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Unformatted text preview: 1 1 1.1. VIBRATION OF A TAUT STRING 1.138J/2.062J/18.376J, WAVE PROPAGATION Fall, 2006 MIT Notes by C. C. Mei Chapter 1. SAMPLE WAVE PROBLEMS To describe a problem in mathematical terms, one must make use of the basic laws that govern the elements of the problem. In continuum mechanics, these are the conser- vation laws for mass and momentum. In addition, empirical constitutive laws are often needed to relate certain unknown variables; examples are equations of state, Hooke’s law between stress and strain, etc. To derive the conservation law one may consider an infinitesimal element (a line segment, area or volume element), yielding a differential equation directly. Alternately, one may consider a control volume (or area, or line segment) of arbitrary size in the medium of interest. The law is first obtained in integral form; a differential equation is then derived by using the arbitrariness of the control volume. The two approaches are completely equivalent. Let us first demonstrate the differential approach. Transverse vibration of a taut string Referring to Figure 1, consider a taut string stretched between two fixed points at x = 0 and x = L . Let the cross-sectional area be S . If there is an initial stretching of ∆ L , the initial tension T must be ∆ L T = ES L by Hooke’s law, where E is Young’s modulus. Now study the lateral displacement of the string from the initial position. By the law of conservation of transverse momentum, the total lateral force on the string element must be balanced by its inertia. Let the lateral displacement be V ( x, t ) and consider a differential element between x and x + dx . The net transverse force due to the difference of tension at both ends of the element is ( T sin α ) x + dx − ( T sin α ) x , 2 1.1. VIBRATION OF A TAUT STRING α V ( T sin α ) x+dx p(x,t) dV x x+dx 0 L- ( T sin α ) x x Figure 1: Deformation of a taut string where ∂V dV ∂x sin α = √ = . dx 2 + dV 2 1 + ( ∂V ) 2 ∂x We shall assume the displacement to be small everywhere so that the slope is also small: ∂V 1. The local value of sin α can then be approximated by ∂x 3 ∂V ∂V + O , ∂x ∂x where the expression O ( δ ) stands for of the order of δ . For any smooth function f , Taylor expansion gives ∂f 2 f ( x + dx ) − f ( x ) = dx + O ( dx ) , ∂x where the derivative is evaluated at x . Hence the net tension is ∂ ∂V T dx + O ( dx ) 2 . ∂x ∂x The instantaneous length ( x, t ) of the string from to x is 2 1 / 2 2 x ∂V ∂V ( x, t ) = dx 1 + = x 1 + O . ∂x ∂x It follows that 2 − x ∂V = O for all < x < L, x ∂x which is of second-order smallness. The string length, hence the tension, is essentially 2 unchanged with an error of O ( ∂V /∂x ) , i.e., T can be taken as constant with a similarly small error. Thus the net tension in the string element is well represented by ∂ 2 V T dx....
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This note was uploaded on 12/04/2011 for the course ESD 18.327 taught by Professor Gilbertstrang during the Spring '03 term at MIT.
- Spring '03