Wavelength and Frequency — Question 1

Question 1

Counting crests requires counting the gaps between them

A snapshot shows six consecutive crests from $x=.20$ to $1.70\,\mathrm{m}$. At a fixed point, the time from the first to the ninth consecutive crest is $.80\,\mathrm{s}$.

1. Infer wavelength, period and frequency.

2. Find the distance advanced by the periodic pattern in one second. Explain both endpoint-counting traps.

Question 1 - Solution

Counting crests requires counting the gaps between them

Method and mathematical working

Label the first event as index zero. Subtract the indices of the two endpoint events to obtain the number of complete intervals.$$\begin{aligned}\lambda&=\frac{x_6-x_1}{6-1},\qquad T=\frac{t_9-t_1}{9-1},\\ f&=T^{-1},\qquad v=\lambda f.\\[4pt]\lambda&=\frac{1.70-0.20}{5}=0.30\,\mathrm m,\quad T=\frac{0.80}{8}=0.10\,\mathrm s,\\f&=10\,\mathrm{Hz},\qquad v=\lambda f=3.0\,\mathrm{m/s}.\end{aligned}$$

(a) Calculation and conclusion

Six crests span five wavelengths, so $\lambda=1.50/5=.30\,\mathrm{m}$. Nine arrivals span eight periods, so $T=.80/8=.10\,\mathrm{s}$ and $f=10\,\mathrm{Hz}$.

(b) Interpretation and limits

One cycle advances one wavelength, giving $v=f\lambda=3.0\,\mathrm{m/s}$ and $3.0\,\mathrm{m}$ in one second. Dividing by six or nine would confuse the number of events with the number of intervals.

Physical check

If only two consecutive crests were recorded, they would span one wavelength. The same endpoint-counting rule must reproduce that simple limit.