Question

Two loudspeakers are separated by a distance of 7.6 m. A listener sits directly in front of one speaker at a distance of 6.6 m so that the two speakers and the listener form a right triangle. Find the lowest frequency for which the path difference from the speakers to the listener is an odd number of half-wavelengths. Assume the speed of sound is 340 m/s.

Find the second lowest frequency for which the path difference from the speakers to the listener is an odd number of half-wavelengths.

Answer 1

Given that :

distance between two speakers, d = 7.6 m

using an equation,

d sin = (1/2) (2m - 1)                                                           { eq.1 }

where, = angle between paths = tan^-1 (6.6 m / 7.6m) = 40.1 degree

(a) The lowest frequency for which the path difference from the speakers to the listener is an odd number of half-wavelengths which is given as :

For m = 1, we have

f = v /    { eq.2 }

where, v = speed of sound = 340 m/s

= (343 m/s) / f

from eq.1,     d sin = (1/2) (2 x 1 - 1)

d sin = (0.5)                                                            { eq.3 }

inserting the value of '' in eq.3,

d sin = (0.5) (343 m/s) / f

f = (0.5) (343 m/s) / [d sin θ]                                                  { eq.4 }

inserting the values in eq.4,

f = (0.5) (343 m/s) / [(7.6 m) sin 40.1⁰]

f = (171.5 m/s) / (4.89 m)

f₁ = 35.1 Hz

For m = 2, we have

from eq.1,      d sin = (1/2) (2 x 2 - 1)

d sin = (1.5)                                                            { eq.5 }

inserting the value of '' in eq.5,

d sin = (1.5) (343 m/s) / f

f = (1.5) (343 m/s) / [d sin θ]                                                  { eq.6 }

inserting the values in eq.6,

f = (1.5) (343 m/s) / [(7.6 m) sin 40.1⁰]

f = (514.5 m/s) / (4.89 m)

f₂ = 105.2 Hz