An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed \(1.30 \mathrm{m}\) apart are powered by a singlefunction generator producing sine waves at \(1200-\mathrm{Hz}\) frequency. A student walks along a line \(12.5 \mathrm{m}\) away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line?

\(λ = \frac{343}{1200} = 0.286 \, m\)#tag_title# Next, find the distance between the central maximum and the first maximum#tag_content# To find the distance between the central maximum and the first maximum, let's consider the path difference causing constructive interference: \[\Delta L = n\lambda\] Where \(\Delta L\) is the path difference, \(n\) is the order of interference, and \(\lambda\) is the wavelength. For the first maximum, \(n = 1\), and we have: \[\Delta L = 1 \cdot 0.286 = 0.286 \, m\] We use trigonometry to find the distance (x) along the line where the student walks: \[\frac{x}{12.5} = \sin{\theta} \implies x = 12.5 \sin{\theta} \] Now, we can relate \(\theta\) to the path difference: \[\Delta L = 1.30 \sin{\theta}\] Substitute the value of \(\Delta L\): \[0.286 = 1.30 \sin{\theta} \implies \sin{\theta} = \frac{0.286}{1.30}\] Finally, we find the distance (x): \[x = 12.5 \sin{\theta} = 12.5 \cdot \frac{0.286}{1.30} = 2.75 \, m\] #Short_Answer# (a) The wavelength of the sound is \(0.286 \, m\). (b) The distance between the central maximum and the first maximum is \(2.75 \, m\).