91Ó°ÊÓ

Frequency of the sound wave that is emitting from the speaker is, $f=1800Hz$

The speed of sound into the room is,$v=340m/s$

Distance between the speakers, $d=2.0m$

The distance of the point of observation from one of the speaker is,$r_1=4.0m$

Consider the two loudspeaker is arranged as given in the figure:

According to the geometry of the set up mentioned above, the distance of observing point from the second speaker, is$$\begin{gathered} r_2=\sqrt{r_1^2+d^2} \\ {r}_2=\sqrt{{\left(4m\right)}^2+{\left(2m\right)}^2} \\ {r}_2=\sqrt{20}m \\ {r}_2=4.472m \end{gathered}$$

Now, the phase difference between the two speakers is ${\mathrm{ϕ}}_{∘}=0$

Path difference between the two sources to reach at the observation point is,

$$\begin{gathered} ∆r=r_2-r_1 \\ ∆r=\left(4.472-4\right)m \\ ∆r=0.472m \end{gathered}$$

The wavelength of the sound wave is,

$$\begin{gathered} \mathrm{λ}=\frac{v}{f} \\ \mathrm{λ}=\frac{343m/s}{1800Hz} \\ \mathrm{λ}=0.1889m \end{gathered}$$

Therefore, the phase difference for the waves coming from the point 4.0 m in front of one of the speakers is,

$$\begin{gathered} \mathrm{ϕ}=\frac{2\mathrm{π}∆r}{\mathrm{λ}}+{\mathrm{ϕ}}_{∘} \\ \mathrm{ϕ}=\frac{2\mathrm{π}×0.472m}{0.1889m}+0 \\ \mathrm{ϕ}=\left(2\mathrm{π}×2.5\right)rad \\ \mathrm{ϕ}=5\mathrm{π}\mathrm{rad} \\ \mathrm{ϕ}=\left(\frac{5}{2}×2\mathrm{π}\right)rad \end{gathered}$$

As the phase difference of the waves at the point mentioned above is an odd multiple $2\mathrm{Ï€}rad$. So, the interference is perfect destructive.