Driven by the same amplifier system, two audio speakers are positioned some distance apart. A person is seated 6.0 m from one speaker and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz , determine the frequencies that produce destructive interference at the listener's location. The speed of sound in air is 320 m s ⁻¹ .

Step-by-step solution

The path difference between the sound waves reaching the listener is x = 6.4 m - 6.0 m = 0.4 m . For destructive interference, the path difference must be an odd multiple of half the wavelength: x = (2n - 1) 2 where n = 1, 2, 3, Using the relation = v f , we get: x = (2n - 1) v 2f Substituting the given values x = 0.4 m and v = 320 m s ⁻¹ : 0.4 = (2n - 1) 320 2f f = 160 0.4 (2n - 1) = 400(2n - 1) Hz Substituting integer values for n : For n = 1 , f = 400 Hz For n = 2 , f = 1200 Hz For n = 3 , f = 2000 Hz For n = 4