{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework3_Solutions

Homework3_Solutions - I HOMEWORK#3 SOLUTIONS 10.1 w(z t =(z...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
I. HOMEWORK #3 SOLUTIONS 10.1) w ( z, t ) = ξ ( z - ct ) 2 w ( z,t ) ∂z 2 = ξ 00 ( z - ct ) 2 w ( z,t ) ∂t 2 = c 2 ξ 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 it’s clear that it satisfies the wave equation. By the same method, we can show that w ( z, t ) = ξ ( z - ct ) + ξ ( 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, I’ll do it the longer of two ways. From Ex. 10.2, we have the following equation for a wave: φ ( z, t ) = (1 - e - ( t - z c ) 5 μs ) e - ( t - z c 5 μs 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. φ max φ 0 = 0 φ 0 = - 1 5 μs e - ( tmax - z c ) 5 μs + 2 5 μs e - 2( tmax - z c ) 5 μs = 0 - 1 + 2 e - ( tmax - z c ) 5 μs = 0 t max - z c 5 μs = ln (2) t max = 5 μsln (2) + z c Plugging in our known values for z and c, we find t = 68 . 4 μs Alternatively, you may have realized that in the example the time is given for then the 1
Image of page 1

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

View Full Document Right Arrow Icon
leading edge of the wave arrives at z=0.1. Therefore, if you take the position independent form φ ( t ) = (1 - e - t 5 μs ) e - t 5 μs and solve for t max , you’ll find that t max = 3 . 5 μs . Now, just add it to the time necessary for the front edge to arrive at z=0.1: 3 . 5 μs + 64 . 9 μs = 68 . 4 μs . 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 μs (aka, when the wave first reaches z=0.1). φ ( z, t ) = (1 - e - t - 64 . 9 μs + z - 0 . 1 c 5 μs ) e - t - 64 . 9 μs + z - 0 . 1 c 5 μs 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, you’ll obtain the following expression: t max = 5 μsln (2) + 0 . 1 c + 64 . 9 μs = 133 . 3 μs .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern