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 .

Complete Question

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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]  

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

See Conceptual Example 6 to review the concepts involved in this problem. A 12.0-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 86.4 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 1.33 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey's arm.

A student must determine the relationship between the inertial mass of an object, the net force exerted on the object, and the object’s acceleration. The student uses the following procedure. The object is known to have an inertial mass of 1.0kg .

Step 1: Place the object on a horizontal surface such that frictional forces can be considered to be negligible.

Step 2: Attach a force probe to the object.

Step 3: Hang a motion detector above the object so that the front of the motion detector is pointed toward the object and is perpendicular to the direction that the object can travel along the surface.

Step 4: Use the force probe to pull the object across the horizontal surface with a constant force as the force probe measures force exerted on the object. At the same time, use the motion detector to record the velocity of the object as a function of time.

Step 5: Repeat the experiment so that the object is pulled with a different constant force.

Can the student determine the relationship using this experimental procedure?

Answer choices:

A) Yes, because Newton’s second law of motion must be used to determine the acceleration of the object.

B) Yes, because the net force exerted on the object and its change in velocity per unit of time are measured.

C) No, because the motion detector should be oriented so that the object moves parallel to the line along which the front of the motion detector is aimed.

D) No, because knowing the net force exerted on the object and its change in velocity per unit of time is not sufficient to determine the relationship.