Energy and Power of a Wave Traveling Along a String — Question 10

Question 10

Power and particle motion together reveal the string parameters

A single sinusoidal traveling wave has mean power $.40\,\mathrm{W}$, maximum transverse particle speed $2.0\,\mathrm{m/s}$, and measured propagation speed $10\,\mathrm{m/s}$.

1. Infer linear density and tension.

2. If its frequency is $25\,\mathrm{Hz}$, infer displacement amplitude and maximum slope. Check whether the small-slope model is reasonably self-consistent.

Question 10 - Solution

Power and particle motion together reveal the string parameters

Method and mathematical working

Eliminate amplitude using the measured maximum material speed. This determines density directly from power and wave speed; recover tension and amplitude afterward.$$\begin{aligned}\overline P&=\tfrac12\mu v u_{\max}^2,\qquad \mu=\frac{2\overline P}{v u_{\max}^2},\\F&=\mu v^2,\quad A=\frac{u_{\max}}{2\pi f},\quad kA=\frac{u_{\max}}v.\\[4pt]\mu&=\frac{2(0.40)}{10(2)^2}=0.020\,\mathrm{kg/m},\quad F=(0.020)(10)^2=2\,\mathrm N,\\A&=\frac2{2\pi(25)}=0.01273\,\mathrm m,\quad kA=2/10=0.20.\end{aligned}$$

(a) Calculation and conclusion

Writing $U=\omega A$, the power law is $\overline P=\mu vU^2/2$. Thus $\mu=2(.40)/(10\times4)=.020\,\mathrm{kg/m}$, and $F=\mu v^2=2.0\,\mathrm{N}$.

(b) Interpretation and limits

$A=U/(2\pi f)=2/(50\pi)=.01273\,\mathrm{m}$. Maximum slope is $kA=U/v=.20$. This is below unity, though not asymptotically tiny; geometric slope corrections are of order $(.20)^2$, so high-precision use would require more caution or smaller amplitudes.

Physical check

The dimensionless ratio $u_{\max}/v=0.20$ is also the maximum slope. It is the appropriate geometric measure for judging the small-slope approximation, rather than amplitude alone.