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To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d = 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .

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okpalawalter8

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  [tex]  v  =  350 \  m/s [/tex]  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by [tex]k  =  60 cm  =  0.6 \  m[/tex]

    The frequency of both speakers is [tex]f =  700 \  Hz[/tex]

Generally the distance of the listener to the first speaker is mathematically represented as

       [tex]L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}[/tex]

       [tex]L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}[/tex]

        [tex]L_1  =   5.39 \  m [/tex]

Generally the distance of the listener to second speaker at its new position is  

          [tex]L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}[/tex]

       [tex]L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}[/tex]

        [tex]L_2  =   5.64  \  m [/tex]  

Generally the path difference between the speakers is mathematically represented as

        [tex]pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}[/tex]

Here [tex]\lambda[/tex] is the wavelength which is mathematically represented as

         [tex]\lambda =  \frac{v}{f}[/tex]

=>    [tex] L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}[/tex]

=>    [tex] L_2 - L_1  =  \frac{n  *  v}{2f}[/tex]  

=>    [tex] L_2 - L_1  =  \frac{n  *  v}{2f}[/tex]  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    [tex]  5.64 - 5.39   =  \frac{1  *  v}{2*700}[/tex]      

=>    [tex]  v  =  350 \  m/s [/tex]  

Ver imagen okpalawalter8
oladayoademosu

The speed of sound in air is 350 m/s

Since the distance between both speakers is 4 m, and the listener is standing 5 m away from halfway between them, the distance L from each loudspeaker to the listener, since the arrangement forms a right-angled triangle, using Pythagoras' theorem,

L = √[(5 m)² + (4/2 m)²]

= √[25 m² + (2 m)²]

= √[25 m² + 4 m²]

= √29 m² = 5.39 m.

Now, when one speaker is moved 60 cm = 0.6 m away from its original position, its distance from the listener is now

L' = √[(5 m)² + (4/2 + 0.6 m)²]

= √[25 m² + (2 m + 0.6 m)²]

= √[25 m² + (2.6 m)²]

= √[25 m² + 6.76 m²]

= √31.76 m²

= 5.64 m.

Now, the path difference when we first have destructive interference is

ΔL = L' - L

= 5.64 - 5.39

= 0.25

Since we have destructive interference for the first time when the speaker is moved, the path difference, ΔL = (n + 1/2)λ where λ = wavelength = v/f where v = speed of sound in air and f = frequency = 700 Hz.

Now, since we have destructive interference for the first time, n = 0.

So,  ΔL = (n + 1/2)λ

ΔL = (0 + 1/2)v/f

ΔL = v/2f

Making v subject of the formula, we have

v = 2fΔL

Substituting the values of the variables into the equation, we have

v = 2fΔL

v = 2 × 700 Hz × 0.25 m

v = 350 m/s

So, the speed of sound in air is 350 m/s

Learn more about interference of sound here:

https://brainly.com/question/1346741

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