# Superposition of waves as a product of y(x) and y(t)

1. Feb 2, 2008

### DodongoBongo

1. The problem statement, all variables and given/known data
Learning Goal: To see how two traveling waves of the same frequency create a standing wave.
Consider a traveling wave described by the formula

$$y_1(x,t) = A \sin(k x - \omega t)$$.

This function might represent the lateral displacement of a string, a local electric field, the position of the surface of a body of water, or any of a number of other physical manifestations of waves.

The principle of superposition states that if two functions each separately satisfy the wave equation, then the sum (or difference) also satisfies the wave equation. This principle follows from the fact that every term in the wave equation is linear in the amplitude of the wave.

Consider the sum of two waves, where $$y_1(x,t)$$ is the wave described in Part A and $$y_2(x,t)$$ is the wave described in Part B. These waves have been chosen so that their sum can be written as follows:

$$y_{\rm s}(x,t) = y_{\rm e}(x) y_{\rm t}(t)$$.

This form is significant because $$y_e(x)$$, called the envelope, depends only on position, and $$y_t(t)$$ depends only on time. Traditionally, the time function is taken to be a trigonometric function with unit amplitude; that is, the overall amplitude of the wave is written as part of $$y_e(x)$$.

Find $$y_e(x)$$ and $$y_t(t)$$. Keep in mind that $$y_t(t)$$ should be a trigonometric function of unit amplitude.
Express your answers in terms of A, k, x, $$\omega$$, and t.

2. Relevant equations
$$y_1(x,t) = A \sin(k x - \omega t)$$

Part A: The wave is traveling in the +x direction.
Part B: $$A \sin (k x + \omega t)$$

3. The attempt at a solution
I tried using the trig identity $$sin(A-B) = sin(A)cos(B) - cos(A)sin(B)$$ to try to break up $$y_1(x,t) = A \sin(k x - \omega t)$$. I also know I need to find y(x) and y(t), so I tried solving for y(x,0) and y(0,t). So now I have:

$$y_1(x,t) = A \sin(k x - \omega t) = A(\sin(k x) \cos(\omega t) - \cos(k x) \sin(\omega t))$$

and Part A + Part B = $$A \sin(k x - \omega t) + A \sin(k x + \omega t)$$

and $$y(x,0) = A \sin(k x - \omega (0) ) = A \sin(k x)$$,
$$y(0,t) = A \sin(k (0) - \omega t) = A \sin(- \omega t)$$

I tried entering: $$2A\sin(k x), \sin(\omega t)$$, but it told me to "Check my trigonometry on term 2" . (It's MasteringPhysics)

I honestly don't know if what I've done so far is a step in the right direction or how to proceed to get y(x,t) = y(x)y(t). Can anyone let me know if I'm doing this right or if I've missed something obvious? I'm also not sure what they mean by "$$y_t(t)$$ should be a trigonometric function of unit amplitude".

I understand the Principle of Superposition; I know that wave functions are added together, I just don't understand how you can break them apart into a product like that.

Last edited: Feb 2, 2008
2. Feb 2, 2008

### DodongoBongo

Solution, but new problem

The last problem's answer was $$\cos(\omega t)$$ for the second part. I have no idea why.

Can someone please explain to me why this works?

3. Feb 2, 2008

### Niles

So we have to find A*sin(kx-wt) + A*(sin(kw+wt).

We get:

A*(sin(kx)*cos(wt) - cos(kx)*sin(wt)) + A*(sin(kx)*cos(wt) + cos(kx)*sin(wt)), which gives us

2*A*sin(kx)*cos(wt).

4. Feb 2, 2008

### DodongoBongo

the new problem

1. The problem statement, all variables and given/known data
Learning Goal: To see how two traveling waves of nearly the same frequency can create beats and to interpret the superposition as a "walking" wave.
Consider two similar traveling transverse waves, which might be traveling along a string for example:

$$y_1(x,t) = A \sin(k_1 x - \omega_1 t)$$ and $$y_2(x,t) = A \sin(k_2 x - \omega_2 t)$$.
They are similar because we assume that $$k_1$$ and $$k_2$$ are nearly equal and also that $$\omega_1$$ and $$\omega_2$$ are nearly equal.

The principle of superposition states that if two waves each separately satisfy the wave equation then the sum (or difference) also satisfies the wave equation. This follows from the fact that every term in the wave equation is linear in the amplitude of the wave.

Consider the sum of the two waves given in the introduction, that is,

$$y_{\rm sum}(x,t) = y_1(x,t) + y_2(x,t)$$.
These waves have been chosen so that their sum can be written as follows:

$$y_{\rm sum}(x,t) = C y_{\rm envelope}(x,t)\$$, $$y_{\rm carrier}(x,t)$$,
where $$C$$ is a constant, and the functions $$y_{\rm envelope}$$ and $$y_{\rm carrier}$$ are trigonometric functions of $$x$$ and $$t$$. This form is especially significant because the first function, called the envelope, is a slowly varying function of both position ($$x$$) and time ($$t$$), whereas the second varies rapidly with both position ($$x$$) and time ($$t$$). Traditionally, the overall amplitude is represented by the constant $$C$$, while the functions $$y_{\rm envelope}$$ and $$y_{\rm carrier}$$ are trigonometric functions with unit amplitude.

Find $$C$$, $$y_{\rm envelope}(x,t)$$, and $$y_{\rm carrier}(x,t)$$.
Express your answer in terms of $$A$$, $$k_1$$, $$k_2$$, $$x$$, $$t$$, $$\omega_1$$, and $$\omega_2$$. Separate the three terms with commas. Recall that $$y_{\rm envelope}$$ (the second term) varies slowly whereas $$y_{\rm carrier}$$ (the third term) varies quickly. Both $$y_{\rm envelope}$$ and $$y_{\rm carrier}$$ should be trigonometric functions of unit amplitude.

3. The attempt at a solution
Okay, I tried going about the problem the same way with the same trigonometric identity. I got the first term right (C = 2A) but MasteringPhysics seems to think that $$k_1$$ and $$k_2$$ are in both terms 2 and 3. This is what I have so far:

$$2A,{\sin}\left(k_{2}x-{\omega}_{2}t\right),{\sin}\left(k_{1}x-{\omega}_{1}t\right)$$

Error Message: Term 2: The correct answer involves the variable $$k_1$$, which was not part of your answer., Term 3: The correct answer involves the variable $$k_2$$, which was not part of your answer.

I really need help understanding how to go about this problem. Does anyone know how to approach this properly?

5. Feb 2, 2008

### Kurdt

Staff Emeritus
When you have added the two together you just have to use a series of trig identities to get the answer. Try adding the two functions first then applying the trig identity you listed above.

EDIT: I had to leave and come back but I see that in those ten minutes I was beaten to it.

6. Feb 2, 2008

### Kurdt

Staff Emeritus
The second problem is tackled in pretty much the same way as the first. Try it out and post your working.

7. Feb 2, 2008

### DodongoBongo

I haven't done trig in a while (I'm in computer science); but $$A \sin(k_1 x - \omega_1 t) + A \sin(k_2 x - \omega_2 t) = 2Asin(k_1 x - \omega_1 t + k_2 x - \omega_2 t)$$ doesn't seem right. In other words, I'm not sure how to add them up before using a trig identity.

I have worked out that $$A \sin(k_1 x - \omega_1 t) + A \sin(k_2 x - \omega_2 t) = A \sin(k_1 x) \cos(- \omega_1 t) - \cos(k_1 x) \sin(- \omega_1 t) + A \sin(k_2 x) \cos(- \omega_2 t) - \cos(k_2 x) \sin(- \omega_2 t)$$, though I'm not sure that really achieves anything.

Last edited: Feb 2, 2008
8. Feb 2, 2008

### Kurdt

Staff Emeritus
Can I clarify something from your second question? IS that the supposed to be the carrier function multiplied by the envelope? If so you're looking for a trig identity that will take a sum of trig functions and turn them into a product.

Have a look at these and see what you can do: http://en.wikipedia.org/wiki/List_of_trigonometric_identities

9. Feb 3, 2008

### DodongoBongo

I tried $$2A,{\sin}\left(\frac{k_{1}x-{\omega}_{1}t+k_{2}x-{\omega}_{2}t}{2}\right),{\cos}\left(\frac{k_{1}x-{\omega}_{1}t-k_{2}x-{\omega}_{2}t}{2}\right)$$, but MasteringPhysics said it was wrong. I am at a loss.

10. Feb 3, 2008

### Kurdt

Staff Emeritus
Can I ask again whether question two states its the carrier multiplied by the envelope? I'm not sure what that bracket and comma mean otherwise?

11. Feb 3, 2008

### DodongoBongo

It does, the "]" was a typo (sorry).

12. Feb 3, 2008

### Kurdt

Staff Emeritus
Ok well your cos function looks like its just a bit of confusion with signs. It should be + omega2 t.

13. Feb 3, 2008

### DodongoBongo

I tried that too, which it also said was wrong.

14. Feb 3, 2008

### Kurdt

Staff Emeritus
Are you putting the functions in the correct order?

15. Feb 3, 2008

### DodongoBongo

I think so, usually MasteringPhysics will tell you to "check your trig" if it sees the wrong trig function. When I did the actual trigonometry to get the terms I did this:

$$A \sin(k_1 x - \omega_1 t) + A \sin(k_2 x - \omega_2 t) = 2A{\sin}\left(\frac{k_{1}x-{\omega}_{1}t+k_{2}x-{\omega}_{2}t}{2}\right){\cos}\left(\frac{k_{1}x-{\omega}_{1}t-k_{2}x+{\omega}_{2}t}{2}\right)$$

MasteringPhysics wants me to break up the terms to get C, $$y_{\rm envelope}(x,t)$$, and $$y_{\rm carrier}(x,t)$$ of

$$y_{\rm sum}(x,t) = y_1(x,t) + y_2(x,t) = y_{\rm sum}(x,t) = C y_{\rm envelope}(x,t)\, y_{\rm carrier}(x,t)$$.

So far, it thinks C is the only part that's right.

16. Feb 3, 2008

### DodongoBongo

I also tried expanding the whole thing into sin(A+B) = sin(A)cos(B) + cos(A)sin(B) and the corresponding one for cos(A+B), but it told me to check my trig. This is due really soon, so help would be really appreciated.

term 1:$$2A$$

term 2:$${\sin}\left(\frac{k_{1}x-{\omega}_{1}t}{2}\right){\cos}\left(\frac{k_{2}x-{\omega}_{2}t}{2}\right)+{\cos}\left(\frac{k_{1}x-{\omega}_{1}t}{2}\right){\sin}\left(\frac{k_{2}x-{\omega}_{2}t}{2}\right)$$

term 3:$${\cos}\left(\frac{k_{1}x-{\omega}_{1}t}{2}\right){\cos}\left(\frac{k_{2}x-{\omega}_{2}t}{2}\right)+{\sin}\left(\frac{k_{1}x-{\omega}_{1}t}{2}\right){\sin}\left(\frac{k_{2}x-{\omega}_{2}t}{2}\right)$$

Last edited: Feb 3, 2008
17. Feb 3, 2008

### DodongoBongo

Apparently $$\cos(((k_1-k_2)*x-(\omega_1-\omega_2)*t)/2),\sin(((k_1+k_2)*x-(\omega_1+\omega_2)*t)/2)$$ is the answer.

I have no idea why or how they got there.

Last edited: Feb 3, 2008
18. Feb 4, 2008

### Kurdt

Staff Emeritus
Well you were almost there. You just needed to tidy up the term in the trig functions. Secondly I suggested that you should swap the sin and cos round earlier because looking at the question again it states a specific order for the entry of the answers. Specifically that the envelope function be input second which is always the function with the $\omega_2-\omega_1$ term.

19. Jan 28, 2009

### linie18

I'm stuck on this same problem. I solved to:
2A,\sin(((k_1+k_2)*x-(\omega_1+\omega_2)*t)/2),\cos(((k_1-k_2)*x-(\omega_1-\omega_2)*t)/2)

but it's saying "There is an error in your submission. Make sure you have formatted it properly."

Can someone help point out the error?

20. Feb 10, 2010

### caitlincui

If the two wave functions are
Y1(x1,t)=A*sin(k1*x-w1*t)
and Y2(x2,t)=A*sin(k2*x-w2*t)
The superposition of these two functions are Y1+Y2=A*sin(k1*x-w1*t)+A*sin(k2*x-w2*t)
according to basic trig equation: sin(a)+sin(b)=2(sin((a+b)/2)*cos((a-b)/2))
so the super position of these two functions became 2A*sin((k1+k2)*x-(w1+w2)*t))/2)*cos(((k1-k2)*x-(w1-w2)*t)/2)
So, since the envelope equation oscillate slower with position and time, we can see that cosine function depends less on time and position, thus, the cosine function is the function for Y envelope, and the sine function is the Ycarrier.
In all, the answer for C, Yenvelope,Ycarrier is 2A, cos(((k1-k2)*x-(w1-w2)*t)/2), sin((k1+k2)*x-(w1+w2)*t)/2)