Wave Speed on a Stretched String — Question 1

Question 1

A pulse measurement determines tension only with the correct linear density

A uniform taut string has mass $.024\,\mathrm{kg}$ over its stretched length $3.0\,\mathrm{m}$. A small transverse pulse travels $1.50\,\mathrm{m}$ in $.050\,\mathrm{s}$. Use $v=\sqrt{F/\mu}$.

1. Infer the linear density and tension.

2. If only half this string is used at the same tension without stretching it further, does the wave speed change?

Question 1 - Solution

A pulse measurement determines tension only with the correct linear density

Method and mathematical working

Calculate mass per unit length from the entire uniform string, and calculate speed from the pulse transit. Then invert the string-speed law.$$\begin{aligned}\mu&=\frac{0.024}{3.0}=0.0080\,\mathrm{kg/m},\\v&=\frac{1.5}{0.050}=30\,\mathrm{m/s},\qquad F=\mu v^2.\\[4pt]F&=(0.0080)(30)^2=7.2\,\mathrm N,\qquad \mu_{\mathrm{half}}=\frac{m/2}{L/2}=\mu.\end{aligned}$$

(a) Calculation and conclusion

$\mu=.024/3=.0080\,\mathrm{kg/m}$ and $v=1.50/.050=30\,\mathrm{m/s}$. Thus $F=\mu v^2=7.2\,\mathrm{N}$. Total mass alone cannot replace linear density in the speed law.

(b) Interpretation and limits

Cutting away half reduces mass and length by the same factor, leaving $\mu$ unchanged. With the same tension the speed remains $30\,\mathrm{m/s}$. End-to-end travel time changes because the propagation distance changes.

Physical check

Cutting a uniform string removes mass and length in the same ratio. Therefore its linear density and speed at the same tension remain unchanged; the transit time over the shorter piece changes.