Wave Speed on a Stretched String — Question 4

Question 4

Equal tension and equal stress are different comparisons

Two strings of the same material density $\rho$ have circular cross sections with radii $r$ and $2r$. Their transverse speed is $v=\sqrt{F/(\rho S)}$, where $S$ is area.

1. Compare their speeds under equal tensions.

2. Compare under equal tensile stresses $\sigma=F/S$. Explain the cancellation in the second comparison.

Question 4 - Solution

Equal tension and equal stress are different comparisons

Method and mathematical working

Express the density through cross-sectional area. Distinguish equal force from equal stress: the latter makes the force increase with area.$$\begin{aligned}\mu&=\rho\pi r^2,\qquad v=\sqrt{\frac{F}{\rho\pi r^2}},\\F=\sigma\pi r^2&\Rightarrow v=\sqrt{\frac{\sigma}{\rho}}.\\[4pt]\frac{\mu_2}{\mu_1}&=\frac{\pi(2r)^2}{\pi r^2}=4,\qquad \frac{v_2}{v_1}=\begin{cases}1/2,&F_2=F_1,\\1,&\sigma_2=\sigma_1.\end{cases}\end{aligned}$$

(a) Calculation and conclusion

The thicker string has four times the cross-sectional area and linear density. At equal tension its speed is half that of the thinner string.

(b) Interpretation and limits

At equal stress, the thicker string also has four times the tension. Since $v=\sqrt{\sigma/\rho}$, both speeds are equal. A statement about diameter dependence is incomplete unless the controlled loading variable is specified.

Physical check

Doubling radius quadruples linear density. At fixed force this halves speed; at fixed stress both force and density quadruple, so their ratio is unchanged.