Question

Two speakers are separated by a distance of $3\text{m}$. A person stands at a distance of $4\text{m}$, directly in front of one of the speakers. If $n$ stands for any integer, find the frequencies (in audible range) for which the listener will hear a minimum sound intensity. Speed of sound in air is $320\text{m/s}$.

• A. $200(2n+1)\text{Hz}$
• B. $160(2n+1)\text{Hz}$
• C. $320(2n+1)\text{Hz}$
• D. $80(2n+1)\text{Hz}$

Answer 1

The correct option is B $160(2n+1)\text{Hz}$


From geometry,
$BP=\sqrt{{\left(AP\right)}^2+{\left(AB\right)}^2}$
From the data given in the question,
$BP=\sqrt{4^2+3^2}=5\text{m}$
Here,
Path difference at $P$, $\Delta =5-4=1\text{m}$

For minimum sound intensity,
Path difference = $(2n+1)\frac{\lambda }{2}$ [Where, $n=0,1,2,\mathrm{3.....}$]

$\Rightarrow 1=(2n+1)\frac{\lambda }{2}$

$\Rightarrow \lambda =\frac{2}{2n+1}$

$\Rightarrow \frac{v}{f}=\frac{2}{2n+1}[\because v=f\lambda ]$

$\therefore f=\frac{2n+1}{2}\times v=\frac{2n+1}{2}\times 320$

$\Rightarrow f=160(2n+1)\text{Hz}$