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

[DodongoBongo]

Homework Statement

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.

Homework 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)

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.

 

[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?

 

[Niles]

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)

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).

 

[DodongoBongo]

the new problem

Homework Statement

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.

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?

 

[Kurdt]

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.

 

[Kurdt]

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

 

[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.

 

[Kurdt]

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

 

[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.

 

[Kurdt]

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?

 

[DodongoBongo]

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

 

[Kurdt]

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

 

[DodongoBongo]

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

 

[Kurdt]

Are you putting the functions in the correct order?

 

[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.

 

[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.

My last answer:

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)

 

[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.

 

[Kurdt]

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.

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.

 

[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?

 

[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 Y_carrier.
In all, the answer for C, Y_envelope,Y_carrier is 2A, cos(((k₁-k₂)*x-(w₁-w₂)*t)/2), sin((k₁+k₂)*x-(w₁+w₂)*t)/2)