Answer to Question #88566 in Classical Mechanics for fun

Question

Sound source A ls located at x = 0, y = 0, and sound source B is located at x = 0, y 3.4 m. The two sources radiate coherently and in phase. An observer at x = 17.4 m, y 0 notes that as he takes a few step5 from y 0 in either the ty or -y direction, the sound intensity increases. What is then the lowest frequency and next to lowest frequency of the sources that can account for this observation? (Enter your answers from smallest to largest.) 1.04 x kHz kHz

Answer 1

path difference between distances to each wave source:

$$L= \sqrt{17.4^2+3.4^2}-17.4 = 0.329\,\text{m}$$

This point is "interference minima". Therefore,

$$L = (n+\frac{1}{2})\lambda, \, n = 0,1,2,\dots$$

Assume speed of sound in air

$$\lambda \cdot \nu = c = 343 \, \text{m/s}$$

Frequency

$$\nu_n = \frac{c}{\lambda_n} =\frac{c }{L}\left(n+\frac12 \right)$$

The lowest and next to lowest frequency corresponds to n=0 nad n=1:

$$\nu_0 =\frac{c }{2L}=521\,\text{Hz}, \,\,\, \nu_1 =\frac{3c }{2L}=1563\,\text{Hz}$$