Wave Speed on a Stretched String — Question 2

Question 2

Stretching changes inertia per unit length as well as tension

A uniform string of conserved mass is stretched from length $L$ to $1.20L$. Its tension increases from $F$ to $1.80F$. Neglect bending rigidity and use linear transverse-wave theory about each stretched state.

1. Find the wave-speed ratio.

2. Find the ratio of end-to-end travel times. Explain why multiplying speed by only $\sqrt{1.80}$ misses part of the change.

Question 2 - Solution

Stretching changes inertia per unit length as well as tension

Method and mathematical working

Conserve the total mass while updating both the tension and the linear density. The travel time also depends on the new length.$$\begin{aligned}\frac{\mu_2}{\mu_1}&=\frac1{1.2},\qquad \frac{v_2}{v_1}=\sqrt{1.8(1.2)},\\\frac{t_2}{t_1}&=\frac{1.2}{\sqrt{2.16}}.\\[4pt]\frac{v_2}{v_1}&=\sqrt{2.16}=1.46969,\qquad\frac{t_2}{t_1}=\sqrt{1.2/1.8}=0.81650.\end{aligned}$$

(a) Calculation and conclusion

The new density is $\mu/1.20$. Thus $v_2/v_1=\sqrt{1.80(1.20)}=\sqrt{2.16}=1.4697$. The stretched material has less inertia per unit current length.

(b) Interpretation and limits

Travel time scales as length divided by speed, so $t_2/t_1=1.20/\sqrt{2.16}=.8165$. Holding density fixed would ignore the stated mass conservation during stretching. Despite the longer path, the travel time decreases here.

Physical check

The speed increase exceeds the length increase, so the travel time decreases. Holding linear density fixed while stretching this same piece would contradict the specified mass conservation.