Two loudspeakers, A and B (Fig. E16.35), are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. What is the lowest frequency for which (a) constructive interference occurs at point Q; (b) destructive interference occurs at point Q?

$\mathrm{v}=\mathrm{\lambda f}$where v is the speed of wave, f is the frequency and$\lambda$ is the wavelength of the wave.

Path difference from points A and B to point Q is $3.0\mathrm{m}-1.0\mathrm{m}=2.0\mathrm{m}$as shown in Figure. Constructive interference implies path difference$=\mathrm{n}\mathrm{\lambda }\mathrm{where}\mathrm{n}=1,2,3...$

Given that,$2.00m=n\lambda$

So,role="math" localid="1664342520874" $\mathrm{\lambda }=\frac{2.00\mathrm{m}}{\mathrm{n}}$

Apply in formula of speed,

role="math" localid="1664342538111" $$\begin{gathered} \mathrm{f}=\frac{\mathrm{v}}{\mathrm{\lambda }} \\ =\frac{\mathrm{vn}}{2.00\mathrm{m}} \\ =\frac{\mathrm{n}\left(344\mathrm{m}/\mathrm{s}\right)}{2.00\mathrm{m}} \\ =\mathrm{n}\left(172\mathrm{Hz}\right) \\ \mathrm{Where}\mathrm{n}=1,2,3,... \end{gathered}$$

Hence, the lowest frequency for which constructive interference occurs is 172 Hz.

Destructive interference implies path difference $=(n/2)\lambda$,$\mathrm{where}\mathrm{n}=1,3,5...$

Given that,$2.00m=(n/2)\lambda$

So,$\mathrm{\lambda }=\frac{4.00\mathrm{m}}{\mathrm{n}}$

Apply in formula for speed,

$$\begin{gathered} \mathrm{f}=\frac{\mathrm{v}}{\mathrm{\lambda }} \\ =\frac{\mathrm{vn}}{4.00\mathrm{m}} \\ =\frac{\mathrm{n}\left(344\mathrm{m}/\mathrm{s}\right)}{4.00\mathrm{m}} \\ =\mathrm{n}\left(86\mathrm{Hz}\right) \\ \mathrm{Where}\mathrm{n}=1,2,3,... \end{gathered}$$

Hence, the lowest frequency for which destructive interference occurs is 86 Hz.