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.

 

[morrobay]

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!

 

[morrobay]

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.