Is Direct Substitution Effective for Solving Linear Wave Equations?

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Homework Help Overview

The discussion revolves around verifying a wave function for a standing wave as a solution to the general linear wave equation through direct substitution. The wave function is given as y = (2A sin kx)cos(ωt), and the equation to be verified is ∂²y/∂x² = (1/v²) ∂²y/∂t².

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of direct substitution, with some expressing uncertainty about how to proceed after calculating the second partial derivatives. Questions arise regarding the relationship between the derived expressions and the wave equation.

Discussion Status

Several participants are exploring the verification process, with some suggesting that the original poster may have inadvertently completed part of the verification. There is an ongoing examination of the relationship between the derived factors and the wave speed, with no explicit consensus reached yet.

Contextual Notes

Participants note the challenge of finding specific relationships in their textbooks, while referencing a relationship provided by a lab teacher. There is mention of the need to clarify definitions of k and ω in terms of wavelength and frequency.

~christina~
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Homework Statement


verify by direct substitution that the wave function for a standing wave given in equation below is a solution to the general linear wave equatin.

[tex]y= (2A sin kx)cos \omega t[/tex]

[tex]\frac{\delta^2y} {\delta x^2}= \frac{1} {v^2} \frac{\delta^2 y} {\delta t^2}[/tex]

Homework Equations

(above)

How do I do this by direct substittution ?

I originally went and just proved it through partial differentiation but I've never heard of direct substitution before.

Help please

Thank you
 
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I have a sneaking suspicion that you've verified the solution by direct substitution without knowing it. The method of direct substitution simply means substituting the proposed solution into the equation and seeing if it works. In other words, taking the second partial derivatives with respect to position and time and seeing if they fit the equation.
 
Yeah, sometimes examiners are really sneaky, and use words which you think are technical, but actually they're just obscure ordinary English. :frown:
 
Hootenanny said:
I have a sneaking suspicion that you've verified the solution by direct substitution without knowing it. The method of direct substitution simply means substituting the proposed solution into the equation and seeing if it works. In other words, taking the second partial derivatives with respect to position and time and seeing if they fit the equation.

well not exactly. I couldn't get them to equal since I wasnt sure what to do after I got:

[tex]\frac{\delta^2 y} {\delta x^2}|_t= -k^2(2A sin kx)cos \omega t)[/tex]

and

[tex]\frac{\delta^2 y} {\delta t^2}|_x= -\omega^2 2A cos\omega t(sin kx)[/tex]

and if that's right...I'm not sure how to make them equal Is there a equation I can use to do that?

[tex](-k^22A sin kx)(cos \omega t)= \frac{1} {v^2} (-\omega^2 cos\omega t)(2A sin kx)[/tex]

Thanks
 
Last edited:
~christina~ said:
well not exactly. I couldn't get them to equal since I wasnt sure what to do after I got:

[tex]\frac{\delta^2 y} {\delta x^2}|_t= -k^2(2A sin kx)cos \omega t)[/tex]

and

[tex]\frac{\delta^2 y} {\delta t^2}= -\omega^2 2A cos\omega t(sin kx)[/tex]

Hi christina!

erm … they're the same! … except for a factor, which you can call v. :smile:
 
tiny-tim said:
Hi christina!

erm … they're the same! … except for a factor, which you can call v. :smile:
Of course they are but I can't say that they are unless I show it first.
The question is to verify by direct substitution...is this right ? (below).

I'm thinking of dividing over the cos and sin and then canceling to just get

[tex]\frac{\omega^2} {k^2} = \frac{1} {v^2}[/tex]

but I'm trying to find a relationship in physics that allows that. If I'm not incorrect.
 
I think I found the relationship after looking at one of my old posts.

It's interesting though how I can't find this equation anywhere in my book but my lab teacher gave it to the class and it works here. ([tex]v= \omega /k[/tex])

[tex](-k^22A)(sin kx)(cos \omega t)= \frac{1} {v^2} (-\omega^22A) (cos\omega t)(sin kx)[/tex]

I'm thinking of dividing over the cos and sin to the left and then canceling to just get

[tex]\frac{\omega^2} {k^2} = \frac{1} {v^2}[/tex]

But I'm confused to as to when I use [tex]v= \omega /k[/tex]

I end up with

[tex]\frac{1} {v^2} = (\frac{k} {\omega})^2[/tex] :confused:

but is there another way to verify?

Thank you
 
Actually I think I found what I was supposed to get( above equation)...:biggrin:...right?
 
Last edited:
~christina~ said:

Homework Statement


verify by direct substitution that the wave function for a standing wave given in equation below is a solution to the general linear wave equatin.

[tex]y= (2A sin kx)cos \omega t[/tex]

[tex]\frac{\delta^2y} {\delta x^2}= \frac{1} {v^2} \frac{\delta^2 y} {\delta t^2}[/tex]

Homework Equations

(above)

How do I do this by direct substittution ?

I originally went and just proved it through partial differentiation but I've never heard of direct substitution before.

Help please

Thank you

Suppose you were asked to show that x= 1 is a solution to x5- 3x2+ 5x- 2= 0. Would you solve the equation? Of course not- you would just substitute 1 for x and show that the equation is true. That's "direct substitution".
 
  • #10
~christina~ said:
It's interesting though how I can't find this equation anywhere in my book but my lab teacher gave it to the class and it works here. ([tex]v= \omega /k[/tex])

Look up the definitions of [itex]k[/itex] and [itex]\omega[/itex] in terms of [itex]\lambda[/itex] and [itex]f[/itex] and substitute them into the equation above. You should get something that looks familiar. :smile:
 
  • #11
HallsofIvy said:
Suppose you were asked to show that x= 1 is a solution to x5- 3x2+ 5x- 2= 0. Would you solve the equation? Of course not- you would just substitute 1 for x and show that the equation is true. That's "direct substitution".

I understand. So this case however, it would be pluging an equation into the linear wave equation for direct substitution.

Thank You HallsofIvy
 

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