0
$\begingroup$

The equation for stationary wave is :

$y(x,t)=A\sin(kx) \cos(\omega t)$. From this equation the amplitude of the stationary wave is said to be $A\sin(kx)$, in other words the displacement dependent function. So why can't it be the other way around. Like why can't the amplitude be $A\cos(\omega t)$ or in other words the time dependent function?

$\endgroup$
0
$\begingroup$

You could have $A\cos(\omega t) \sin(kx)$ and talk about the ‘amplitude’ of the time dependent function $\cos$ if you wanted to. It’s a matter of interpretation, the result is the same.

To see this, take out the time dependence (scale it to 1, the same as considering the wave at time $t=0,2\pi,4\pi,...$). Now you just have a wave in space.

As you change $A$ bigger or smaller, you make the wave in space taller or shorter. If you set $A$ negative, then the wave ‘inverts’ and what was positive becomes negative and vice versa.

Now imagine that you move $A$ back and forth, growing and shrinking your wave and inverting it, with a regular motion that can be described in time by $\cos(wt)$. This function is $A\cos(\omega t)$. So then we have added time dependence back in.

What the function $\cos(\omega t)$ does is it scales each point in space by a factor (between -1 and 1) that depends on time in a certain way. So each point in space ‘waves’ up and down between its maximum value and its minimum value.

If you thought of $A$ attached to $\cos$ rather than $\sin$, then the time dependent function scales each point in space between $-A$ and $A$. So the scaling with time becomes stronger if $A$ is large, and weaker if $A$ is small.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The time-dependent cosine makes oscillations in time, it is "a wave". What is its amplitude? It is a factor standing at the time dependent cosine. For each point in space the amplitude is different. "Hand waving" implies constant $x$ (you) and time-dependent hand motion (a wave).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I liked your example. However you gave an example of space dependent amplitude. So can you explain what will it be for a time dependent amplitude? $\endgroup$ – Aslan Jan 25 '19 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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