Question

Two in-phase loudspeakers emit identical 1000 Hz sound waves along the x-axis.

What distance should one speaker be placed behind the other for the sound to have an amplitude 1.90 times that of each speaker alone?

Answer 1

hat depends on the speed of sound, which is not given.
Let's say we have 20°C, then speed of sound is c = 343m/s

The wavelength of the sound wave is:
λ = c/f = 343m/s / 1000s⁻¹ = 0.343m

First speaker emits a sinusoidal sound wave propagating along the x-axis. The displacement of the wave is can be written as:
y₁(x,t) = ŷ∙sin( 2∙π∙(x/λ - f∙t) + φ)
ŷ is the amplitude φ the phase

Let the second speaker stand ∆x behind the first. It emits similar waves
shifted by ∆x. So replace x in equation above x+∆x an you get the equation for this wave.
y₂(x,t) = ŷ∙sin( 2∙π∙((x+∆x)/λ - f∙t) + φ)

According to principle of superposition the two waves add up to form the resulting sound wave.
y = y₁ + y₂
= ŷ∙ [ sin( 2∙π∙(x/λ - f∙t) + φ) + sin( 2∙π∙((x+∆x)/λ - f∙t) + φ) ]
use sum to product identity to simplify
sin(a) + sin(b) = 2∙cos((a-b)/2)∙sin((a+b)/2)
=>
y = 2∙ŷ ∙ cos(π∙∆x/λ) ∙ sin( 2∙π∙(x/λ - f∙t) + π∙∆x/λ + φ)

The sine-term represents the propagation of the wave. The cosine is constant factor. Hence 2∙ŷ ∙ cos(π∙∆x/λ) is the amplitude of the wave.

The second speaker ought to be placed so, that this amplitude is 1.7 time that of each speaker. That means
2∙ŷ ∙ cos(π∙∆x/λ) = 1.9∙ ŷ
solve for ∆x
∆x = λ ∙arcos(1.9/2) / π
= 0.343m ∙arcos(0.95) / π
= 0.0347m
= 3.47cm