In the figure, two loudspeakers, separated by a distance of d;at the position of a listener, who is d, = 3.81 m directly in front of one of the speakers. Consider the audible range for normal hearing, 20 Hz to 20 kHz. (a) What is thelowest frequency that gives the minimum signal (destructive interference) at the listener's ear? (b) What is the lowest frequency that gives the maximum signal(constructive interference) at the listener's ear? (Take the speed of sound to be 343 m/s.)2.74 m, are in phase. Assume the amplitudes of the sound from the speakers are approximately the same%DdiSpeakersListener(a) NumberUnits(b) NumberUnits

Answered: In the figure, two loudspeakers,…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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In the figure, two loudspeakers, separated by a distance of d;
at the position of a listener, who is d, = 3.81 m directly in front of one of the speakers. Consider the audible range for normal hearing, 20 Hz to 20 kHz. (a) What is the
lowest frequency that gives the minimum signal (destructive interference) at the listener's ear? (b) What is the lowest frequency that gives the maximum signal
(constructive interference) at the listener's ear? (Take the speed of sound to be 343 m/s.)
2.74 m, are in phase. Assume the amplitudes of the sound from the speakers are approximately the same
%D
di
Speakers
Listener
(a) Number
Units
(b) Number
Units

Expert Solution

Step 1

Given:

$d_{1} = 2.74 m d_{2} = 3.81 m v = 343 m / s$

Distance between speaker and listener be d₃:

$d_{3} = \sqrt{{(d_{1})}^{2} + {(d_{2})}^{2}} = \sqrt{{(2.74)}^{2} + {(3.81)}^{2}} = \sqrt{7.5 + 14.5} = \sqrt{22} = 4.7 m$

$∆ L = d_{3} - d_{2} = 4.7 - 3.81 = 0.89 m$

$v_{s o u n d} = f λ$

(a)

For destructive interference,

$\frac{∆ L}{λ} = 0.5, 1.5, 2.5 \frac{∆ L}{v_{s o u n d} / f_{m i n}} = 0.5, 1.5, 2.5 f_{m i n} = \frac{v_{s o u n d}}{∆ L} \times 0.5, 1.5, 2.5 = \frac{343}{0.89} \times 0.5, 1.5, 2.5$

Lowest frequency that gives minimum signal at listener ear;

$f_{m i n, 1} = \frac{343}{0.89} \times 0.5 = 192.7 H z$

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