How to find potential energy per unit volume of a simple harmonic transversal wave in a string?

How to derive the formula for potential energy per unit volume of a simple harmonic wave transversal in a string?

My book just states the formula as $\frac{1}{2} \rho v^2 (\frac{dy}{dx})^2$ without the proof. $\rho$ stands for density and $v$ for velocity.

• Can you clarify the notations what they stand for? – user36790 Dec 16 '15 at 19:24
• Do you know the expression for particle velocity and that of the total energy of a SHW? – SchrodingersCat Dec 16 '15 at 19:24
• @Aniket yes I know that $\frac{1}{2}mw^2A^2$ is total energy in SHM – user74370 Dec 16 '15 at 19:31
• done @user36790 – user74370 Dec 16 '15 at 19:36

Consider a short string of length $dx\;.$ We will use small-scale approximation so that the magnitude of tension $T$ is not changed by the transverse displacement of the string. Potential energy can be calculated by finding how much the string becomes long when it is displaced transversely. say, it has a length of $ds$ when it is deformed. therefore the work done on the deformation is the required potential energy as $$\text{PE}= U= T(ds-dx)\;.$$ Now, \begin{align}ds&= \sqrt{(dx^2 + dy^2)}\\&= dx\sqrt{(1+(\delta_x y)^2)}\;.\end{align} Using Taylor series & neglecting higher-order terms of $\delta_x y^2$ (since, by small-scale approximation, $\delta_x y \ll 1$), we get $$ds - dx\approx \frac{1}{2} {(\delta_x y)}^2 \;dx.$$ Therefore, potential energy $$\text{PE}\approx \frac{1}{2} T {(\delta_x y)}^2\; dx\;.$$ Therefore, potential energy density \begin{align}\frac{dU}{dx}&\approx \frac{1}{2} T {(\delta_x y)}^2\\&= \frac{1}{2} \rho v^2 \left(\frac{\partial y}{\partial x}\right)^2\end{align} using the definition of velocity $v= \left(\frac{T}{\rho}\right)^{1/2}\;.$
• Could you just explain why $ds=\sqrt{{dx}^2+{dy}^2}$ – user74370 Dec 17 '15 at 0:18