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

[PsychonautQQ]

Homework Statement

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

Homework Equations

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

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.

 

[TSny]

The Attempt at a Solution

...the stretched length equals
Δx(1+1/2(dy/dx)^2)

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

Should that be F(Δx(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)?

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.

 

[PsychonautQQ]

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.

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

 

[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?

 

[Nickie]

I would like to clarify that potential energy per unit length in a string is not equal to kx^2, as stated in the Homework Equations section. The potential energy in a string is actually given by the equation U = (1/2)k(x^2 + (dy/dx)^2), where k is the string's tension and x is the displacement from equilibrium.

Now, moving on to the given problem, the potential energy per unit length can be calculated by dividing the total potential energy (U) by the stretched length (Δx). So, the correct equation would be U/Δx = (1/2)k((x^2 + (dy/dx)^2)/Δx). This can be simplified to U/Δx = (1/2)k((Δx(1+1/2(dy/dx)^2))/Δx). Note that the Δx terms in the numerator and denominator cancel out, leaving us with (1/2)k(1+1/2(dy/dx)^2). This is the same as the given equation, 1/2F(dy(x,t)/dx)^2, where F is the force applied to the string.

To answer your question about the -Δx in the numerator, it is simply a way of expressing the stretched length (Δx) in terms of the original length of the string. This makes it easier to compare the potential energy per unit length of different strings with different original lengths. Without the -Δx term, the equation would only give us the potential energy per unit length for a string with a stretched length equal to its original length.

I hope this explanation helps to clarify your concerns. Remember, it is always important to use the correct equations and units when solving scientific problems.