How Do Different Equations Affect the Initial Direction of a Traveling Wave?

[songoku]

Homework Statement

This is not homework

I want to ask about equation of traveling wave. I read in my book the derivation of equation of traveling wave. The explanation starts from equation $y=A \sin (kx)$ then change into $y=A \sin (kx - \omega t)$ as the wave moves to the right. If the wave moves to the left, the term inside the sin would be $(kx + \omega t)$

Relevant Equations

$y=A \sin (kx + \omega t)$

$y=A \sin (-kx - \omega t)$

But in the notes from teacher, the equation is $y=A \sin (kx - \omega t)$ for wave traveling to the right and $y=A \sin (-kx - \omega t)$ for wave traveling to the left

When I transform the equation of the wave traveling to the left using trigonometry:
$$y=A \sin (-kx - \omega t)$$
$$y=-A \sin (kx + \omega t)$$

The issue is in the practice question, there is question asking the initial direction of movement of the wave, will it move upwards or downwards. Equation $y=A \sin (kx + \omega t)$ states that the wave initially moves upwards but $y=-A \sin (kx + \omega t)$ states the wave moves downwards initially. It means that the answer to the question will differ, depends on which equation I use.

In this youtube video,
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There is another equation for wave traveling to the right, which is $y=A \sin (\omega t - kx)$ which is equal to $y=-A \sin (kx - \omega t)$. This means that the initial vertical movement for wave traveling to the right will differ from two previous equations.

How to address the difference? Thanks

 

[Frabjous]

A traveling sine wave should always look “identical”, just shifted in time and space. The value of the amplitude is irrelevant.
Look at sin(f(x,t)). It is “identical” for f(x,t)=constant.
For kx=±constant+ωt, as t increases, x increases⇒right going
For kx=±constant-ωt, as t increases, x decreases⇒left going

 

[haruspex]

The issue is in the practice question, there is question asking the initial direction of movement of the wave, will it move upwards or downwards.

How exactly does the question specify the wave? If it only tells you which way it is moving as a whole in respect of left/right you cannot tell whether it is moving up or down at the origin at t=0.

A traveling sine wave should always look “identical”, just shifted in time and space. The value of the amplitude is irrelevant.
Look at sin(f(x,t)). It is “identical” for f(x,t)=constant.
For kx=±constant+ωt, as t increases, x increases⇒right going
For kx=±constant-ωt, as t increases, x decreases⇒left going

I don't think that's what you meant. If f(x,t)=constant then sin(f(x,t)) is a horizontal line. And kx=±constant+ωt doesn't work either, since those are necessarily independent variables.
How about, given sin(kx+ωt),
ω/k <0 moves right,
ω/k>0 moves left
ω/k=0 is static?

But none of that relates the question being asked in post #1.

 

[Frabjous]

How exactly does the question specify the wave? If it only tells you which way it is moving as a whole in respect of left/right you cannot tell whether it is moving up or down at the origin at t=0.

I don't think that's what you meant. If f(x,t)=constant then sin(f(x,t)) is a horizontal line. And kx=±constant+ωt doesn't work either, since those are necessarily independent variables.
How about, given sin(kx+ωt),
ω/k <0 moves right,
ω/k>0 moves left
ω/k=0 is static?

But none of that relates the question being asked in post #1.

I guess that I didn’t explain it clearly. Take a feature at time t₀; there is an x₀ associated with it. We now have an f(x₀,t₀) for that feature. At a later time, t₁, the feature is at x₁ at which point you can use the equations that I wrote to determine x₁ and the direction of travel.

 

[haruspex]

I guess that I didn’t explain it clearly. Take a feature at t₀; there is an x₀ associated with it. We now have an f(x₀,t₀) for that feature. At a later time, t₁ the feature is at x₁ at which point you can use the equations that I wrote to determine x₁ amd the direction of travel.

So you did not intend specifically a sine function, you meant it to be general.
If f(x,t) is a traveling wave of some arbitrary shape described by y=g(x) and its horizontal motion is given by h(t) then f=g(x-h(t)).

 

[Frabjous]

Ok, let me try to clean my comments up while sitting at a computer. Hopefully, third time is a charm. My apologies.

We have f(kx±ωt)=f(φ)
For a traveling wave, we are interested in the location x_a of a specific φ₀ at a time, t_a.
So we set kx_a±ωt_a=φ₀
So kx_a=φ₀$\mp$ωt_a (1)

What happens to x_a as t_a increases for a specific φ₀ in (1)?

For f(kx-ωt), x_a increases (moves to the right); i.e., a right traveling wave.

For f(kx+ωt), x_a decreases (moves to the left); i.e., a left traveling wave.

 

[Frabjous]

Also the A is arbitrary in Asin(φ). You can write Asin(-φ)=-Asin(φ)=Bsin(φ) where B is arbitrary. Once you actually fit the parameter, things are determined uniquely.

 

[haruspex]

Ok, let me try to clean my comments up while sitting at a computer. Hopefully, third time is a charm. My apologies.

We have f(kx±ωt)=f(φ)
For a traveling wave, we are interested in the location x_a of a specific φ₀ at a time, t_a.
So we set kx_a±ωt_a=φ₀
So kx_a=φ₀$\mp$ωt_a (1)

What happens to x_a as t_a increases for a specific φ₀ in (1)?

For f(kx-ωt), x_a increases (moves to the right); i.e., a right traveling wave.

For f(kx+ωt), x_a decreases (moves to the left); i.e., a left traveling wave.

Yes, I think that matches post #5, except that I allowed a generalised motion, not a constant velocity.
Anyway, as I wrote, this isn't relevant to the question in post #1. @songoku is concerned with the sign of $\dot y$ at (0, 0). The two forms of the basic equation cited effectively differ in terms of the sign of the amplitude, or, equivalently, a half wavelength shift in phase. These forms will yield opposite answers to the question posed.
I must say the usual form I'm aware of is $A\sin(kx\pm \omega t)$, where all constants are assumed positive - or the same with an arbitrary phase shift, or with cos instead of sin. I'd never seen it given as $A\sin(\omega t\pm kx)$.

 

[Frabjous]

Yes, I think that matches post #5, except that I allowed a generalised motion, not a constant velocity.
Anyway, as I wrote, this isn't relevant to the question in post #1. @songoku is concerned with the sign of $\dot y$ at (0, 0). The two forms of the basic equation cited effectively differ in terms of the sign of the amplitude, or, equivalently, a half wavelength shift in phase. These forms will yield opposite answers to the question posed.
I must say the usual form I'm aware of is $A\sin(kx\pm \omega t)$, where all constants are assumed positive - or the same with an arbitrary phase shift, or with cos instead of sin. I'd never seen it given as $A\sin(\omega t\pm kx)$.

They don’t give different answers. You pick up two negative signs which cancel. One from -A and one from -Φ when you take the derivative using the chain rule.

In your equations, the two A’s are not identical for a specific wave.

 

[haruspex]

They don’t give different answers. You pick up two negative signs which cancel. One from -A and one from -Φ when you take the derivative using the chain rule.

With $y=A\sin(kx-\omega t)$, $\dot y=-A\omega\cos(kx-\omega t)$. The sign of $\dot y$ at (0,0) depends on $A\omega$. Change the sign of one only and the sign of $\dot y$ changes.
The two forms @songoku quotes effectively have opposite signs for A but the same $\omega$, or you could view them as the same sign for A and opposite signs for both k and $\omega$. Either way, they yield opposite signs for $A\omega$.

 

[Frabjous]

You need to use the value kx=π/2 to fit A. Everything is uniquely defined. The sign of A is determined by whether you use φ or -φ. You cannot arbitrarily choose it.

 

[haruspex]

You need to use the value kx=π/2 to fit A. Everything is uniquely defined. The sign of A is determined by whether you use φ or -φ. You cannot arbitrarily choose it.

I would say A is, by convention, nearly always taken to be positive (likewise k and $\omega$), but I was not suggesting flipping the sign of A. My observation was that the difference between the $A\sin(kx-\omega t)$ and $A\sin(\omega t-kx)$ forms is equivalent to flipping the sign of A and hence leads to opposite signs of $\dot y(0,0)$.

 

[Frabjous]

I would say A is, by convention, nearly always taken to be positive, but I was not suggesting flipping the sign of A. My observation was that the difference between the $A\sin(kx-\omega t)$ and $A\sin(\omega t-kx)$ forms is equivalent to flipping the sign of A and hence leads to opposite signs of $\dot y(0,0)$.

A is arbitrary. Changing the functional form will change it’s value.
Without loss of generality we can write
Asin(kx-ωt) and Bsin(ωt-kx)
let t = 0.
assume the maximum occurs at x=π/2k
let the maximum value of the wave be C(a positive number)
then
Asin(π/2)=A=C
Bsin(-π/2)=-B=C; B=-C
You are not free to arbitrarily choose both signs. The data have a say in the matter.

 

[haruspex]

A is arbitrary. Changing the functional form will change it’s value.
Without loss of generality we can write
Asin(kx-ωt) and Bsin(ωt-kx)
let t = 0.
assume the maximum occurs at x=π/2k
let the maximum value of the wave be C(a positive number)
then
Asin(π/2)=A=C
Bsin(-π/2)=-B=C; B=-C
You are not free to arbitrarily choose both signs. The data have a say in the matter.

I think all we are arguing about is how to express the error in (or misleading nature of) the video.
If A is positive in the video the equation adduced does not much the picture at the start. There is a half wavelength phase shift.
Your view, then, is that to match the picture the video effectively makes A negative, so $\dot y(0,0)=A\omega$ is negative.

 

[songoku]

How exactly does the question specify the wave? If it only tells you which way it is moving as a whole in respect of left/right you cannot tell whether it is moving up or down at the origin at t=0.

The question gives equation of traveling wave, something like $y=3 \sin (-4x-5t)$ then asks about the amplitude, frequency, wavelength, speed, direction of travel of wave (left or right), direction of movement of wave initially (up or down)

 

[haruspex]

The question gives equation of traveling wave, something like $y=3 \sin (-4x-5t)$ then asks about ... direction of movement of wave initially (up or down)

Direction of movement where, at the origin? Then it's simple. Evaluate $\dot y(0,0)$.

 

[Frabjous]

Direction of movement where, at the origin? Then it's simple. Evaluate $\dot y(0,0)$.

You can also think about the answer. You have a left going wave. Look at the value of x that is a little bit greater than zero. That point will soon be at 0.

 

[songoku]

Direction of movement where, at the origin? Then it's simple. Evaluate $\dot y(0,0)$.

The question does not specify but I think it is at origin

You can also think about the answer. You have a left going wave. Look at the value of x that is a little bit greater than zero. That point will soon be at 0.

You mean I put t = 0 and then some random value of x close to zero (e.g. 0.01 m) to the equation to get the sign of y? If y is positive, then the wave travels down. If y negative then the wave travels up?

Thanks

 

[hutchphd]

For a transverse wave yes

Direction of movement where, at the origin? Then it's simple. Evaluate y˙(0,0).

Why limit yourself to the origin? It will also be true at any x evaluate $$\dot f (x,t)\left. \right |_{t=0}$$

 

[Frabjous]

You mean I put t = 0 and then some random value of x close to zero (e.g. 0.01 m) to the equation to get the sign of y? If y is positive, then the wave travels down. If y negative then the wave travels up?

You have a left going wave. The wave to the right of you is positive negative. This means …

 

[songoku]

You have a left going wave. The wave to the right of you is positive. This means …

Sorry I don't understand your hint

 

[haruspex]

For a transverse wave yes

Why limit yourself to the origin? It will also be true at any x evaluate $$\dot f (x,t)\left. \right |_{t=0}$$

At any time, parts of the wave will be moving one way in the transverse direction, parts the other.

 

[haruspex]

The wave to the right of you is positive.

Not according to post #15.

 

[Frabjous]

Not according to post #15.

Corrected.

 

[Frabjous]

Sorry I don't understand your hint

A wave is coming towards a point. What does that mean for the point?

 

[songoku]

A wave is coming towards a point. What does that mean for the point?

The point will be disturbed so it will have same displacement as the wave

 

[Frabjous]

Yes.

 

[songoku]

Thank you very much for the help and explanation caz, haruspex, hutchphd