Find Resultant Wave of y1 & y2: Get Hint Now!

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SUMMARY

The discussion focuses on finding the resultant wave from two given wave functions: y1 = 3sin(kx - wt) and y2 = 4cos(kx + wt). The correct resultant wave is derived using trigonometric identities and the amplitude addition formula, resulting in y = 7cos(wt + 45)sin(kx + 45) being identified as incorrect. Participants emphasize the importance of correctly applying the addition formula for waves with different amplitudes and ensuring the arguments of the sine and cosine functions are consistent.

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moham_87
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Hiiiiiiiiiii everyone
I've these two waves
y1=3sin(kx-wt)
y2=4cos(kx+wt)
I need to find the resultant wave (y1+y2)
I got that answer:

y= 7 cos(wt+45) sin(kx+45)
is that right?? please if not give me a hint

==================My efforts=============================
I added the amplidtude mathematically
then using trigonometric rules i added the two "sin" and "cos" functions
but i still need to know if my answer is true

bYYyYyyyYYyYYyYE
 
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The addition formula that I have for the sum of waves with different amplitudes is:

A cos(x) + B sin(x) = \sqrt {A^2 + B^2} cos(x \pm \delta)
tan( \delta) = \frac {sin \delta} {cos \delta} = \pm \frac B A
 
Integral said:
The addition formula that I have for the sum of waves with different amplitudes is:

A cos(x) + B sin(x) = \sqrt {A^2 + B^2} cos(x \pm \delta)
tan( \delta) = \frac {sin \delta} {cos \delta} = \pm \frac B A

What is that segma?
 
and what about if...

What about if "x" is not equal in both equations??
 
Integral,

The arguments for the sin and cos in your identity are both x. I may be missing a simplification you're seeing, but he's got one wave going left and one going right, so his arguments aren't the same.

I tried using the identities for the sum and difference of two angles, but nothing cancelled, so it just got messy. But like I said, I may be missing something.
 
yeah it gets a little messy but:

3 \sin {(kx-\omega t)}= 3( \sin{kx} \cos{\omega t }- \sin{\omega t} \cos {kx} )
and
4\cos{(kx+\omega t)}= 4(\cos{kx} \cos {\omega t}+ \sin{kx} \sin{\omega t})
add these together to get

\cos {kx} (4 \cos {\omega t} - 3 \sin {\omega t})+ \sin{kx}(3 \cos {\omega t}+ 4 \sin{\omega t})

now apply the formula in my first post to the terms in parentheses.

the \delta (thats a low case delta) is defined in the second line of my first post.
 
Last edited:
Integral, nicely done!

Ok moham87, it doesn't look like your answer's right. But you've got your hint, so have at it! And watch those signs, or they'll kill you.
 

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