Perhaps more like a mathematical issue here...

When we consider a wave propagating from medium 1 to a medium 2, boundary located at $x=0$, we have the following wave equations for the incident, reflected and transmitted waves:

$$y_i (x,t) = A \sin (k_1 x - \omega t)$$

$$y_r (x,t) = B \sin (k_1 x + \omega t)$$

$$y_t (x,t) = C \sin (k_2 x + \omega t)$$

And we have $y(x,t) = y_i + y_r$ for $x<0$ and $y(x,t) = y_t$ for $x>0$

The boundary conditions are: $y(0^-,t) = y(0^+,t)$ and $\frac{\partial y(0^-,t)}{\partial x}=\frac{\partial y(0^+,t)}{\partial x}$.

The first boundary conditions gives:

$$ (A-B) \sin(\omega t) = C \sin (\omega t) $$

while the second gives:

$$(A+B) k_1 \cos (\omega t) = C k_2 \cos (\omega t)$$

In all derivations I found, they cancelled the sines and cosines, concluding that:

$$ A - B = C \tag{1}$$


$$(A+B) k_1 = C k_2.\tag{2}$$

However, mathematically speaking, these conclusions are only valid for those values of $t$ in which the sines/cosines are $\ne 0$, correct?

Why can we conclude the last two equations in general? And, if not, what can we physically interpret about the points of time where either sine or cosine = $0$?


2 Answers 2


Right, but the constants don't depend on $t$, so in particular OP's relations (1) & (2) hold for all $t$ anyway.


Let's look at $(A-B) \sin(\omega t) = C \sin (\omega t)$. This equation needs to be valid for all times $t$. When $t=n\pi/\omega$ the equation is certainly still valid. The equation becomes $0=0$. Therefore at times when $\sin(\omega t)=0$ we don't have any issues. $A-B=C$ doesn't suddenly make the equation invalid when $\sin(\omega t)=0$

Another way to see this is to just not think about it as dividing by $\sin(\omega t)$. Certainly $\sin(\omega t)=\sin(\omega t)$, so we need $A-B=C$ in order for the equation to remain true. If $A-B\neq C$, then our equation is true only when $\sin(\omega t)=0$, which is not true for all times and is not what we want.


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