Question

A source emitting sound of frequency $90\text{Hz}$ is placed in front of a wall at a distance $3\text{m}$ from it. A detector is also placed in front of the wall at the same distance from it as shown in the figure. Find the minimum distance between the source and the detector for which the detector detects a maximum of sound. Speed of sound in air = $360{\text{ms}}^{-1}$.

• A. $2\text{m}$
• B. $4\text{m}$
• C. $6\text{m}$
• D. $8\text{m}$

Answer 1

The correct option is D $8\text{m}$


For obtaining maxima, the sound waves from the source and the reflector should interfere constructively at $\text{D}$.
Condition for constructive interference,
Path difference $\Delta l=n\lambda$ , where $n=1,2,3,\mathrm{4....}$

Let $x$ be the minimum seperation between source and detector, for which two sound waves interfere constructively.
Since the wave undergoes phase reversal (of $\pi$) upon reflection, (equivalent to path difference $\lambda /2$)
thus, the first maxima corresponds to path difference,
$\Delta l=\frac{\lambda }{2}$
From the given figure,

$\{2\times {\left[\right(3{)}^2+\frac{x^2}{4}]}^{1/2}-x\}=\frac{\lambda }{2}..........\left(1\right)$

From the data given in the question,

$\lambda =\frac{v}{f}=\frac{360}{90}=4\text{m}..........\left(2\right)$
Substituting $\left(2\right)$ in $\left(1\right)$, we get

${[36+x^2]}^{1/2}-x=2\Rightarrow [36+x^2]=(x+2{)}^2$

$\Rightarrow 36+x^2=x^2+4+4x$
$\Rightarrow 4x=32\Rightarrow x=8\text{m}$

Thus, option (d) is the correct answer.