Is Direct Substitution Effective for Solving Linear Wave Equations?

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The discussion focuses on verifying the wave function for a standing wave, y = (2A sin kx)cos(ωt), as a solution to the linear wave equation through direct substitution. Participants clarify that direct substitution involves substituting the proposed solution into the wave equation and checking if the resulting expressions are equivalent. The user expresses confusion about equating the second partial derivatives obtained from the substitution and seeks guidance on the relationship between ω, k, and v. Ultimately, the conversation emphasizes understanding how to manipulate the equations to demonstrate that the wave function satisfies the linear wave equation. The verification process hinges on recognizing the relationship v = ω/k to establish the equality.
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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.

y= (2A sin kx)cos \omega t

\frac{\delta^2y} {\delta x^2}= \frac{1} {v^2} \frac{\delta^2 y} {\delta t^2}

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:

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

and

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

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

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

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:

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

and

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

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

\frac{\omega^2} {k^2} = \frac{1} {v^2}

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. (v= \omega /k)

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

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

\frac{\omega^2} {k^2} = \frac{1} {v^2}

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

I end up with

\frac{1} {v^2} = (\frac{k} {\omega})^2 :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.

y= (2A sin kx)cos \omega t

\frac{\delta^2y} {\delta x^2}= \frac{1} {v^2} \frac{\delta^2 y} {\delta t^2}

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. (v= \omega /k)

Look up the definitions of k and \omega in terms of \lambda and f 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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