Direct substitution problem

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In summary, direct substitution means plugging the proposed solution into the equation and checking if it satisfies the equation. In this conversation, the participants are discussing how to verify a wave function for a standing wave as a solution to the general linear wave equation. The method of direct substitution is suggested, and the participants work through the substitution and show that it is a valid solution. They also discuss the relationship between the variables in the equation and how to use it to simplify the verification process.
  • #1
~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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  • #2
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.
 
  • #3
Yeah, sometimes examiners are really sneaky, and use words which you think are technical, but actually they're just obscure ordinary English. :frown:
 
  • #4
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
 
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  • #5
~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:
 
  • #6
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.
 
  • #7
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
 
  • #8
Actually I think I found what I was supposed to get( above equation)...:biggrin:...right?
 
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  • #9
~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
 

What is a direct substitution problem?

A direct substitution problem is a type of mathematical problem where you are given an equation with a variable, and you are asked to find the value of the variable by plugging in a specific value for it.

When do I use direct substitution?

Direct substitution is used when you have a function that has a specific value for the variable, and you need to find the corresponding output value. It is commonly used in algebra and calculus problems.

How do I solve a direct substitution problem?

To solve a direct substitution problem, you simply replace the variable in the equation with the given value and then solve the resulting equation. Make sure to follow the order of operations and simplify the expression if necessary.

What are some common mistakes in direct substitution problems?

One common mistake is forgetting to follow the order of operations, which can lead to incorrect solutions. Another mistake is not simplifying the expression after substituting the value, which can also result in an incorrect answer.

Can I use direct substitution for any type of equation?

No, direct substitution can only be used for linear equations, where the variable has a power of 1. It cannot be used for equations with variables that have powers greater than 1 or for equations with multiple variables.

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