Two loudspeakers are 2.63 m apart. I am standing 1.55 m from one of them and 3.5

Question

Two loudspeakers are 2.63 m apart. I am standing 1.55 m from one of them and 3.58 m from the other. The two speakers are driven by a single oscillator. If the frequency of the oscillator is swept from 100 Hz to 1000 Hz, find the lowest frequency (Hz) at which I will hear an enhancement of the sound intensity due to constructive interference of the waves from the two speakers. Use 343 m/s for the speed of sound. (DO NOT use the double-slit equation, d sin ? = m?. This equation is only valid for observations far from the two slits, compared to the distance between the two slits.)

Explanation / Answer

path diff = 3.58- 1.55 = 2.03
the lowest frequency when = path diff = 2.03

frequency = V /

= 343 / 2.03

* the lowest frequency = 168.96 169 Hz

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