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.

Suppose I have a wave traveling to the right described by $e^{iw(t-{x\over c})}$. (It obeys the 1D wave equation). AND at $x=0$, there is a mass $M$ fixed to the string such that we have $M{d^2y\over dt^2}|_{x=0}= T{\partial y\over \partial x}|_{x=0_+}-T{\partial y\over \partial x}|_{x=0_-}$.

How on earth do I find the reflected and transmitted waves??

Thanks.

share|improve this question

1 Answer 1

On the incoming, $x<0$, side of the mass there is the incident wave and the reflected wave so $$y=e^{i\omega(t-x/c)}+Ae^{i\omega(t+x/c)}$$ where the incident wave has an amplitude of unity as described in the problem statement and the reflected wave has an unknown amplitude, $A$, and is traveling in the opposite direction. On the outgoing, $x>0$, side of the mass there is only the transmitted wave so $$y=Be^{i\omega(t-x/c)}\text{.}$$

The string must me continuous at $x=0$ so $A=B-1$. Continuity also implies that there can't be a phase change other than 0 or $\pi$ at $x=0$, which is why I haven't bothered to include phases in the three terms of $y$.

The boundary condition at $x=0$ (note that it doesn't matter which whether we pick the $x<0$ or $x>0$ version of $y$ for the left hand side, we'll get the same answer) gives

$$ -BM\omega^2e^{it\omega}=\frac{iT\omega}{c}e^{it\omega}(1-A-B) $$

Applying continuity makes this

$$ -BM\omega^2e^{it\omega}=\frac{2iT\omega}{c}e^{it\omega}(1-B)\text{.} $$

Solving for $B$ gives

$$B=\frac{2T}{2T+icM\omega}=\frac{4T^2}{4T^2-c^2M^2\omega^2}-\frac{2iTcM\omega}{4T^2-c^2M^2\omega^2}$$

which in turn gives $$A=\frac{-icM\omega}{2T+icM\omega}=\frac{c^2M^2\omega^2}{4T^2-c^2M^2\omega^2}-\frac{2iTcM\omega}{4T^2-c^2M^2\omega^2}\text{.}$$

Strictly speaking, we now know everything about the reflected and transmitted waves. To visualize the motion of the string, we can take the real part of our solution to the wave equation.

For the $x<0$ side this gives

$$ y=\cos[\omega(t-x/c)]+\frac{cM\omega}{4T^2-c^2M^2\omega^2}\left ( cM\omega\cos[\omega(t-x/c)]+2T\sin[\omega(t-x/c)] \right ) $$

For the $x>0$ side we have

$$ y=\frac{2T}{4T^2-c^2M^2\omega^2}\left ( 2Tcos[\omega(t-x/c)]+cM\omega\sin[\omega(t-x/c)] \right ) $$

share|improve this answer
    
Thanks! But part of the condition uses the value at $x=0$ which wave should I use for that? –  Raj Nov 1 '11 at 20:04
    
The general one you've already expressed in your question. Apply the boundary conditions and use the resulting equations. –  sevenofdiamonds Nov 1 '11 at 20:23
    
Thanks again, sevenofdiamonds. Sorry for my being daft. Could you have a look at the following? : By the BC, $-M\omega^2 e^{i\omega(t-{x\over c})}=-T|\tau|{\omega\over c}\cos(\omega t +\phi_\tau)+T|R|{\omega\over c}\cos(\omega t +\phi_R)$ where $|\tau|, |R|$ are amplitudes of the transmitted and reflected waves respectively. But then what happens? –  Raj Nov 1 '11 at 21:01
    
I'm not going to be able to get back to this for a while, but you definitely don't need the phases since the string has to be continuous at x=0. –  sevenofdiamonds Nov 2 '11 at 0:05
    
I've now posted a full answer. –  sevenofdiamonds Nov 2 '11 at 14:38

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.