Question

The two speakers are placed 37.2 cm apart. A single oscillator makes the speakers vibrate in phase at a frequency of 2.02 kHz. At what angles, measured from the perpendicular bisector of the line joining the speakers, would a distant observer hear maximum sound intensity? Minimum sound intensity? (Take the speed of sound to be 340 m/s.)

maximum intensities: (List smallest angle first.)

θ1max	=  °
θ2max	=  °
θ3max	=  °

minimum intensities: (List smallest angle first.)

θ1min	=  °
θ2min	=  °

Answer 1

from the given

   d = 37.2 cm

      = 0.372 m

   f = 2.02 kHz

     = 2020 Hz

   v = 340 m / s

   from the theory we know that

   v = f λ

   wavelength will be

   λ = v / f

  = (340 m / s) / (2020Hz)  

λ = 0.1683 m

   we have the equation

   d sin θ = m λ

   so the angle for maxima is

   θ = sin^-1[m(λ / d)]

   (λ / d) = 0.1683/0.372= 0.4524

   where m = 0, 1, 2, ........

     θ = sin^-1[m (λ / d)]

    First we find

      θ = sin^-1(0.4524) =26.897˚

   The angle for minima is

   θ = sin^-1[(2 m + 1) (λ / 2d)]

= sin^-1[(2 m + 1) (0.1683) / (2x0.372)]

   solve for m = 0 and 1

m=0,

θ= sin^-1[(0.1683) / (2x0.372)]

=13.07 ˚

m=1,

= sin^-1[(2 + 1) (0.1683) / (2x0.372)]

θ =42.73 ˚