{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HWsol05 - Physics 2214 Fall 2008 Solutions to Problem Set#5...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
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
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. 2
Background image of page 2
(a) Use the principle of superposition along with the boundary conditions to draw the left end of the string ( 0 x 4 m ) at t = 0 . 6 s . Solution The left end is fixed The principle of superposition means we need to consider the following: since y (0 , t ) = 0 y ( x, t ) = F t + x v - F t - x v (15) Where F ( t + x v ) was the incident pulse. To plot - F ( t - x v ) (the reflected pulse), invert the pulse with respect to the x direction ( x/v → - x/v ) and invert the pulse with respect to the y direction F → - F . At t = 0 . 6 m this look like this: Of course, in this class of problems, don’t take negative x axis seriously, it’s just the means to figure out how the boundary condition impose the reflected waves.
Background image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}