Two in-phase loudspeakers are placed along a wall and are separated by a distance of 4. 00 m. They emit sound with a frequency of 514 hz. A person is standing away from the wall, in front of one of the loudspeakers. What is the closest nonzero distance from the wall the person can stand and hear constructive interference? the speed of sound in air is 343 m/s.

Only when the path difference between the two interfering sound waves is an odd multiple of half of the sound wave's wavelength can constructive interference occur. The path difference for destructive interference will be the integer multiple of the sound's wavelength.

What is interference, exactly?

In physics, interference is the combined effect of two or more wave trains traveling in parallel or opposite directions. At each point that is affected by more than one wave, the effect is the addition of the amplitudes of the individual waves.

The distance of partition of the speaker is d=4.00m.

λ = v / f = 343 m/s / 514 Hz = 0.67 m.

Pitheagoras' theorem can be applied in the following manner as x, h, and the distance d between speakers define a square triangle:

(2) (h+x)(h-x) = d2 When we divide both sides in (3) and (1), we get: h2 = d2 + x 2 h2 – x2 = d2

We have: h + x = d2 / n (4) by subtracting both sides in (4) and (1).

2 x = (d2/n) - n = (d2 – n2 2) / 2 n (5) We must select the maximum value for n that is greater than or equal to x>0:

When we solve for x and substitute the values of d, and n in (5), we get: nmax = d / = 5.7 nmax = 5.

x = 16 – 25. (0.67)2 / 2.5.(0.67) = 0.729 m

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