Wave Speed on a Stretched String — Question 10

Question 10

A uniform-speed design requires tension and density to track each other

A slowly varying taut string has local tension $F(x)>0$ and local linear density $\mu(x)>0$. The desired local wave speed is the constant $v_0$. Treat local wave propagation as valid.

1. Find the necessary relation between $F(x)$ and $\mu(x)$.

2. If $F(x)=F_0(1+x/L)$ over $0\le x\le L$, design $\mu(x)$ and find total mass. Does constant local speed alone guarantee zero reflection?

Question 10 - Solution

A uniform-speed design requires tension and density to track each other

Method and mathematical working

Solve the local speed condition for the required density at every point, then integrate this density to find total mass.$$\begin{aligned}\mu(x)&=\frac{F_0}{v_0^2}\left(1+\frac{x}{L}\right),\\m&=\int_0^L\mu(x)\,dx=\frac{3F_0L}{2v_0^2}.\\[4pt]m&=\frac{F_0}{v_0^2}\left[x+\frac{x^2}{2L}\right]_0^L=\frac{3F_0L}{2v_0^2}.\end{aligned}$$

(a) Calculation and conclusion

From $v_0^2=F(x)/\mu(x)$, the required density is $\mu(x)=F(x)/v_0^2$. Both parameters must vary in the same proportion.

(b) Interpretation and limits

Here $\mu=F_0(1+x/L)/v_0^2$ and $M=\int_0^L\mu\,dx=3F_0L/(2v_0^2)$. Constant local speed fixes the ratio, but not all properties controlling reflection. Spatial variations can still scatter waves; zero reflection does not follow from this ratio alone.

Physical check

The density rises from $F_0/v_0^2$ to twice that value, so the mass equals length times their arithmetic mean. Constant local speed does not by itself eliminate scattering from varying impedance.