Two stereo speaks `S_(1) and S_(2)` are separated by a distance of `2.40 m`. A person (P) is at a distance of `2.40 m`. A person `(P)` is at a distance of `3.20 m` directly in front of one of the speakers as shown in Fig. 7.8. Find the frequencies in the audible range `( 20 - 20,000 Hz)` for which the listener will hear a minimum sound intensity . Speed of sound in air = 320 m//s`.

Explanation

Waves from `S_(1)` to `S_(2)` reach at point `P`.
Therefore , the path differnce between waves reaching from `S_(1)` to `S_(2)` will be `Delta = S_(2) P - S_(1) P`
Given ` S_(1)P = 3.20 m`
`S_(2) P = sqrt (( 2.40)^(2) + ( 3.20)^(2)) = 4.0 m`
`:. Delta = 4.0 - 3.20 = 0.80`
For minima , path difference
` Delta = ( 2 r - 1) ( lambda)/(2) , r = 1, 2, 3 , ....`
We have ` lambda = v//n` , where `n` is the frequency
`Delta = ( 2 r - 1 ) (v)/( 2 n)`
`0.80 = ( 2 r - 1) (v)/( 2n)`
Frequency `n = (( 2 r - 1) v)/( 2 xx 0.80) = ( 2r - 1) ( 320)/( 1.6) Hz = ( 2 r - 1) 200 Hz`
Therefore , the required frequencies are `n = ( 2 r - 1) xx 200` with ` r = 1, 2 , .... 50`.
Putting the values of ` r` in the above equation , we get
` n = 200 Hz , 600 Hz , 1000 Hz , 1400 Hz , 1800 Hz ,..... ,19800 Hz`