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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.

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  • $\begingroup$ Can you clarify the notations what they stand for? $\endgroup$
    – user36790
    Dec 16, 2015 at 19:24
  • $\begingroup$ Do you know the expression for particle velocity and that of the total energy of a SHW? $\endgroup$ Dec 16, 2015 at 19:24
  • $\begingroup$ @Aniket yes I know that $\frac{1}{2}mw^2A^2$ is total energy in SHM $\endgroup$
    – user74370
    Dec 16, 2015 at 19:31
  • $\begingroup$ done @user36790 $\endgroup$
    – user74370
    Dec 16, 2015 at 19:36

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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.

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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 $v= \left(\frac{T}{\rho}\right)^{1/2}\;.$

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  • $\begingroup$ Could you just explain why $ds=\sqrt{{dx}^2+{dy}^2} $ $\endgroup$
    – user74370
    Dec 17, 2015 at 0:18
  • $\begingroup$ Okay, I've found a pic; maybe that can sought out your confusion. $\endgroup$
    – user36790
    Dec 17, 2015 at 2:30

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