Standing Waves — Question 2

Question 2

Node spacing and antinode timing recover the component wave speed

Adjacent permanent nodes on a string are at $x=.20$ and $.50\,\mathrm{m}$. At an antinode, successive positive displacement maxima are $.050\,\mathrm{s}$ apart.

1. Find wavelength, frequency and traveling-component speed.

2. If the observed maximum antinode displacement is $8\,\mathrm{mm}$, find the amplitude of each of the two equal traveling components.

Question 2 - Solution

Node spacing and antinode timing recover the component wave speed

Method and mathematical working

Use node separation for spatial periodicity and successive positive antinode peaks for temporal periodicity. The standing antinode amplitude is twice either equal traveling amplitude.$$\begin{aligned}\lambda&=2(0.50-0.20)=0.60\,\mathrm m,\\f&=\frac1{0.050}=20\,\mathrm{Hz},\qquad v=f\lambda.\\[4pt]v&=(20)(0.60)=12\,\mathrm{m/s},\qquad A_{\mathrm{traveling}}=8/2=4\,\mathrm{mm}.\end{aligned}$$

(a) Calculation and conclusion

Node spacing $.30\,\mathrm{m}$ is half a wavelength, so $\lambda=.60\,\mathrm{m}$. The antinode period is $.050\,\mathrm{s}$, giving $f=20\,\mathrm{Hz}$ and $v=f\lambda=12\,\mathrm{m/s}$.

(b) Interpretation and limits

Antinode amplitude is twice a component amplitude, so each traveler has amplitude $4\,\mathrm{mm}$. Neither the node spacing nor the antinode amplitude equals the corresponding full traveling-wave quantity without its factor of two.

Physical check

Using adjacent zero crossings in time would yield half a period, but the stated repeated positive peaks give a full period. The distinction prevents a factor-of-two error.