How is x-vt a rightward moving wave?

[Jaccobtw]

Homework Statement

How is x-vt a rightward moving wave?

Relevant Equations

f(x-vt) = disturbance

The only way I can see x-vt being a rightward moving wave is if x-vt = some position the wave had initially. As t increases, x-vt gets smaller, despite the fact that it is a rightward moving wave. For example, if x = 10 m and v = 1 m/s, as t increases, x-vt describes a position the wave had in the past. Is this correct intuition?

 

[PeroK]

It's a rightward moving wave for $v > 0$. At $t =0$, the wave takes the form of $y = f(x)$. Choose a function $f$. Something simple. For example $f(x) = x$. Draw that initial state of the wave on a graph.

Then, for $t = 1$, say, the wave takes the form $y = f(x -v) = x -v$. Draw that state of the wave on a graph. Take $v = 1$ as well, so that $y = x - 1$.

Has the wave moved left or right?

 

[kuruman]

Another way of looking at this is from the standpoint of the wave.

For positive $x$ at a particular instant in time $t$ and at position $x$, the function $f(x-vt)$ return a specific value, say $3$ units. Time always increases. At a later time $t'>t$ must $x$ increase, decrease or stay the same for the function to return the same value of $3$ units?

 

[Jaccobtw]

It's a rightward moving wave for $v > 0$. At $t =0$, the wave takes the form of $y = f(x)$. Choose a function $f$. Something simple. For example $f(x) = x$. Draw that initial state of the wave on a graph.

Then, for $t = 1$, say, the wave takes the form $y = f(x -v) = x -v$. Draw that state of the wave on a graph. Take $v = 1$ as well, so that $y = x - 1$.

Has the wave moved left or right?

f(x) = 10 for t = 0

For t = 1, and v = 1

10-(1)(1) =9

We began at 10 and now we’re at 9, so hasn’t the wave moved to the left?

 

[PeroK]

f(x) = 10 for t = 0

For t = 1, and v = 1

10-(1)(1) =9

We began at 10 and now we’re at 9, so hasn’t the wave moved to the left?

That's not two graphs! This is what the graphs of $y = x$ and $y = x-1$ look like.

 

[PeroK]

PS I've shown two points on the wave (P and Q) at times 0 and 1.

 

[jtbell]

Another example: consider the function $f(x)=x^2$, whose graph is a parabola opening upwards, with its vertex at $(x,y)=(0,0)$.

For $v=+1$, $f(x-vt)=(x-t)^2$.

At $t=0$ we have the original function.

At $t=1$, $f(x-vt)=(x-1)^2$. It's still a parabola. Where is the vertex?

At $t=2$, $f(x-vt)=(x-2)^2$. It's still a parabola. Where is the vertex?

 

[PeroK]

Note that, in general, if $a > 0$ and $g(x) = f(x-a)$, then $g(x)$ is $f(x)$ translated to the right. One way to see this is to consider a point $x_0$ where $f(x_0) = 0$. The corresponding zero for the function $g(x)$ is at $x_1 = x_0 + a$. As $g(x_1) = f(x_1 - a) = f(x_0 + a - a) = f(x_0) = 0$.

From this, we can see that the zeroes of the function $g(x)$ are the zeroes of the function $f(x)$ translated to the right by $a$. And, everything else is translated to the right as well,

 

[Jaccobtw]

PS I've shown two points on the wave (P and Q) at times 0 and 1.

That's not two graphs! This is what the graphs of $y = x$ and $y = x-1$ look like.

Thank you, I think I’ve got it down. (x, f(x)) is where we want to find the disturbance. x-vt is the coordinate in the moving reference frame. It says, what is the disturbance at (x, f(x)) when the wave has moved a distance vt.

 

[PeroK]

Thank you, I think I’ve got it down. (x, f(x)) is where we want to find the disturbance. x-vt is the coordinate in the moving reference frame. It says, what is the disturbance at (x, f(x)) when the wave has moved a distance vt.

Perhaps, but $y = f(x - vt)$ is a function of two variables. You need either an animation, or to have an additional $t$ axis. Perhaps with $t$ going into the page. It's not enough, IMO, to try to think of a wave as a static function in 2D.

 

[Orodruin]

I’ll try to offer yet another perspective. The function $f$ describes the shape of the wave. We can consider a point on the wave with a particular constant value of $f$, call it $f_0$. Now consider how the point (or more generally, points) with this value are moving with time. We have $f(x-vt) = f_0$ so - assuming $f$ is locally invertible - we must have $x-vt = c_0$ for some constant $c_0$ such that $f(c_0) = f_0$. Solving for $x$ we find $$x= vt + c_0.$$
Therefore, the $x$ value for which the wave takes the value $f_0$ increases (assuming $v > 0$) with time. The wave moves to the right.

 

[Steve4Physics]

This somewhat repeats what has already been said, but there is an important underlying point that's maybe worth making explicitly...

Suppose you have any function $y=f(x)$ and some (say positive) constant $a$.

The graphs of $y=f(x)$ and $y = f(x-a)$ are the same shape - but the graph of $y=f(x-a)$ is shifted in the +ve x direction by an amount $a$.

(Correspondingly, the graphs of $y=f(x)$ and $y = f(x+a)$ are the same shape - but the graph of $y=f(x+a)$ is shifted in the -ve x direction by an amount $a$.)

For example:

If the ‘constant’ $a$ increases uniformly over time, this means the graph steadily shifts in the +ve x direction.

 

[Jaccobtw]

Perhaps, but $y = f(x - vt)$ is a function of two variables. You need either an animation, or to have an additional $t$ axis. Perhaps with $t$ going into the page. It's not enough, IMO, to try to think of a wave as a static function in 2D.

What if I phrase it like this: We want to find the disturbance at $(x, f(x))$ at time $t$ when the wave travels at velocity $v$. Subtraction shifts the graph to the right. $x - vt$ is smaller than $x$ because it gives the coordinate where we want to find the disturbance in the moving reference frame.

 

[PeroK]

What if I phrase it like this: We want to find the disturbance at $(x, f(x))$ at time $t$ when the wave travels at velocity $v$. Subtraction shifts the graph to the right. $x - vt$ is smaller than $x$ because it gives the coordinate where we want to find the disturbance in the moving reference frame.

Or, let the mathematics speak for itself. See post #12 by Steve.