Wave Speed on a Stretched String — Question 5

Question 5

Equal transit times can reveal unequal string lengths

Two taut string sections carry the same tension. Their linear densities are $\mu$ and $4\mu$. A pulse takes the same time to traverse each section, and the total length is $3.0\,\mathrm{m}$.

1. Find the length of each section.

2. If tension in both sections is multiplied by four, what happens to their equal transit time and to the length ratio?

Question 5 - Solution

Equal transit times can reveal unequal string lengths

Method and mathematical working

Use the common tension to obtain a speed ratio, then impose equal transit times and the total-length constraint.$$\begin{aligned}v_2&=\frac{v_1}{2},\qquad \frac{L_1}{v_1}=\frac{L_2}{v_2},\\L_1&=2L_2,\qquad L_1+L_2=3\,\mathrm m.\\[4pt]L_2&=3/(2+1)=1\,\mathrm m,\quad L_1=2\,\mathrm m,\qquad F\to4F\Rightarrow t\to t/2.\end{aligned}$$

(a) Calculation and conclusion

The denser section has half the speed: $v_2=v_1/2$. Equal times require $L_2/L_1=v_2/v_1=1/2$. With $L_1+L_2=3$, the lengths are $2.0$ and $1.0\,\mathrm{m}$.

(b) Interpretation and limits

Each speed doubles, so each transit time halves. Their ratio and the required length ratio remain unchanged. The common tension sets the overall time scale but cancels from this comparison.

Physical check

The slower string must be shorter for the travel times to match. Multiplying both tensions by four doubles both speeds without changing the length ratio.