has the form of E(t) = K(t) + V (t), where K(t)/V (t) is the kinetic/potential energy, respectively, and this can be interpreted as the total energy of the entire string. et T, u = u(, t), ρ be, respectively, the tension, displacement and density in the string. If we denote the distributed eternal forces per unit mass in the positive u-direction by q, then de/dt can

2 1656 Hwajoon Kim be epressed as de dt = T u u t ] + ρ qu t d, and this means that the rate of change of the total energy of the string is equal to the rate at the eternal forces working on the string. Normally, the total of the eternal forces acting on the segment is determined as T (b, t)sinθ(b, t) T (a, t)sinθ(a, t) = b a T sinθ d, (3) where θ(, t) is the angle of inclination of the graph of the displacement function and T is the tension in the string. If the distributed eternal forces(frictional damping/gravity) act on the string, then the above equality must be replaced with b b a (T sinθ) d + ρ q d, a where q is the distributed eternal forces per unit mass in the positive u- direction. This leads to the PDE u ρ t = T u + ρ q. Using this equation, we would like to check the variants of energy integral. On the other hand, let us take a look into the preceding researches for related topics. Several researches have been pursued for energy equation[1, 5-6, 1, 15]. [1] has proposed mathematical modeling of wind turbine in a wind energy conversion system, [5] has dealt with a simple proof of a result conjectured by Onsager on the energy conservation for weak solutions of the 3D incompressible Euler s equation, and [6] has shown that the energy function is a non-increasing and the solution is unique. [1] has checked the representation of energy equation by aplace transform[-4, 7-11, 13-14], and [15] is checking the energy decay for a wave equation with nonlinear boundary memory and damping source. In this article, we would like to eplore the variants of energy equation by using the above equality. The PDE modeling of the vibrating string First, let us consider a uniform string stretching between fied end points whose no eternal forces act on it and it is at rest. Net, let us release the string. Then it will vibrate in the plane. We would like to find the conditions determining the purely transverse plane vibrations of a perfectly fleible uniform string. et a function u = u(, t) call the displacement function, and let s(, t), ρ(, t),

3 The variants of energy integral induced by the vibrating string 1657 T (, t) be, respectively, the arc length, density and tension in the string. Since the mass of the segment above the interval [a, b] is constant, we have for arbitrary constants a and b. Hence d b b s ρds = (ρ dt a a t )d = s (ρ )d =. t This implies the given string is in an equilibrium position at some time, and so, we can put ρ s = ρ. The forces acting in the masses of a system consist of internal forces, which are forces acting between particles of the system, and eternal forces. et us consider the segment of the string lying the interval [a, b]. Then the mass of this segment is s(b,t) s(a,t) ρ ds = ρ (b a). (1) The u-coordinate of the center of mass of the segment is 1 s(b,t) ρu ds = ρ (b a) s(a,t) 1 b ρ u d = 1 b u d ρ (b a) a b a a and the u-component of the acceleration of the center of mass is d 1 b u d = 1 b u dt tt d. () b a a b a a Generally, the sum of eternal forces can be epressed by T (b, t)sinθ(b, t) T (a, t)sinθ(a, t) = b a T sinθ d (3) for θ(, t) is the angle of inclination of the graph of the displacement function. Equating the product of the mass (1) and the acceleration () to the force (3), we have ρ u tt = T sinθ. Similarly, since the -component of the acceleration of the center of mass is zero, we have = T cosθ.

4 1658 Hwajoon Kim Since for fied t, θ(, t) is the angle of the inclination of the curve u = u(, t) whose slope is u (, t), we have u = tanθ, and so sinθ = u, cosθ = 1. Implies, ρ u tt = [T u ] = T u + u [ T ], and from we have [ T ] =, ρ u tt = T u. (4) Since the equation (4) is nonlinear, to avoid this difficulty, we restrict our consideration to vibration which is small in the sense that the relative change in arc length of the string at each point is small. That is s/ = is approimately equal to 1. Since the change in the magnitude T of the tension from its equilibrium value T depends on the relative change in arc length, T varies little form T. Thus we can put T as T. Implies, we have the equation u tt = c u < <, t > where the constant c is given by c = T /ρ. This is the famous onedimensional wave equation. The displacement function must satisfies the boundary conditions u(, t) =, u(, t) =. To find the initial conditions, we consider the motion of the center of mass of a segment. To specify this motion, we must specify its initial position and initial velocity as u(, ) = f(), u t (, ) = g().

5 The variants of energy integral induced by the vibrating string The variants of energy integral induced by the vibrating string As in the preceding introduction let us write q is the distributed eternal forces acting on the string. As it is mentioned in introduction, the total of the eternal forces acting on the segment is determined as u ρ t = T u + ρ q. (5) With the aid of this equation, let us check the following theorem. The details with respect to the energy integral E(t) appear in [1]. Theorem 3.1 Suppose that the equilibrium position of the string lie along the -ais with the fied ends at = and =. When we choose the u-ais perpendicular to the -ais and in the plane of vibration, the position of the string at any time t sames to the graph of the displacement function u = u(, t). et ρ, u, T be, respectively, the density, displacement and tension in the string. Then the followings are obtained from the equation (5). (a) T u = ρ (u tt q(, t)) 1 (b) lim h k ρ u 1 t h = ρ u t d dk (c) dt = ρ qu t T o u u t d + T [u (, t)u t (, t) u (, t)u t (, t)] d (d) ρ qu t d = [ (e) dt (1 ρ u t ) T u u t ] d ρ qu t = de dt T [u (, t)u t (, t) u (, t)u t (, t)]. where, the subinterval [ k, k + h] [, ] and K is the kinetic energy of the string. Proof. Since the equation (5) can be rewritten as ρ (u tt q(, t)) = T u, (a) follows. et us check the (b). Since a segment of the string on the subinterval [ k, k +h] has mass ρ h and velocity u t, its kinetic energy K(t) is 1 ρ u t h. Applying the definition of integral, we easily obtain the result. On the other hand, (c) is followed from dk dt = d 1 dt ρ u t d

6 166 Hwajoon Kim = = ρ qu t ρ qu t d d dt 1 T u d + [T u u t ] T u u t d + T [u (, t)u t (, t) u (, t)u t (, t)]. In the above equation, we have interchanged the order of differentiation and integration by the result of []. Net, (d) means the rate at which work is done on the string by the distributed eternal forces. et us multiply the equation (5) by u t and integrate with respect to from to. Then ρ u tt u t d = T u u t d + ρ qu t d. (6) By the way, since ρ u tt u t d = d 1 dt ρ u t d (7) d ρ qu t d = [ dt (1 ρ u t ) T u u t ] d follows. Finally, (e) is followed by the equation (6) and (7). References [1] Bradshaw, Po, D. H. Ferriss, and N. P. Atwell, Calculation of boundarylayer development using the turbulent energy equation, J. of Fluid Mech. 8 (1967), [] Hc. Chae and Hj. Kim, The validity checking on the echange of integral and limit in the solving process of PDEs, Int. J. of Math. Anal., 8 (14), [3] Ig. Cho and Hj. Kim, The solution of Bessel s equation by using integral transforms, Appl. Math. Sci., 7 (13), [4] Ig. Cho and Hj. Kim, The aplace transform of derivative epressed by Heviside function, Appl. Math. Sci., Vol. 9 (13), [5] Constantin, Peter, E. Weinan, and Edriss S. Titi, Onsager s conjecture on the energy conservation for solutions of Euler s equation, Comm. in Math. Phys. 165 (1994), [6] P. Haarsa and S. Pathat, Nonincreasing energy function on inhomogeneous diffusion equation with Neumann B.C., Appl. Math. Sci., 8 (14),

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