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Unformatted text preview: I. HOMEWORK #3 SOLUTIONS 10.1) w ( z,t ) = ( z ct ) 2 w ( z,t ) z 2 = 00 ( z ct ) 2 w ( z,t ) t 2 = c 2 00 ( z ct ) Wave Equation: 2 w ( z,t ) z 2 = 1 c 2 2 w ( z,t ) t 2 Plug in the two equations from above, and its clear that it satisfies the wave equation. By the same method, we can show that w ( z,t ) = ( z ct ) + ( z + ct ) is also a solution. The physical meaning of w ( z,t ) is a wave with components traveling in both the forward and backward directions along the z axis. 10.3) There are a couple of ways to do this problem. Here, Ill do it the longer of two ways. From Ex. 10.2, we have the following equation for a wave: ( z,t ) = (1 e ( t z c ) 5 s ) e ( t z c 5 s where c = 1540 m/s a) z = 0.1 m We can find when the peak arrives at the point of interest using the first derivative to find the value of t where the pressure is maximized. max = 0 = 1 5 s e ( tmax z c ) 5 s + 2 5 s e 2( tmax z c ) 5 s = 0 1 + 2 e ( tmax z c ) 5 s = 0 t max z c 5 s = ln (2) t max = 5 sln (2) + z c Plugging in our known values for z and c, we find t = 68 . 4 s Alternatively, you may have realized that in the example the time is given for then the 1 leading edge of the wave arrives at z=0.1. Therefore, if you take the position independent form ( t ) = (1 e t 5 s ) e t 5 s and solve for t max , youll find that t max = 3 . 5 s . Now, just add it to the time necessary for the front edge to arrive at z=0.1: 3 . 5 s + 64 . 9 s = 68 . 4 s . b) Now we simply need to adjust our original form into a backward traveling wave with appropriate time delay, since the wave begins at z=0.1 at time t = 64 . 9 s (aka, when the wave first reaches z=0.1). ( z,t ) = (1 e t 64 . 9 s + z . 1 c 5 s ) e t 64 . 9 s + z . 1 c 5 s c) Again, there is a long way and a short way to do this problem. The long way: Use the first derivative on the equation from part b to find t max by setting z=0. Doing this, youll obtain the following expression: t max = 5 sln (2) + . 1 c + 64 . 9 s = 133 . 3 s ....
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 Fall '10
 MacFall

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