Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $\partial D /\partial x$.

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
I guess the question is, why isn't that allowed? Since there is a discontinuity in the linear density of the rope, it's not immediately intuitively obvious that a kink can't develop at that point. (I think it can be derived from energy conservation, though.) –  Nathaniel May 16 '13 at 4:02
    
@Nathaniel I made an addition to address your points. –  joshphysics May 16 '13 at 4:36
    
thanks, that's a very elegant argument. –  Nathaniel May 16 '13 at 8:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.