Wavelength and Frequency — Question 10

Question 10

Wavelength bounds propagate through a reciprocal relation

A nondispersive wave speed is known exactly as $v=12\,\mathrm{m/s}$. Its frequency is bounded by $19\le f\le21\,\mathrm{Hz}$; these are hard bounds, not standard deviations.

1. Find exact wavelength bounds.

2. Compare them with the symmetric first-order estimate about $f=20\,\mathrm{Hz}$. Why are the exact bounds asymmetric?

Question 10 - Solution

Wavelength bounds propagate through a reciprocal relation

Method and mathematical working

Use monotonicity to map the endpoints of the frequency interval. Keep exact hard bounds distinct from a differential estimate of the deviations.$$\begin{aligned}\frac{d\lambda}{df}&=-\frac v{f^2}<0,\qquad \frac{d^2\lambda}{df^2}=\frac{2v}{f^3}>0.\\[4pt]\frac{12}{21}\le\lambda\le\frac{12}{19}&\Rightarrow0.57143\le\lambda\le0.63158\quad(\mathrm m),\\\lambda_0&=0.600\,\mathrm m,\qquad |\delta\lambda|\simeq\frac{12}{20^2}(1)=0.030\,\mathrm m.\end{aligned}$$

(a) Calculation and conclusion

$\lambda=v/f$ decreases with frequency, so $12/21\le\lambda\le12/19$, or $.57143\le\lambda\le.63158\,\mathrm{m}$. The larger frequency yields the smaller wavelength.

(b) Interpretation and limits

The central value is $.600\,\mathrm{m}$ and $|d\lambda|\approx(v/f^2)|df|=.030\,\mathrm{m}$. The estimate is $.600\pm.030\,\mathrm{m}$. The reciprocal function is curved, making exact upper and lower deviations unequal; hard bounds should use the endpoint calculation.

Physical check

The positive curvature explains the asymmetric exact interval about the central value. Combining the lowest frequency with the largest wavelength is a necessary ordering check.