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Two loudspeakers are 2.5 m apart. A person stands 2.5 m from one speaker and 4.0 m from the other. Let T = 20°

Two loudspeakers are 2.5 m apart. A person stands 2.5 m from one speaker and 4.0 m from the other. Let T = 20°C. (a) What is the lowest frequency at which destructive interference will occur at this point? (b) Calculate two other frequencies that also result in destructive interference at this point (give the next two lowest). (second lowest) & (3rd lowest)

Public Comments

1. The lowest frequency at which destructive interference could occur is if 1/2 a wavelength = 1.5 m (the difference between the two) The speed of sound can be found with the following formula v = 331 m/s + (0.6 m/s/C)•T So at 20C you get a speed of 343 m/s. We know that (1/2)*lambda = 1.5 m, so lambda = 3.0m The wavelength must be 3 m. Using the universal wave equation v = f*lambda f = v/lambda f = (343m/s)/(3 m) = 114 Hz The lowest freq would be 114 Hz The next two lowest would be a 3lambda/2 = 1.5 and 5lambda/2 = 1.5 so lambda = 1m and lambda = 0.6m Those give f = 343 Hz and 572 Hz