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

Given:
Distance between the two speakers d = 2.40 m
Speed of sound in air v = 320 ms⁻¹
Frequency of the two stereo speakers f = ?

As shown in the figure, the path difference between the sound waves reaching the listener is given by :

\[∆ x =  S_2 L -  S_1 L\]

\[∆ x = \sqrt{(3 . 2 )^2 + (2 . 4 )^2} - 3 . 2\]

Wavelength of either sound wave:

\[= \left( \frac{320}{f} \right)\]

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

\[\therefore    ∆ x = \frac{(2n + 1)\lambda}{2}\]

So,

\[\sqrt{(3 . 2 )^2 + (2 . 4 )^2} - 3 . 2 = \frac{(2n + 1)}{2}\left( \frac{320}{f} \right)\] 

\[ \Rightarrow \sqrt{16} - 3 . 2 = \frac{\left( 2n + 1 \right)}{2}\left( \frac{320}{f} \right)\] 

\[ \Rightarrow 0 . 8 \times 2f = \left( 2n + 1 \right) \times 320\] 

\[ \Rightarrow   1 . 6f = \left( 2n + 1 \right) \times 320\] 

\[ \Rightarrow f =   200(2n + 1)\]

On putting the value of n = 1,2,3,...49, the person can hear in the audible region from 20 Hz to 2000 Hz.