Finding the slop of a string (Waves)

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Homework Help Overview

The discussion revolves around finding the slope of a transverse wave described by the equation y(x,t) = Asin(kx - ωt). Participants are exploring the differentiation of this wave function with respect to position x and time t, focusing on the mathematical implications of the derivative.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to differentiate the wave equation to find the slope, indicating a need to apply the chain rule. Some participants question the correctness of the derivative obtained and suggest checking for unit consistency. Others inquire about the treatment of variables during differentiation.

Discussion Status

The discussion is ongoing, with participants providing guidance on checking the derivative's consistency and clarifying the treatment of variables in the differentiation process. Multiple interpretations of the differentiation process are being explored.

Contextual Notes

There is an indication that the original poster may have misunderstood the application of the chain rule or the treatment of constants and variables in the differentiation process.

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Homework Statement



Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itself).

Consider a transverse wave traveling in a string. The mathematical form of the wave is

[tex]y(x,t) = Asin(kx - \omega t)[/tex]

Find the slope of the string [tex]\frac{dy(x,t)}{dx}[/tex] as a function of position x and time t.

Homework Equations





The Attempt at a Solution



So to find the slope of that equation, I would just need to take it's derivative. x is the variables, and [tex]\omega[/tex], A, t are constants. I would also need to use the chain rule when doing the derivative of sin.

So I get: [tex]\frac{dy(x,t)}{dx} = (Acos(kx - \omega t))*(k - \omega t)[/tex]

but it says I'm wrong. Any suggestions where I went wrong?
 
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Check your result for consistency of units.
 
[tex]\frac d {dx} ( kx- \omega t)[/tex] = ?
 
Integral said:
[tex]\frac d {dx} ( kx- \omega t)[/tex] = ?

would I treat the x as a variable and t as a constant?
 

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