Question

Two stereo speakers are separated by a distance of 2.4 m. A person stands at a distance of 3.2 m as shown directly in front of one of the speakers. The frequencies in audible range for which the listener will hear a minimum sound intensity are (Speed of the sound in air is $320m{s}^{-1}$)

• A. 160 $(2n+1)$
• B. 320 $(2n+1)$
• C. 200 $(2n+1)$
• D. 100 $(2n+1)$

Answer 1

The correct option is C 200 $(2n+1)$

Path $AP=3.2m$

Path $BP=\sqrt{(3.2{)}^2+(2.4{)}^2}=sqrt16=4m$

Path difference between AP and BP: $\Delta =BP-AP=4-3.2=0.8m$

For minimum sound intensity at a point path difference needs to be odd integer multiple of half wavelength.

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

$\therefore 0.8=(2n+1)\frac{1}{2}\frac{v_{sound}}{f_{sound}}$

$\Rightarrow f_{sound}=\frac{320}{1.6}(2n+1)=200(2n+1)$ where $n=0,1,2,\mathrm{3......}$