Question

Two stereo speakers are separated by a distance of 2.40 m. A person stands at a distance of 3.20 m directly in front of one of the speakers as shown in figure. Find the frequencies in the audible range (20-2000 Hz) for which the listener will hear a minimum sound intensity. Speed of sound in air = 320 m s⁻¹.

Answer 1

Given:
Distance between the two speakers d = 2.40 m
Speed of sound in air v = 320 ms⁻¹
Frequency of the two stereo speakers f = ?

As shown in the figure, the path difference between the sound waves reaching the listener is given by:
$∆x=S_{2}L-S_{1}L$
$∆x=\sqrt{(3.2{)}^2+(2.4{)}^2}-3.2$

Wavelength of either sound wave:
$=\left(\frac{320}{f}\right)$
We know that destructive interference will occur if the path difference is an odd integral multiple of the wavelength.
$\therefore ∆x=\frac{(2n+1)\lambda }{2}$
So,
$$\begin{gathered} \sqrt{(3.2{)}^2+(2.4{)}^2}-3.2=\frac{(2n+1)}{2}\left(\frac{320}{f}\right) \\ \Rightarrow \sqrt{16}-3.2=\frac{\left(2n+1\right)}{2}\left(\frac{320}{f}\right) \\ \Rightarrow 0.8\times 2f=\left(2n+1\right)\times 320 \\ \Rightarrow 1.6f=\left(2n+1\right)\times 320 \\ \Rightarrow f=200(2n+1) \end{gathered}$$

On putting the value of n = 1,2,3,...49, the person can hear in the audible region from 20 Hz to 2000 Hz.