Simple wave questions- am I right?

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SUMMARY

The discussion focuses on the mathematical analysis of two waves, y1(x,t) and y2(x,t), represented by the equations y1(x,t) = Asin(kx - ωt) and y2(x,t) = Asin(kx + ωt). The sum of these waves, expressed as Ys = Ye(x) * Yt(t), reveals that the envelope function Ye depends solely on position while Yt depends solely on time. The first displacement at x=0 is confirmed to be 0, and the first time t1 when the string is perfectly straight is calculated as t1 = π/(2ω).

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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
 
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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?
 

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