# Odd constraint problem: Reflected and Transmitted Power of String

1. Mar 31, 2014

### N00813

1. The problem statement, all variables and given/known data
Given that a string is constrained such that dy/dx = 0 at x = 0 and unconstrained otherwise, what is the reflected and transmitted power?
y is the deflection of the string from the x-axis. y_1 is incident wave, y_r is reflected and y_t is transmitted.

2. Relevant equations

Reflected power, transmitted power have already been derived in terms of impedances.
$$Impedance Z = \frac{Driving Force}{string element velocity}$$
Continuity of y and dy/dx.

3. The attempt at a solution
Knowing that y and dy/dx are continuous, I wrote $\frac{\partial y_1}{\partial x} +\frac{\partial y_r}{\partial x} = \frac{\partial y_t}{\partial x} =0$ at x = 0.

Substituting in the general solution $y_1 = e^{-ikx+iwt}, y_r = re^{+ikx+iwt}, y_t = te^{-ikx+iwt}$;

I got 1 + r = t and 1 - r = t = 0 at x = 0.

The latter suggests all reflection, no transmission, which isn't correct because it doesn't satisfy the first equation.

Last edited: Apr 1, 2014
2. Apr 1, 2014

### BvU

It does not satisfy which equation precisely ?

3. Apr 1, 2014

### N00813

The given constraint (and continuity of gradient) means that -ik(1-r) = -ik(t) = 0.
Continuity of string means that 1 + r = t.

If t = 0, then 1-r = 0 so r = 1. But then 1 + r = 2 =/= t = 0.

4. Apr 1, 2014

### BvU

Yes, so there must be something wrong at a more elementary level: even if you only impose continuity of y and dy/dx on these "general solutions" you end up with r = 0 ! This way you have three equations with only t and r as unknowns. Are you sure the y are general enough ?

5. Apr 1, 2014

### N00813

I know the string satisfies a wave equation. Evanescent waves would be my next guess, but beyond that I'm stuck.

6. Apr 1, 2014

### dauto

The solution for that problem requires incoming waves from both directions at the same time. (producing a standing wave with an anti-node at the origin)

Last edited: Apr 1, 2014
7. Apr 1, 2014

### N00813

There'd be no reflected and transmitted power then, would there? Since the wave is standing?

8. Apr 1, 2014

### dauto

Good point. I'm not sure what to make of that. The question doesn't seem to make much sense since there is incoming energy from both ends.

9. Apr 1, 2014

### N00813

Earlier in the question, it did say to use a complex exponential for y (=A exp(ikx-iwt)) to prove that time-averaged power = 1/2 Zw^2 A^2. I assumed that carried forwards.

I suppose I'll have to ask my supervisor for this, then. Perhaps the answer really is zero.