Asked by josh
                Two identical loudspeakers are 2 m apart. A person stands 4.6 m from one speaker and 3.4 m from the other. What is the lowest frequency at which destructive interference will occur at this point?
            
            
        Answers
                    Answered by
            drwls
            
    Destructibe interference will occur if the difference between the distances to the two speakers is n + 1/2, where n is an integer zero or higher. When n = 0, you get the longest wavelength and lowest frequency with destructive interference.
Therefore 4.6 - 3.4 = 1.2 = (wavelength)/2
wavelngth = 2.4 m
freq
    
Therefore 4.6 - 3.4 = 1.2 = (wavelength)/2
wavelngth = 2.4 m
freq
                    Answered by
            drwls
            
    Destructive interference will occur if the difference between the distances to the two speakers is n + 1/2, where n is an integer equal to zero, or higher. When n = 0, you get the longest wavelength and lowest frequency with destructive interference.
Therefore 4.6 - 3.4 = 1.2 = (wavelength)/2
wavelength = 2.4 m
frequency = (speed of sound)/2.4 m
= 140 Hz (approx.)
    
Therefore 4.6 - 3.4 = 1.2 = (wavelength)/2
wavelength = 2.4 m
frequency = (speed of sound)/2.4 m
= 140 Hz (approx.)
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