What exactly does 'x-vt' mean in the wave equation?

[Jayxl]

So the equation is
f(x,t) = f(x-vt,0) = A sine k(x-vt).

Now Since x is displacement of wave from origin, t is time taken for that,wouldn't 'x-vt' always be equal to 0 ?
If x-vt is 0,
Y = A sine k(x-vt) will also be zero.
So am I writing the equation wrong?

 

[davenn]

Hi there Jayxl
welcome to PF

Now Since x is displacement of wave from origin, t is time taken for that,wouldn't 'x-vt' always be equal to 0 ?

no, as you said x is the displacement = velocity "v" times t (time) so it can only be 0 if there is no wave propagation

the - sign represents a forward ( rightward ) wave propagation of the wave (x-vt)
if a + sign then its a leftward ( backwards) propagating wave (x+vt)Dave

 

[DrDu]

Now Since x is displacement of wave from origin, t is time taken for that,wouldn't 'x-vt' always be equal to 0 ?

What kind of waves are you considering?
x isn't the displacement of the wave whatever, but the point in space whence the displacement is measured.

 

[davenn]

I could get out of my depth very quickly here
not really my field of interest ... am not familiar with the 0 in the f(x-vt,0), am wondering if that defines the
displacement at the origin time which really means f(x-vt,0) = f(x) ?

hope some one more knowledgeable chimes in :)Dave

 

[davenn]

x isn't the displacement of the wave whatever, but the point in space whence the displacement is measured.

which is the same as the displacement from origin @ time t, when velocity =v

you can't measure the displacement of a point without referring to where it has been displaced from

 

[DrDu]

That's why I was asking about which waves we are talking. For example, I could consider transversal waves on a string. Then if the extension of the string is in x direction and the tranversal displacement is in y direction then f=y (if the undisplaced string is at y=0).

 

[davenn]

That's why I was asking about which waves we are talking. For example, I could consider transversal waves on a string.

ahh ok

I may be wrong ( as said, not my forte) I was understanding that this was for longitudinal waves traveling through/along a medium
ie, along your guitar string, not the traverse motion if the string ... can it be used to describe both ?D

 

[DrDu]

Of course it can. But in general, x refers to the undisplaced position $x=x_\mathrm{undisplaced}$ and f would be $f=x_\mathrm{displaced}-x_\mathrm{undisplaced}$.

 

[davenn]

But in general, x refers to the undisplaced position x=x undisplaced x=x_\mathrm{undisplaced} and f would be f=x displaced −x undisplaced f=x_\mathrm{displaced}-x_\mathrm{undisplaced}.

OK so how does that fit in with

To displace any function f(x) to the right, just change its
argument from x to x-x₀, where x₀ is a positive number.

If we let x₀ = v t, where v is positive and t is time, then the displacement
increases with increasing time

am always willing to learn

D

 

[willem2]

So the equation is
f(x,t) = f(x-vt,0) = A sine k(x-vt).

Now Since x is displacement of wave from origin, t is time taken for that,wouldn't 'x-vt' always be equal to 0 ?
If x-vt is 0,
Y = A sine k(x-vt) will also be zero.
So am I writing the equation wrong?

x is not the displacement of the wave from the origin. x is just the x coordinate of the point where we are measuring the displacement of the waving medium from its normal value. This could be a displacement in the same direction of the x-coordinate (longitudal waves), or in a different direction (transverse waves), or even a pressure (sound waves).

The solution of the wave equation f(x,t) = Asin (k (x-vt)) gives you the values of the displacement for all values of x and t, so saying x-vt has to be 0 makes no sense

 

[Jack Davies]

For a given position $x$ and time $t$, you evaluate the function $f(x,t)$ at those values to determine the height of the wave.

This is different from $v$, the speed of propagation of the wave. Consider yourself moving to the right with displacement defined by $x(t)=v' t +x_0$. Then, plugging this into $f(x,t)$ you can see that $f(x(t),t)$ is constant in time if and only if $v' = v$. Thus we conclude that the speed of the wave is indeed $v$.

 

[atyy]

So the equation is
f(x,t) = f(x-vt,0) = A sine k(x-vt).

Now Since x is displacement of wave from origin, t is time taken for that,wouldn't 'x-vt' always be equal to 0 ?
If x-vt is 0,
Y = A sine k(x-vt) will also be zero.
So am I writing the equation wrong?

f(x,t) means f is a function of x and t, eg. x²t³.

However, that is too general for a solution to the wave equation, which is why you wrote f(x,t) = f(x-vt,0), which means that f could be any function of x-vt, eg. (x-vt)³ + exp(x-vt).

If you have a function f(x), then f(x+a) is the shape shifted to the left by a.

So f(x-vt,0) mean that as t increases, vt increases, and the shape shifts to the right by vt. In other words the shape travels to the right as time increases.

 

[jtbell]

am not familiar with the 0 in the f(x-vt,0),

I've never seen the general function written that way, either. I've always seen it as just f(x-vt). Perhaps the OP could give us a link to where he saw it, or maybe someone else recognizes it.

 

[Point Conception]

With y = A sin (kx - ωt) the calculator should be on radians for sin , correct ?
I just made up some numerical values for : A =1. λ =3m. k = 2.09 rad/unit distance , frequency = 4 cy/sec , v = 12m/s . ω = 25.12 rad/s
At t₀
x₀ = 10m
At t₁ = 4 sec
x₁ = 58 m
So y = Asin(kx-ωt) = y = sin (121.2-100.4)
With sin in radians = .955
With sin in degrees = .355

 

[jtbell]

y=f(x-vt) simply means that x and t appear only in the combination x-vt, possibly with other constants "surrounding" that combination (loosely speaking). Here's a simple example: y=(x-2t)². To see that 2 is really a velocity, set t to a couple of different values, and make graphs of y versus x for each of them.

Setting t=0 gives y=x². The graph is a parabola with its vertex at the origin (x=0).

Setting t=1 gives y=(x-2)². The graph is a parabola with the same shape as before, but with its vertex on the x-axis at x=2.

Setting t=2 gives y=(x-4)². The graph is a parabola with the same shape as before, but with its vertex on the x-axis at x=4.

Clearly y=(x-2t)² represents a parabola moving in the +x direction with speed v=2.

What about $y=A \sin (kx-\omega t)$, a common description of a sinusoidal wave? Rewrite it a bit: $$y=A \sin \left[ k \left( x-\frac{\omega}{k}t \right) \right] \\
y = A \sin [k(x-vt)]$$ where $v=\omega/k$. So this is also in the form y=f(x-vt).

Now put in some numbers. Let A=1, $k=\pi/5$ and $\omega=0.4\pi$, so $$y = \sin \left( \frac{\pi}{5} x - 0.4 \pi t \right)$$ As we did for the parabola above, make three graphs of y versus x, setting t=0, t=1 and t=2 respectively. You should get three sinusoidal graphs, with wavelength $\lambda = 2\pi/k = 10$, shifted from one to the next in the +x direction by v = ω/k = 2 units.

 

[robphy]

Amplifying jtbell's comments...

Check out https://www.desmos.com/calculator/bjt6dleg5h and the screenshot below.

Let u=x-vt, with v>0.
When t=0, then u=x (for all x).
Consider the snapshot of a disturbance in space at t=0 (the dotted graph).
At a later time t=1, that disturbance has advanced to the right by a displacement vt.
(I tried to show this on a faux 3-d plot. Imagine a stack of xy-planes for each time t, with the disturbance function drawn on it...each plane drawn slightly displaced to minimize overlap.)

Pick any value of U (say U=0)... and consider the value of the function at f(0), call it A.
So, A is the disturbance at u=0 (that is when x-vt=0 or when x=vt).
At time t=0, A is the disturbance at x=0 (since u=0).
At time t=1, A is the disturbance at x=v*1 (since u=0).
At time t=2, A is the disturbance at x=v*2 (since u=0).
That is to say, the "disturbance at x=0 when t=0" is advancing to the right with velocity v.

Now pick another value of U (say U=0.1)...
and consider the value of the function at f(0.1), call it B.
So, B is the disturbance at u=0.1 (that is when x-vt=0.1 or when x=0.1+vt).
At time t=0, B is the disturbance at x=0.1 (since u=0.1).
At time t=1, B is the disturbance at x=0.1+v*1 (since u=0.1).
At time t=2, B is the disturbance at x=0.1+v*2 (since u=0.1).
That is to say, the "disturbance at x=0.1 when t=0" is advancing to the right with velocity v.

And so on for each value of U that you choose (that is, for each location x at time t=0).

Thus, with v>0,
"f(x-vt)" describes the disturbance pattern in space (i.e. at locations x) moving to the right (as t increases, x must increase to keep x-vt=constant).
and
"f(x+vt)" describes the disturbance pattern in space (i.e. at locations x) moving to the left (as t increases, x must decrease to keep x+vt=constant).

https://www.desmos.com/calculator/bjt6dleg5h

 

[DrDu]

Quite interesting discussion. A pitty that the OP didn't show up again!

 

[Point Conception]

Quite interesting discussion. A pitty that the OP didn't show up again!

I agree and together with : https://www.physicsforums.com/threads/wave-number-k-2pi-lambda-radians-unit-distance.836845/
I now understand the concept of the wavenumber k, 2π/λ
One cycle of a wavelength is equal to an angular displacement of 2π radians. So if x = 6m and λ = 2m then k x = 6π radians.