Two 80-W loudspeakers are located 8 meters apart on the stage of an auditorium.

Question

Two 80-W loudspeakers are located 8 meters apart on the stage of an auditorium. A listener is seated 40m from one and 48m from the other. A signal generator drives the
two speakers in phase with the same amplitude and frequency. The frequency is swept through the audible range from 20-20,000 Hz.
A)Assuming the amplitudes of the waves at the listener are essentially the same, what are the two highest frequencies at which the listener will hear a minimum signal because of destructive interference?
B)What is the intensity of the sound at the listener from the 40m distant speaker only, assuming the speaker is a point source?

A) I got the frequencies to be 465 and 463. m=465.9 using destructive and using 20k Hz for f.
B) I need help on, not sure what to use.

Explanation / Answer

I believe your part A works out fine, part B is actually the easier part.

You'll need the equation for Intensity, which is I=P/A

P is Power and is given as 80-W.

A is your area. Since we are assuming a point source, the sound goes out in all directions, creating a sphere. You will use the area of a sphere for A, which is A= 4r2

Your radius will be 40 m since the total radius of the created sphere at that point would be 80 m.

You'll be left with 80W/4(40m)2

This should give you about 3.98 mW/m2

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