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chem3322_hwk1_soln

# chem3322_hwk1_soln - Chem 3322 homework#1 solutions out of...

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Chem 3322 homework #1 solutions, out of 18 marks Problem 1 – 8 marks Prove that, if u 1 ( x, t ) and u 2 ( x, t ) are, separately, solutions to the classical wave equation with propagation speeds of v 1 and v 2 respectively, then the function u ( x, t ) = au 1 ( x, t ) + bu 2 ( x, t ) is also a solution, where a and b are real numbers, only if v 1 = v 2 . Make it clear what happens if v 1 6 = v 2 . Solution: We are told that 2 u 1 ∂x 2 = 1 v 2 1 2 u 1 ∂t 2 (1) and 2 u 2 ∂x 2 = 1 v 2 2 2 u 2 ∂t 2 (2) Now, let us consider the function u ( x, t ) = au 1 ( x, t ) + bu 2 ( x, t ): 2 u ∂x 2 = a 2 u 1 ∂x 2 + b 2 u 2 ∂x 2 (3) (here we used the linearity property of the derivative operator) = a 1 v 2 1 2 u 1 ∂t 2 + b 1 v 2 2 2 u 2 ∂t 2 (4) using Eqns. (1) and (2). Now, for u to solve the wave equation, we must have 2 u ∂x 2 = 1 v 2 2 u ∂t 2 (5) for some propagation speed v . Clearly this holds if v 1 = v 2 , since we can factor a common velocity out of the right hand side of Eqn. (4). On the other hand, if v 1 6 = v 2 , we cannot factor a common velocity out of Eqn. (4) and hence we cannot rearrange this equation into

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