# Homework Help: Wave Equation for a Progressive Wave

1. Jul 27, 2016

### Taniaz

1. The problem statement, all variables and given/known data
In some places, the wave equation is y=Asin(wt-kx) and in other places they have y = Asin(kx-wt) and they treat them as if they are equal. How are they equal? They also had y=-Asin(wt-kx). What is the difference between all 3?

2. Relevant equations
As above

3. The attempt at a solution

2. Jul 28, 2016

### andrewkirk

The third one is equal to the second one, because $\sin(-x)=-\sin x$.
The first one is almost the same as the other two, with the only difference being a phase shift of 180 degrees (half a cycle), because $\sin(x+180^\circ)=-\sin x$. In many calculations phase does not matter. Where it does, it would be important to pick one of the above and stick with it throughout the calculation.

3. Jul 28, 2016

### Taniaz

In the book I have, I was reading up on the equation of a standing wave with a reflection at a fixed end.
This is what they said: Consider a stretched rop fixed at O, and consider a point P, at a distance x from O.
Let O receive (from the right) an incident train of waves of equation: yo = A sin wt
Being a fixed end, O receives simultaneously a reflected wave (to the right) of the form: y'o= -A sin (wt)
which results in O having zero displacement at all times.
Point P receives reflected waves with a time lag. The equation of its displacement becomes: y1= - Asin(wt-kx)
Simultaneously, P receives the incident waves ahead of O. The equation is: y2= A sin(wt+kx)
The resultant displacement of P is y=y1+y2
So y=2Asin (kx) cos (wt)

In a video I saw, he was using the incoming wave as y1= A sin (kx-wt) and y2= A sin (wt+kx)
And the answer is obviously going to be the same.

What I didn't understand was in the book, they took the upside down reflection equation as - A sin(wt-kx). Does the minus sign in this case come from the negative amplitude of the upside down reflected wave? Why didn't the video take this into account?

I do know that A sin(wt-kx) is when the wave is travelling in the positive x-direction and A sin(wt+kx) is when the wave is travelling in the negative x-direction.

4. Jul 28, 2016

### pasmith

Sine is an odd function : $\sin (kx - \omega t) \equiv - \sin(\omega t - kx)$.

5. Jul 28, 2016

### Taniaz

Thanks pasmith. Could you please take a look at the post#3?

6. Jul 30, 2016

### Taniaz

n the book I have, I was reading up on the equation of a standing wave with a reflection at a fixed end.
This is what they said: Consider a stretched rop fixed at O, and consider a point P, at a distance x from O.
Let O receive (from the right) an incident train of waves of equation: yo = A sin wt
Being a fixed end, O receives simultaneously a reflected wave (to the right) of the form: y'o= -A sin (wt)
which results in O having zero displacement at all times.
Point P receives reflected waves with a time lag. The equation of its displacement becomes: y1= - Asin(wt-kx)
Simultaneously, P receives the incident waves ahead of O. The equation is: y2= A sin(wt+kx)
The resultant displacement of P is y=y1+y2
So y=2Asin (kx) cos (wt)

In a video I saw, he was using the incoming wave as y1= A sin (kx-wt) and y2= A sin (wt+kx)
And the answer is obviously going to be the same.

What I didn't understand was in the book, they took the upside down reflection equation as - A sin(wt-kx). Does the minus sign in this case come from the negative amplitude of the upside down reflected wave? Why didn't the video take this into account?

I do know that A sin(wt-kx) is when the wave is travelling in the positive x-direction and A sin(wt+kx) is when the wave is travelling in the negative x-direction.

7. Jul 31, 2016

### mukundpa

I think a lot of confusion is there. I simply understand it by putting x = 0 and t = 0 as well as I think the phase decreases in the direction of wave motion. Thus

y = A sin (wt - kx) wave moving in positive x direction and initially wave displacement (y) at origin is zero and increasing in positive y direction with time.
y = A sin (wt + kx) wave moving in negative x direction and initially wave displacement (y) at origin is zero and increasing in positive y direction with time.
y = A sin (kx - wt) wave moving in negative x direction and initially wave displacement (y) at origin is zero and increasing in negative y direction with time.

8. Jul 31, 2016

### conscience

Wrong .

Wave represented by y = A sin (kx - wt) moves in positive x direction .

9. Jul 31, 2016

### mukundpa

May be, but I think at a particular moment (t = constant) if we increase x the phase angle increases and as per I think with at a moment of time phase angle decrease in the direction of wave motion. Please let me now how you thinks that it is moving in positive x direction. .

10. Jul 31, 2016

### conscience

Sorry I do not understand your point .

Here is some food for thought for you . If you believe y = A sin ( wt - kx ) travels in positive x-direction , then what makes you think that y = A sin (kx - wt) moves in negative x-direction .

What is the difference between y = A sin ( wt - kx ) and y = A sin (kx - wt) ?

11. Jul 31, 2016

### mukundpa

Thanks but where is the food? I need it. I do not believe, I try to make logic and try to understand. In first equation x is negative and with increase in x the phase angle (wt - kx) decrease while in second x is positive and with increase in x phase angle (kx - wt) increases (considering snapshot at time t)

12. Jul 31, 2016

### Taniaz

y=Asin (kx +wt) or A sin (wt+kx) travels in the negative directiion and
y=A sin(kx-wt) or y = A sin(wt-kx) travel in the positive x direction

13. Jul 31, 2016

### conscience

Correct .

14. Jul 31, 2016

### conscience

When you take a snapshot at time t , you basically fix 't' . And as soon as you do that the wave is not moving anymore . Now , you cannot determine whether the wave is moving in positive or negative direction . Both waveforms look same .

Last edited: Jul 31, 2016
15. Aug 1, 2016

### mukundpa

16. Aug 1, 2016

### Taniaz

"If it is cosine A(wt-kx), a crest is at wt-kx=0. This means that at t= 0 the crest is at x=0 and at a later time t, it is at x= k/w*t. This is the same equation that holds for a body moving along the x axis with uniform velocity v=k/w in the positive direction (from left to right)."

I found this on another post by a scientific adviser on the forum. I don't understand the part where he says at a later time t, x =k/w *t
how did he get this??

17. Aug 1, 2016

### Taniaz

I get it he wrote v as k/w when it should be w/k then it follows because w=2pi (f) and
k =2pi/lambda so v= 2pi f / 2pi/lambda which gives v= f (lambda) so yeah.