(a) A source of sound `S` and a detector `D` are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line `SD` as shown in figure. It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of `20 cm.` Find the frequency of the sound emitted. Velocity of sound in are is `336 m//s.`

(b) A source emitting sound of frequency `180 Hz` is placed in front of a wall at a distance of `2 m` from it. A detector is also placed in front of the wall at the same distance from it. find the minimum distance between the source and the detector for which the detector detects a maximum of sound . Speed of sound in air `= 360 m//s.`

Explanation

(a)
`Delta x = (S DP + PD) - SD = n lambda (i)`
`Delta x' = (S DP + 2x + PD) - SD = (2n + 1) (lambda)/(2) (ii)`
`(ii) - (i)`
`2x = (lambda)/(2) = (v)/(2f)`
`f = (v)/(4x) = (336)/(4 xx 0.2) = 420 Hz`
(b)
Path difference `Delta x = SPD - SD = 2 SP - SD`
`= [2(2^(2) + x^(2)/(4))^(1//2) - x ]`
For maximum intensity
`Delta x = n lambda = (nv)/(f)`
`[2(2^(2) + x^(2)/(4))^(1//2) -x] = n((360)/(180))`
For minimum value of `x,n = 1`
`[2(4 + x^(2)/(4))^(1//2) -x] = 2`
`2(4 + x^(2)/(4))^(1//2) = x + 2 rArr (4 + x^(2)/(4))^(1//2)`
`= (1 + x/(2))`
Squaring both sides
`4 + (x^(2))/(4) = 1 + (x^(2))/(4) + x`
`x = 3 m`