I have a brief question regarding the formula for wave displacement that I've just encountered. My textbook says:

For a simple plane wave, we have, for a simple harmonic with displacement $u$:

$$u = A \cos(\kappa x - \omega t)$$

where $\omega$ is the angular frequency, $\kappa$ is the wavenumber ($\omega$/wave-velocity), $A$ is the maximum amplitude, and $t$ is time.

Here's my question. If $\kappa$ is defined as $\omega/v$, then I get, by plugging into the formula:

$$u = A \cos(\frac{\omega}{v}x - \omega t)$$

$$= A \cos(\omega (\frac{x}{v} - t))$$

But since $t = \frac{x}{v}$,we get:

$$u = A \cos(\omega(t - t)) = A$$

So by my reasoning (which is obviously wrong), I always end up with $u = A$, which basically makes the whole $\cos$-term redundant. If anyone can explain to me what is wrong with my reasoning, I would greatly appreciate it!


1 Answer 1


It is not the case that $x=vt$. The $x$ and $t$ in your expression for $u$ represent a position $x$ along the wave and a time $t$ of your choosing at which one might want to determine the amplitude of the wave. One can choose $t$ and $x$ completely independently of one another; there is no functional relationship between them, and in particular, they are not directly related to the wave propagation speed.

For example, you could ask "what is the amplitude of the wave at the origin at time zero?" This would correspond to determining the amplitude at $x=0$ and $t=0$, and in this case the expression $$ u = A\cos(\kappa x - \omega t) $$ would give $u=A$. On the other hand, you could ask "what is the amplitude of the wave at the origin but at time $t_0>0$?" This would correspond to $x=0$ and $t=t_0$ which would give amplitude $$ u = A\cos(\omega t_0) $$

  • $\begingroup$ Awesome! Thanks a lot for your very clear explanation. I really appreciate it. $\endgroup$
    – user12277
    Jun 28, 2013 at 19:04
  • $\begingroup$ @user12277 Sure thing. Have fun wavin' $\endgroup$ Jun 28, 2013 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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