1 d select the cell containing vibstring and animate

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Unformatted text preview: x, t) = f (x + ct) is a solution to the partial differential equation ∂u ∂u −c = 0. ∂t ∂x (Hint: Use the “chain rule.”) b. Show that if g : R → R is any well-behaved function of one variable, u(x, t) = g (x − ct) is a solution to the partial differential equation ∂u ∂u +c = 0. ∂t ∂x 97 c. Show that for any choice of well-behaved functions f and g , the function u(x, t) = f (x + ct) + g (x − ct) is a solution to the differential equation ∂2u ∂2u ∂ ∂ − c2 2 = +c 2 ∂t ∂x ∂t ∂x ∂u ∂u −c ∂t ∂x = 0. Remark: This gives a very explicit general solution to the equation for the vibrations of an infinitely long string. d. Show that f (x + ct) + f (x − ct) 2 is a solution to the initial value problem u(x, t) = ∂2u ∂2u − c2 2 = 0, 2 ∂t ∂x u(x, 0) = f (x), ∂u (x, 0) = 0. ∂t 4.4.2. Show that if the tension and density of a string are given by variable functions T (x) and ρ(x) respectively, then the motion of the string is governed by the equation ∂2u 1∂ ∂u = T (x) . 2 ∂t ρ(x) ∂x ∂x 4.5 The initial value problem for the vibrating...
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This document was uploaded on 01/12/2014.

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