# How does stress of a string translate into transverse motion?

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?

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 $$w$$.