The Wave Equation — Question 10

Question 10

Variable tension adds a term that a local speed alone misses

A string has prescribed stationary tension $F(x)$ and density $\mu(x)$. External agents maintain the longitudinal equilibrium; transverse forces come from the string tension. Slopes are small.

1. Derive the transverse equation and expand its spatial derivative.

2. If $F(x)/\mu(x)=v_0^2$ is constant, does the equation reduce to $y_{tt}=v_0^2y_{xx}$? State the extra condition needed.

Question 10 - Solution

Variable tension adds a term that a local speed alone misses

Method and mathematical working

Form the force difference before differentiating. With position-dependent tension, the product rule introduces an extra first-derivative term.$$\begin{aligned}\mu(x)y_{tt}&=\partial_x\!\left(F(x)y_x\right)\\&=F(x)y_{xx}+F^{\prime}(x)y_x.\\[4pt]F/\mu=v_0^2&\Rightarrow y_{tt}=v_0^2y_{xx}+\frac{F^{\prime}}\mu y_x,\\\text{uniform-wave reduction}&\Rightarrow F^{\prime}y_x=0.\end{aligned}$$

(a) Calculation and conclusion

The force difference is $\partial_x(Fy_x)dx$, giving $\mu(x)y_{tt}=\partial_x[F(x)y_x]=F y_{xx}+F^{\prime}y_x$. The second term arises because the magnitude of tension varies as well as its direction.

(b) Interpretation and limits

Dividing by $\mu$ gives $y_{tt}=v_0^2y_{xx}+(F^{\prime}/\mu)y_x$. A constant ratio alone does not remove the extra term. Uniform tension, or a particular solution whose extra term vanishes, is needed for the standard homogeneous form. This explains why local speed alone does not establish identical propagation behavior.

Physical check

Even if $F(x)/\mu(x)$ is constant, $F^{\prime}(x)y_x/\mu(x)$ generally remains. The constant local speed ratio therefore does not reduce this equation to the uniform-string equation.