Two identical coherent sound sources R and S with frequency f are 5   m apart. An observer standing equidistant from the source and at a perpendicular distance of 12   m from the line R S hears maximum sound intensity. When he moves parallel to R S , the sound intensity varies and is a minimum when he comes directly in front of one of the two sources. Then, a possible value of f is close to (the speed of so

Step-by-step solution

For a minima at P , path difference of sounds reaching P must be an odd multiple of half wavelength. So, S P - R P = 2 n + 1 λ 2 . . . ( i ) where, n = 0 , 1 , 2 , 3 … From above figure, S P = R P 2 + R S 2 S P = 12 2 + 5 2 = 13 So, path difference, S P - R P = 13 - 12 = 1 m Hence, from Eq. ( i ) , 1 = 2 n + 1 v 2 f or frequency of sound, f = 2 n + 1 2 v Possible values of frequency of sound are for n = 0 , f 1 = v 2 = 330 2 = 165 Hz For n = 1 , f 2 = 3 v 2 = 495 Hz , … … , etc. Hence, option ' a ' matches with f 2