Standing Waves — Question 1

Question 1

Equal opposing travelers separate into a spatial pattern and a time factor

Two waves on an infinite string are $A\cos(kx-\omega t)$ and $A\cos(kx+\omega t)$, where $\omega=vk$.

1. Find the standing-wave form and locate nodes and antinodes.

2. Find the spacing of adjacent nodes and the distance from a node to the nearest antinode. Is the antinode amplitude $A$ or $2A$?

Question 1 - Solution

Equal opposing travelers separate into a spatial pattern and a time factor

Method and mathematical working

Apply the cosine sum identity to the two counterpropagating waves. The spatial factor sets the local amplitude, while the temporal factor controls their common oscillation.$$\begin{aligned}y&=A\cos(kx-\omega t)+A\cos(kx+\omega t)\\&=2A\cos kx\cos\omega t,\\\cos kx=0&\Rightarrow kx=(n+\tfrac12)\pi.\\[4pt]x_{{\mathrm{node}},m}&=(m+1/2)\lambda/2,\quad x_{{\mathrm{anti}},m}=m\lambda/2,\\\Delta x_{\mathrm{node-node}}&=\lambda/2,\quad\Delta x_{\mathrm{node-anti}}=\lambda/4.\end{aligned}$$

(a) Calculation and conclusion

The sum is $y=2A\cos(kx)\cos(\omega t)$. Nodes obey $kx=(n+1/2)\pi$; antinodes obey $kx=n\pi$. The local amplitude is $2A|\cos kx|$.

(b) Interpretation and limits

Adjacent nodes are $\pi/k=\lambda/2$ apart, and node-to-nearest-antinode distance is $\lambda/4$. Antinode amplitude is $2A$ because both traveling contributions reinforce there. The standing pattern has no translating crest train.

Physical check

Adjacent zeros of the spatial factor differ by $\pi/k=\lambda/2$. A neighboring node and antinode differ by half that distance; a zero of the temporal factor is not a permanent node.