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Unformatted text preview: Wave equation in Cartesian coordinates • The wave equation is given in one spatial dimension ∂ 2 u ∂ x 2 = 1 v 2 ∂ 2 u ∂ t 2 • We again use separation of variables u ( x , t ) = X ( x ) T ( t ), and then we can write the wave equation as, X 00 X = 1 v 2 T 00 T = k 2 • Hence we wind up with two Helmholtz equations to solve, X 00 + k 2 X = 0 T 00 + k 2 v 2 T = 0 Wave equation in Cartesian coordinates, continued • We find solutions T ( t ) = sin ω t , T ( t ) = cos ω t , X ( x ) = sin kx , and X ( x ) = cos kx • Here the angular frequency ω = kv , and k = 2 π/λ • From y ( x , t ) = X ( x ) T ( t ), we have four different basic solutions y ( x , t ) = sin kx sin ω t y ( x , t ) = sin kx cos ω t y ( x , t ) = cos kx sin ω t y ( x , t ) = cos kx cos ω t • We can have linear combinations of solutions of this kind depending on the boundary conditions and initial conditions (superposition principle!) Example: Waves on a string with fixed ends • Imagine a string (e.g. a guitar string) fixed at the ends, so thatImagine a string (e....
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 Spring '03
 Staff
 Sin, Vibrating string, Boundary value problem, Partial differential equation, sin kx sin

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