Interference — Question 4

Question 4

A sliding acoustic branch doubles the path-length adjustment

A two-path acoustic interferometer recombines equal-amplitude waves. Moving a U-shaped branch by $x$ lengthens one path by $2x$. At $x=0$ the signals reinforce. Successive minima are separated by $0.170\,\mathrm m$ of branch motion; the driver frequency is $1000\,\mathrm{Hz}$.

1. Infer wavelength and sound speed.

2. Find the smallest positive motion from the initial maximum to a minimum. Explain why treating branch motion as path difference gives the wrong speed.

Question 4 - Solution

A sliding acoustic branch doubles the path-length adjustment

(a) Method and calculation

Successive minima require a relative phase change of $2\pi$, hence a full wavelength of added path:$$2\Delta x=\lambda,\quad\lambda=2(0.170)=0.340\,\mathrm m,\quad c=f\lambda=340\,\mathrm{m/s}.$$

(b) Interpretation and further calculation

From a maximum to the nearest minimum, the added path is $\lambda/2$, so$$2x_{\min}=\lambda/2,\qquad x_{\min}=\lambda/4=0.0850\,\mathrm m.$$Using $\lambda=\Delta x$ would give $170\,\mathrm{m/s}$, half the correct result. Both parallel legs lengthen when the U branch moves.

Physical check

Only changes in branch length are needed; an unknown integer number of initial wavelengths does not affect the fringe spacing.