Question

Two loudspeakers, $\mathit{A}$and $\mathbf{B}$, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is $\mathbf{172}\mathbf{Hz}$. You are $\mathbf{8}\mathbf{.}\mathbf{00}\text{\hspace{0.17em}}\mathbf{m}$from role="math" localid="1655809995061" $\mathit{A}$. What is the closest you can be to $\mathit{B}$and be at a point of destructive interference?

Short answer

The closet distance from $B$ is $1\mathrm{m}$.

Step-by-step solution

Given Data

Frequency of sound emitted: $f=172\text{\hspace{0.17em}Hz}$

The distance of observer from A:$l=8\text{\hspace{0.17em}m}$

Condition of destructive interference

Condition of destructive interference is $\mathbf{\Delta l}\mathbf{=}\frac{n}{2}\mathbf{\lambda }$ , $\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{7}\mathbf{,}\mathbf{9}\mathbf{,}...\mathbf{)}$where$\mathbf{\Delta l}$ is the path difference and$\mathit{\lambda }$ is the wavelength of the sound emitted from the speakers.

Use the condition

Given that$f=172\text{\hspace{0.17em}Hz}$ and we know that the speed of sound in air is$v=344\text{\hspace{0.17em}m/s}$ .

So,

The two speakers are $8\mathrm{m}$apart. Let the distance from B is $x$. So, we can write $\Delta l=8-x$.

Now,

From the given values of $n$, for $n=7$, we get the closest distance.

Hence, the closest distance from $B$ is $1m$.