Question

Two loudspeakers in a ${20}^{o}C$room emit 686 Hz sound waves

along the x-axis.

a. If the speakers are in phase, what is the smallest distance between the speakers for which the interference of the sound

waves is maximum destructive?

b. If the speakers are out of phase, what is the smallest distance

between the speakers for which the interference of the sound

waves is maximum constructive?

Short answer

a. The smallest distance between the speakers for which the interference of the sound waves is maximum destructive is $25cm.$

b. If the two speakers are out of phase, the smallest distance between the speakers for which the interference of the sound

waves is maximum constructive is$25cm.$

Step-by-step solution

Part a Step 1: Given information

Two loudspeakers are in phase. Therefore, the phase difference between the sound waves produced by these is ${\varphi }_{\circ }=0$

Consider the smallest distance between the speakers for which the interference of the sound waves is maximum destructive is $d$.

The frequency of the sound waves $f=686Hz$

At ${20}^{\circ }C$, the speed of sound in air isrole="math" localid="1650377361718" $v=343m/s.$

Determination of the smallest distance between the speakers

For the condition of destructive interference produced by the sound waves emitted by the two speakers, the phase difference between the waves is,

$∆\varphi =\frac{2\pi ∆x}{\lambda }...\left(1\right)$

Here, the path-difference between the two waves is, $∆x=d$

For destructive interference, the phase difference can be written as,

role="math" localid="1650379152001" $∆\varphi =2\left(n+\frac{1}{2}\right)\pi ...\left(2\right)$where $n=0,1,2...$

Now, for smallest path difference, $n=0$. So, from the above equations 1 and 2,we can write,

role="math" localid="1650379605442" $$\begin{gathered} 2\pi \frac{d}{\lambda }=2\left(0+\frac{1}{2}\right)\mathrm{\pi } \\ 2\mathrm{\pi }\frac{\mathrm{d}}{\mathrm{\lambda }}=\mathrm{\pi } \\ \mathrm{d}=\frac{\mathrm{\lambda }}{2} \\ \mathrm{d}=\frac{\mathrm{v}}{2\mathrm{f}}\left[\because \mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}}\right] \\ \mathrm{d}=\frac{343\mathrm{m}/\mathrm{s}}{\left(2\times 686\right)\mathrm{Hz}} \\ \mathrm{d}=0.25\mathrm{m} \\ \mathrm{d}=25\mathrm{cm} \end{gathered}$$

Part b Step 1: Given information

When two speakers are out of phase, the phase difference between the sound waves produced by these is ${\varphi }_{\circ }=\pi$

Consider the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is $d\text{'}$

Determination of the smallest distance between the speakers in out of phase condition

Consider the condition of constructive interference, the path difference between the two waves are

$∆\varphi =2n\pi ...\left(1\right)$where $n=0,1,2....$

Now, the phase difference between the waves can be written in terms of path-difference as,

$$\begin{gathered} ∆\varphi =\frac{2\pi ∆x}{\lambda }+{\varphi }_{\circ } \\ ∆\varphi =\frac{2\pi d\text{'}}{\lambda }+\mathrm{\pi }...\left(2\right) \end{gathered}$$

Consider $n=1,$for the constructive interference pattern, we can write using the two equations 1 and 2 as,

role="math" localid="1650379635522" $$\begin{gathered} \frac{2\pi d\text{'}}{\lambda }+\mathrm{\pi }=2\mathrm{\pi } \\ \frac{2\mathrm{\pi d}\text{'}}{\mathrm{\lambda }}=\mathrm{\pi } \\ \mathrm{d}\text{'}=\frac{\mathrm{\lambda }}{2} \\ \mathrm{d}\text{'}=\frac{\mathrm{v}}{2\mathrm{f}}\left[\because \mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}}\right] \\ \mathrm{d}\text{'}=\frac{343\mathrm{m}/\mathrm{s}}{\left(2\times 686\right)\mathrm{Hz}} \\ \mathrm{d}\text{'}=0.25\mathrm{m} \\ \mathrm{d}\text{'}=25\mathrm{cm} \end{gathered}$$