Question

Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. Part A What is the lowest possible frequency of sound for which this is possible

Answer 1

Answer:

By the Pythagorean Theorem the distances from the speakers os

5 and 5.5 (rounding) meters   - let y be the wavelength in the solution

n y = 5      n is number of wavelengths from speaker

(n + m) y = 5.5      m must be integral for constructive interference

m y = .5       subtracting equations

m = 2      and y = ,25   for the above conditions

(n + 2) y = 5.5     substituting for m

n = 5.5 / .25 - 2 = 20

f = v / y     using frequency of sound

f = 340 / .25 = 1360 / sec    for lowest frequency

Check: D1 = y n  = ,25 * 20 = 5

and D2 = .25 * 22 = 5.5     for the distances traveled