Question

Two stereo speakers are separated by a distance of 2.40 m. A person stands at a distance of 3.20 m directly in front of one of the speakers as shown in figure. Find the frequencies in the audible range (20-2000 Hz) for which the listener will hear a minimum sound intensity. Speed of sound in air = 320 m s−1.

Answer 1

Explanation:

Two stereo speakers are separated by a distance of 2.40 m. A person stands at a distance of 3.20 m directly in front of one of the speakers as shown in figure (16-E3). Find the frequencies in the audible range (20-2000 Hz) for which the listener will hear a minimum sound intensity. Speed of sound in air = 320 m/s. Read more on Sarthaks.com - https://www.sarthaks.com/43433/two-stereo-speakers-are-separated-by-a-distance-of-2-40-m-a-person-stands-at-a-distance-of-3-20

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Answer 2

The frequencies in the audible range (20-2000 Hz) for which the listener will hear a minimum sound intensity is $f=\frac{(2 n+1) 320}{1.6}$

Explanation:

Gap between the two speakers d = 2.40 m

Speed of the sound in air v = 3.20 m/sec

Two Stereo speaker frequency f =?

As shown in the figure, the difference in trajectory between the sound waves that enter the listener is given by:

$\Delta x=\sqrt{3.2^{2}+2.4^{2}}-3.2$

Every sound wave has wavelength:

$=\frac{320}{f}$

We know that destructive interference will occur if the path difference is an odd integral multiple of the wavelength.

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

So,  

$\sqrt{3.2^{2}+2.4^{2}}-3.2=\frac{(2 n+1) \lambda}{2}$

$\sqrt{3.2^{2}+2.4^{2}}-3.2=\frac{(2 n+1)}{2} \frac{320}{f}$

$\sqrt{10.24+5.76}-3.2=\frac{(2 n+1)}{2} \frac{320}{f}$

$\sqrt{16}-3.2=\frac{(2 n+1)}{2} \frac{320}{f}$

$4-3.2=\frac{(2 n+1)}{2} \frac{320}{f}$

$0.8=\frac{(2 n+1)}{2} \frac{320}{f}$

$1.6 f=(2 n+1) 320$

$f=\frac{(2 n+1) 320}{1.6}$

The person will hear in the audible region from 20 Hz to 2000 Hz if he puts the value of $n=1,2,3, \dots 49$