Potential energy per unit length in a string (sin wave)

1. Aug 11, 2013

PsychonautQQ

1. The problem statement, all variables and given/known data

Given that the stretched length of a string is Δx(1+1/2(dy/dx)^2) show that the potential energy per unit length is equal to

1/2F(dy(x,t)/dx)^2

2. Relevant equations
potential energy = kx^2
cos(kx-wt)
idk really...

3. The attempt at a solution
The fact that the stretched length equals
Δx(1+1/2(dy/dx)^2)

can be derived from the fact that the strings stretched length is equal to (x^2+x(dy/dx)^2)^1/2
and then simplified with binomial expansion.

according to my answer key, the first step to the solution of solving potential energy per unit length is understanding that it is equal to
(FΔx(1+1/2(dy/dx)^2)-Δx) / Δx
The part about this step that I don't understand is why is the -Δx term in the numerator? Wouldn't the work per unit length just equal (Force * the amount it is stretched / Δx)? which would equal
FΔx(1+1/2(dy/dx)^2) / Δx.

yeah.. I don't understand why the -Δx is in the numerator i guess sums up my concerns.

2. Aug 11, 2013

TSny

Should that be F(Δx(1+1/2(dy/dx)^2)-Δx) / Δx?

Yes, that's right. But "the amount it is stretched" is not the same as "the stretched length". For example, suppose I have a spring that has an unstretched length of 20 cm. (This is the length from one end of the spring to the other end when the spring is not stretched.) Then I stretch it until it has a "stretched length" of 30 cm. (This is the length from one end of the spring to the other when it is stretched.) The "amount it is stretched" would be 10 cm.

3. Aug 11, 2013

PsychonautQQ

Cool, i'm still a little confused how that is represented in the equation by the term -Δx

4. Aug 11, 2013

TSny

Δx is the length of a section of string before it was stretched.

Δx(1+y'2/2) is the length of the same section after it has been stretched.

So, how would you write an expression for the amount the section has been stretched?