# Potential Energy of a string (Wave Equation PDEs)

This may turn out to be an elementary question, but I'm having trouble understanding where this actually comes from.

In my PDEs course, we're learning about the wave equation. We're given the following problem on our current homework set.

Consider a taught string $$0 ≤ x ≤ L$$ with constant density $$\rho$$ and constant tension $$T$$ whose vertical displacement at time $$t$$ and position $$x$$ is described by the function $$u(x, t)$$.

(a) Write down the integral representing the total kinetic energy in the segment $$a ≤ x ≤ b$$.

(b) Write a formula for the total potential energy in the segment $$a ≤ x ≤ b$$ assuming that the potential energy is proportional to the diﬀerence between the length of the stretched string on the interval $$[a, b]$$ and the equilibrium length of the string on the interval $$[a, b]$$, and explain using dimensional analysis why the constant of proportionality is the tension.

(c) Approximate the square root in your answer to (b) using Taylor’s theorem.

(d) Derive the wave equation from conservation of energy.

My question is regarding part (b). I've found a source online that says that

$$dl = \sqrt{dx^2 + dy^2} - dx \approx \frac{1}{2} (\frac{\partial y}{\partial x})^2 dx$$

I understand that the change in length of the string is represented by its magnitude minus the change in position, i.e.

$$dl = \sqrt{dx^2 + dy^2} - dx$$

But, I'm having trouble showing its equivalence to

$$\frac{1}{2} (\frac{\partial y}{\partial x})^2 dx$$

My attempt at it is

$$\sqrt{dx^2 + dy^2} - dx = \sqrt { 1 + (\frac{dy}{dx})^2 } dx -dx \neq \frac{1}{2} (\frac{\partial y}{\partial x})^2 dx$$

Any help towards the correct answer would be appreciated. I understand why you multiply by $$T$$ to get the potential energy, it's just showing the work up until that point.

Thank you.

• Hint: Taylor expansion.
– mcd
Commented Nov 3, 2016 at 23:45
• If I understand where you're going with this, the answer to part (b) is $T \sqrt{dx^2 + dy^2} - dx$? Where part (c) is the Taylor expansion leading to the potential energy listed above?
– NoVa
Commented Nov 3, 2016 at 23:46
• I guess I didn't list it, whoops. $PE = \frac{T}{2}(\frac{\partial y}{\partial x})^2 dx$
– NoVa
Commented Nov 3, 2016 at 23:54

$$(1 + x)^p \approx 1 + px$$
for $x \ll 1$. From this
$$\sqrt{dx^2 + dy^2} - dx = dx\sqrt{1 + \left(\frac{\partial y}{\partial x} \right)^2} - dx \approx dx\left[ 1 + \frac{1}{2}\left(\frac{\partial y}{\partial x} \right)^2\right] - dx = \frac{1}{2}\left(\frac{\partial y}{\partial x} \right)^2dx$$