# Two speakers driven in a phase by the same amplifier at a frequency of 640 Hz, are 7 m apart. A...

## Question:

Two speakers driven in a phase by the same amplifier at a frequency of {eq}640 \ Hz {/eq}, are {eq}7 \ m {/eq} apart. A listener originally at the position of one of the speakers starts walking away from that speaker in a direction that is perpendicular to the line connecting the two speakers. How far do they have to walk to find their first sound minimum or destructive interference? The speed of sound is {eq}340 \ m / s. {/eq}

## The interference of sound waves from two synchronous sources as it moves away from them.

During the movement, the listener will observe a decrease in sound intensity to zero (destructive interference)

at those points where the difference in distance from the sources to these points R will be an odd number of half-waves.

## Answer and Explanation:

In this case

{eq}R=\sqrt{7^2+x^2}-x {/eq} (1)

x-distance between the listener and the first speaker from which he began to move.

Conditions for destructive interference

{eq}R=\lambda*(1/2+n),(2); \lambda=v/f=340/640m=0.531m - wavelength; n=0,1,2,3...; v,f-\text{speed and frequency of sound} {/eq}

According to (1) and (2), we get

{eq}7^2+x^2=(x+0.531(1/2+n))^2=x^2+(0.531)^2*(1/2+n)^2+2x*0.531*(1/2+n) {/eq} (3)

After simplification and transformations of expression (3), we get

{eq}x=(49-(0.531)^2*(1/2+n)^2)/(2*0.531(1/2+n)) {/eq} (4)

Since the conditions of the problem, x>0,

then the numerator in the fraction (4) more than zero

and therefore

{eq}n<7/0.531-1/2=12.68 {/eq}

So, the first sound minimum (n=12) will be if

{eq}x=(49-0.281*156.25)/13.275=0.38m {/eq}

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from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5