Simple wave questions- am I right?

  • Thread starter Thread starter bcjochim07
  • Start date Start date
  • Tags Tags
    Wave
bcjochim07
Messages
366
Reaction score
0

Homework Statement


Consider the sum of two waves , y1(x,t) + y2(x,t) where y1(x,t) is the wave described by y1(x,t) = Asin(kx-[tex]\omega[/tex]t) and y2 is a wave of the same amplitude moving in the opposite direction. These waves have been chosen so that their sum can be written as follows: Ys= Ye(x)*Yt(t).

This form is significant because Ye , called the envelope, depends only on position, and Yt 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 Ye.

1) At the position x=0 what is the string's displacement (assuming the standing wave Ys is present) ? Express your answer in terms of the parameters given in the problem introduction.

2) At some times, the string will be perfectly straight. Find the first time t1 when this is true. Express t1 in terms of [tex]\omega[/tex], k, and necessary constants.




Homework Equations





The Attempt at a Solution



1) So for wave 1 the displacement eqn. is y1= Asin(kx-[tex]\omega[/tex]t) & for wave 2 it is y2=Asin(kx+[tex]\omega[/tex]t). When I add them together, I get:
Ys= Asin(kx-[tex]\omega[/tex]t) + Asin(kx+[tex]\omega[/tex]t)

and using a trig identity,
I get Ys= 2Asin(kx)*cos([tex]\omega[/tex]t)

so my question is: For part 1 do I just plug in x=0 to the x dependent part of the eqn? If so I get the displacement is 0.

For part 2 I set up this equation using the time dependent part of the equation:

0=cos[tex]\omega[/tex]t [tex]\pi[/tex]/2 = [tex]\omega[/tex]t and I got t1= [tex]\pi[/tex]/(2[tex]\omega[/tex])

Am I interpreting the equation correctly?

Thanks
 
Last edited:
bcjochim07 said:
1) So for wave 1 the displacement eqn. is y1= Asin(kx-[tex]\omega[/tex]t) & for wave 2 it is y2=Asin(kx+[tex]\omega[/tex]t). When I add them together, I get:
Ys= Asin(kx-[tex]\omega[/tex]t) + Asin(kx+[tex]\omega[/tex]t)

and using a trig identity,
I get Ys= 2Asin(kx)*cos([tex]\omega[/tex]t)

so my question is: For part 1 do I just plug in x=0 to the x dependent part of the eqn? If so I get the displacement is 0.
Looks good to me.
bcjochim07 said:
For part 2 I set up this equation using the time dependent part of the equation:

0=cos[tex]\omega[/tex]t [tex]\pi[/tex]/2 = [tex]\omega[/tex]t and I got t1= [tex]\pi[/tex]/(2[tex]\omega[/tex])

Am I interpreting the equation correctly?

Thanks
Again, looks good to me.
 
bcjochim07 said:

Homework Statement


Consider the sum of two waves , y1(x,t) + y2(x,t) where y1(x,t) is the wave described by y1(x,t) = Asin(kx-[tex]\omega[/tex]t) and y2 is a wave of the same amplitude moving in the opposite direction. These waves have been chosen so that their sum can be written as follows: Ys= Ye(x)*Yt(t).

[snip]

... for wave 2 it is y2=Asin(kx+[tex]\omega[/tex]t).

There is another function that also fits the condition "y2 is a wave of the same amplitude moving in the opposite direction".

Is there any more information that you have perhaps omitted?
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
27
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K