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Interference Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 12.0 m to the right of speaker A. The frequency of the waves emitted by each speaker is 688 Hz. You are standing between the speakers, along the line connecting them, and are at a point of constructive interference. How far must you walk toward speaker B to move to a point of destructive interference

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Answer:

0.125 m

Explanation:

From wave equations, we know that;

Wavelength;λ = f/v

Where v is speed of sound in air with a constant value of 344 m/s

Frequency;f = 688 Hz

λ = v/f

λ = 344/688

λ = 0.5 m

Constructive interference will occur when the two waves differ by a distance of nλ and when they have the same wavelength.

At constructive interference point, my distance from both speakers will be x.

Since the distance between speaker A and B is 12 m, then;

difference in path is;

nλ = 12 - x - x

nλ = 12 - 2x

Making x the subject, gives;

x = 6 - nλ/2

Plugging in 0.5 for λ gives;

x = 6 - n(0.5)/2

x = 6 - n/4

So, since it's at point 1,we'll label it accordingly.

Thus;

x_1 = 6 - 0.25(n_1)

Now, we know that destructive interference will occur when the two waves differ by a distance of;(½ + n)λ and when they have the same wavelength.

Thus,

12 - 2x = (½ + n)λ

Plugging in 0.5 for λ to give;

12 - 2x = ¼ + n/2

Divide through by 2 to give;

6 - x = ⅛ + n/4

x = 6 - ⅛ - n/4

Since second point, then;

x2 = 5.875 - 0.25(n_2)

To find out how far I must walk toward speaker B to move to a point of destructive interference, it will be;

x1 - x2 which gives;

6 - 0.25(n_1) - 5.875 + 0.25(n_2)

This gives;

0.125 - (0.25(n_1) - 0.25(n_2))

This means that 0.125 m is the distance since I have to subtract difference of 0.25 multiplied by the difference of n_1 and n_2 from 0.125

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