This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Review for Final Exam The first section covers reflection and transmission in a string, while the second section covers how to derive and explain the Doppler effect, both of which were mentioned to me in conjunction with your final exam. However, theres much more to study. For help in learning how to do the RC circuit problem on your final, I strongly encourage you guys to go over the last two problems in 1Bch26.pdf. Also, study 1Bch27.pdf for help in computing magnetic forces. And, of course, study everything else that Dr. Coroniti mentioned in relation to the final. 1 Reflection and Transmission in a String Suppose you have two strings tied together in a knot, which is located at x = 0. String 1, on the left ( x < 0), has linear mass density 1 , while string 2, on the right ( x > 0), has linear mass density 2 . A traveling wave with amplitude A i and wave number k 1 moves to the right through string 1, and is incident on the knot from the left. We wish to solve for the amplitude A t of the traveling wave transmitted into string 2 and the amplitude A r of the wave reflected back into string 1. Suppose each wave has angular frequency , and that both strings are under the same tension F T . Lets start by writing the general equations for the waves in strings 1 and 2. for string 1, we have y 1 ( x,t ) = y i ( x,t ) + y r ( x,t ) = A i cos( k 1 x- t ) + A r cos( k 1 x + t ) The term with subscript i is the incident wave and the term with subscript r is the reflected wave. So we see that string 1 contains a superposition of an incident and a reflected wave. In string 2, we have y 2 ( x,t ) = A t cos( k 2 x- t ) Notice how we use kx- t for waves traveling to the right, and kx + t for waves traveling to the left. So, here are are general solutions in each string: y 1 ( x,t ) = A i cos( k 1 x- t ) + A r cos( k 1 x + t ) y 2 ( x,t ) = A t cos( k 2 x- t ) The spatial derivatives of these solutions are y 1 x ( x,t ) =- k 1 A i sin( k 1 x- t )- k 1 A r sin( k 1 x + t ) y 2 x ( x,t ) =- k 2 A t sin( k 2 x- t ) 1 In order to solve for A r and A t , we must impose the following conditions at the knot, at x = 0: y 1 (0 ,t ) = y 2 (0 ,t ) y 1 x (0 ,t ) = y 2 x (0 ,t ) The first condition says that the waves must be continuous at the knot at all times. The second condition says that the waves must be smooth at the knot at all times. Smoothness means that the spatial derivative must exist (i.e., be well-defined), which means that itmeans that the spatial derivative must exist (i....
View Full Document
- Winter '07