How does stress of a string translate into transverse motion?

Question

If you have a tentioned string, the extra stretch of each segment is given by the strain ($ε$):

$ε = \frac{\text{segment length}}{\text{rest length}} -1$

Imagining each segment's response to strain as that of a simple elastic spring, the stress force ($σ$) of this extra strain is given by:

$σ = Eε$, where $E$ is Young's Modulus.

The simplest wave equation for transverse ($w$) waves of a string would be, where subscripts are derivatives of time ($t$) and length ($x$):

$ρw_{tt} = \frac{T}{A} w_{xx}$

($T$ is tension, $ρ$ is mass density, and $A$ is cross sectional area.)

If you want to incorporate the stress force into these basic wave equations in a way that is reasonable, how would you do so?

I read one article that suggests for transverse waves, you can phrase it as:

$ρw_{tt} = \frac{T}{A} w_{xx} + (σw_x)_x = \left(\frac{T}{A} + σ\right)w_{xx} + σ_xw_x$

I'm not sure how they got the $(σw_x)_x$ term. Is that correct?

Answer 1

I am not sure where it comes from, but the physics seem quite clear (it should be possible to rederive it again from variational principles). A line element from $x$ to $x+dx$ undergoes two transverse surface forces:

• The tension force resulting from the large static uniform string tension, say $T/A$ which is a constant;
• The spatially-varying tension force resulting from small longitudinal deformations, say $\sigma = E \varepsilon$ which isn't necessarily constant.

Thus, following the same steps as in these notes, the balance of momentum reads $$ \rho w_{xx} = \left[\left(\frac{T}{A} + \sigma\right) w_x\right]_x = \frac{T}{A}w_{xx} + \left(\sigma w_x\right)_x $$ which is the desired equation of motion for the transverse displacement $w$.