{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}