Simple wave questions- am I right?

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The discussion centers on the sum of two waves, y1 and y2, which are described by sine functions moving in opposite directions. The resulting wave, Ys, can be expressed as the product of a position-dependent envelope Ye and a time-dependent function Yt. For part one, the displacement at x=0 is determined to be zero, confirming the interpretation of the equation. In part two, the first time the string is perfectly straight is calculated as t1 = π/(2ω), which is also validated by participants. Overall, the interpretations and calculations presented align correctly with the wave equations.
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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-\omegat) 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 \omega, k, and necessary constants.




Homework Equations





The Attempt at a Solution



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

and using a trig identity,
I get Ys= 2Asin(kx)*cos(\omegat)

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\omegat \pi/2 = \omegat and I got t1= \pi/(2\omega)

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-\omegat) & for wave 2 it is y2=Asin(kx+\omegat). When I add them together, I get:
Ys= Asin(kx-\omegat) + Asin(kx+\omegat)

and using a trig identity,
I get Ys= 2Asin(kx)*cos(\omegat)

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\omegat \pi/2 = \omegat and I got t1= \pi/(2\omega)

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-\omegat) 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+\omegat).

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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