HWsol05 - Physics 2214 Fall 2008 - Solutions to Problem Set...

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Unformatted text preview: Physics 2214 Fall 2008 - Solutions to Problem Set #5 1. Use the chain rule to show that the wave equation, ∂ 2 y ( x,t ) ∂t 2 = v 2 ∂ 2 y ( x,t ) ∂x 2 (1) can be rewritten as ∂ 2 y ( u + ,u- ) ∂u + ∂u- = 0 (2) where u + ≡ x + vt and u- ≡ x- vt . Solution • First, the derivative of u + ,u- in terms of x,t : ∂u + ∂x = ∂ ( x + vt ) ∂x = 1 (3) ∂u- ∂x = ∂ ( x- vt ) ∂x = 1 (4) ∂u + ∂t = ∂ ( x + vt ) ∂t = v (5) ∂u- ∂t = ∂ ( x- vt ) ∂t =- v (6) • Now using the chain rule for a derivative of x : ∂y ( u + ,u- ) ∂x = ∂y ( u + ,u- ) ∂u + 1 ∂u + ∂x + ∂y ( u + ,u- ) ∂u- 1 ∂u- ∂x = ∂y ( u + ,u- ) ∂u + + ∂y ( u + ,u- ) ∂u- (7) and for a derivative of t : ∂y ( u + ,u- ) ∂t = ∂y ( u + ,u- ) ∂u + v ∂u + ∂t + ∂y ( u + ,u- ) ∂u-- v ∂u- ∂t = v ∂y ( u + ,u- ) ∂u +- ∂y ( u + ,u- ) ∂u- (8) • After finding the first derivative, we can find the second derivatives: ∂ 2 y ( u + ,u- ) ∂t 2 = v ∂ ∂y ( u + ,u- ) ∂u +- ∂y ( u + ,u- ) ∂u- ∂t = v ∂ ∂y ( u + ,u- ) ∂u +- ∂y ( u + ,u- ) ∂u- ∂u + v ∂u + ∂t + ∂ ∂y ( u + ,u- ) ∂u +- ∂y ( u + ,u- ) ∂u- ∂u-- v ∂u- ∂t = v 2 ∂ 2 y ( u + ,u- ) ∂u 2 +- ∂ 2 y ( u + ,u- ) ∂u + ∂u-- ∂ 2 y ( u + ,u- ) ∂u- ∂u + + ∂ 2 y ( u + ,u- ) ∂u 2- (9) 1 ∂ 2 y ( u + ,u- ) ∂x 2 = ∂ ∂y ( u + ,u- ) ∂u + + ∂y ( u + ,u- ) ∂u- ∂x = ∂ ∂y ( u + ,u- ) ∂u + + ∂y ( u + ,u- ) ∂u- ∂u + 1 ∂u + ∂x + ∂ ∂y ( u + ,u- ) ∂u + + ∂y ( u + ,u- ) ∂u- ∂u- 1 ∂u- ∂x = ∂ 2 y ( u + ,u- ) ∂u 2 + + ∂ 2 y ( u + ,u- ) ∂u + ∂u- + ∂ 2 y ( u + ,u- ) ∂u- ∂u + + ∂ 2 y ( u + ,u- ) ∂u 2- (10) Remember that we assume the solutions are twice differentiatable, thus ∂y ( u + ,u- ) ∂u- ∂u + = ∂y ( u + ,u- ) ∂u + ∂u- , and we find ∂ 2 y ( u + ,u- ) ∂t 2 = v 2 ∂ 2 y ( u + ,u- ) ∂u 2 +- 2 v 2 ∂ 2 y ( u + ,u- ) ∂u + ∂u- + v 2 ∂ 2 y ( u + ,u- ) ∂u 2- (11) ∂ 2 y ( u + ,u- ) ∂x 2 = ∂ 2 y ( u + ,u- ) ∂u 2 + + 2 ∂ 2 y ( u + ,u- ) ∂u + ∂u- + ∂ 2 y ( u + ,u- ) ∂u 2- (12) • Finally, we see that the wave equation ∂ 2 y ∂t 2- v 2 ∂ 2 y ∂x 2 =0 (13) gives 0 = ∂ 2 y ( u + ,u- ) ∂t 2- v 2 ∂ 2 y ( u + ,u- ) ∂x 2 = v 2 ∂ 2 y ( u + ,u- ) ∂u 2 +- 2 v 2 ∂ 2 y ( u + ,u- ) ∂u + ∂u- + v 2 ∂ 2 y ( u + ,u- ) ∂u 2-- v 2 ∂ 2 y ( u + ,u- ) ∂u 2 + + 2 ∂ 2 y ( u + ,u- ) ∂u + ∂u- + ∂ 2 y ( u + ,u- ) ∂u 2- ! =- 4 v 2 ∂ 2 y ( u + ,u- ) ∂u + ∂u- (14) thus “regular” wave equation is true if and only if ∂ 2 y ( u + ,u- ) ∂u + ∂u- = 0 2. At t = 0 , the wave pulses shown below are moving away from one another. The wave speed is 10 m / s . The right end ( x = 25 m ) is free (massless slip-ring boundary condition) and the left end ( x = 0 ) is fixed....
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This note was uploaded on 12/24/2008 for the course PHYS 2214 taught by Professor Giambattista,a during the Fall '07 term at Cornell University (Engineering School).

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HWsol05 - Physics 2214 Fall 2008 - Solutions to Problem Set...

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