Two loudspeakers as shown in fig. below separated by a distance $3 \mathrm{m}$, are in phase. Assume that the amplitudes of the sound from the speakers is approximately same at the position of a listener, Who is at a distance $4.0 m$ in front of one of the speakers. For what frequencies does the listener hear minimum signal ? Given that the speed of sound in air is $330 m{s}^{-1}$

The distance of the listener from the second speaker $= \sqrt{{\left(3\right)}^2+{\left(4\right)}^2 } =\sqrt{25} = 5m$

path difference $= \left(5-4.0\right)m = 1m$

For fully destructive inference $1 \mathrm{m} = \frac{\left(2m+1\right)\lambda }{2}$

Hence          $\mathrm{\lambda } = \frac{2}{\left(2\mathrm{m}+1\right)}\mathrm{m}$

The corresponding frequencies are given by:

                          $\mathrm{n} = \left[\frac{330\times \left(2\mathrm{m}+1\right)}{2}\right] {\mathrm{s}}^{-1}$, for $m=0,1,2,3,4,.....................$

                              $=165\left(2\mathrm{m}+1\right) {\mathrm{s}}^{-1}$ , for $m = 0,1,2,3,4,.......................$

Therefore the frequencies for which the listener would hear a minimum intensity $165 \mathrm{Hz}, 495 \mathrm{Hz}, 825 \mathrm{Hz}.....$