A source emitting sound of frequency 180 Hz is alced in front of a wall at a distance of 2 m from it. A detector is also placed in front of the weal 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 `=360ms^-1`.

Answer

3m

Explanation

For a maxima minimum path difference should be `lambda`

`lambda = (v)/(n) = (360)/(180) = 2m`
If S and D are separted by d, we use path difference
`Delta = 2sqrt(4 + (d^2)/(4)) -d`
for maxima wa use
`Delta = lambda = 2m`
`2sqrt(4+ (d^2)/(4)) -d = 2`
`4(4+ (d^(2))/(4)) = 4 + d^(2) - 4d`
16 = 4 - 4 d
d = 3
New path difference after displacing the obstacle should become
`Delta = (3lambda)/(2)`

`Delta = 2sqrt((1.5)^(2) + (2 + x )^(2)) - 3 = 3`
`4(2.25 + 4 + x^2 + 4x)` = 36
`x^(2) + 4x - 2.75 = 0`
`x = (- 4 + sqrt(16 +11))/(2) ` = 0.598.