Thursday, 26 March 2020

newtonian mechanics - Boundary conditions on wave equation



wave equation question


I am having trouble understanding the boundary conditions.


From the solutions, the first is that $D_1(0, t) = D_2(0, t)$ because the rope can't break at the junction.


The second is that $\dfrac{\partial D_1}{\partial x} D_1(0, t) = \dfrac{\partial D_2}{\partial x}(0, t)$. How can I interpret this physically? I'm not quite sure how to think about $\partial D /\partial x$.



Answer



The second condition is saying that there is no discontinuity in the slope of the rope at the junction. In other words, there is no "kink" in the rope.


Imagine if this assumption were to fail in the following way: $$ \frac{\partial D_1}{\partial x}(0,t) = -1, \qquad \frac{\partial D_2}{\partial x}(0,t) = 1 $$ Then near the origin, the rope would look like the function $f(x) = |x|$ does at the origin; there would be a "triangular kink" in the rope facing upward.


Addendum. Why can't there be a kink? In response to Nathaniel's response, here's why there can't be a kink. We argue by way of contradiction.


Suppose that there were a kink, and consider a small mass element centered on the junction. In the presence of a kink, the tensions on either side of the junction would point in different directions, and there would therefore be a net force on the small mass element. Now consider taking the size of that mass element to zero. There will be a net force on the mass element as we take the limit of its size to zero, but its mass will vanish, which yields a contradiction to Newton's second law.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...