Question

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 distance from the wall the person can stand and hear constructive interference? The speed of sound in air is 343 m/s.
Multiple choice:
1.64 m
1.15 m
0.344 m
0.729 m

Answer 1

Answer: 0.729 m

Explanation:

In order to be able to hear a constructive interference, the difference between the distances from both speakers to the person must be a multiple of the wavelength of the sound.

We can find out the wavelength as we know the value of the frequency and the sound speed, as follows:

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

So, if we call x to the distance perpendicular to the wall, right front of the left speaker, and h to the distance to the right speaker, we can write the following expression:

h – x = n λ (1)

As x, h, and the distance d between speakers determine a square triangle, we can apply Pitheagoras ‘theorem, as follows:

h2 = d2 + x 2

h2 – x2  = d2 (2)

(h+x)(h-x) = d2 (3)

Dividing both sides in (3) and (1), we get:

h + x = d2 / n λ (4)

Substracting both sides in (4) and (1), we have:

2 x = (d2/ n λ) - n λ = (d2 – n2 λ2) / 2 n λ  (5)

We must choose the maximum value for n that satisfies x>0, as follows:

nmax = d / λ = 5.7 → nmax = 5

Solving for x, and replacing for the values of d, λ and n in (5), we get:

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

Explanation: