Loudspeakers Two
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Two in-phase loudspeakers emit identical 1000 Hz sound waves along the x-axis.What distance should one speaker?
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 1.70 times that of each speaker alone
That 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) = 1.7∙ ŷ
solve for ∆x
∆x = λ ∙arcos(1.7/2) / π
= 0.343m ∙arcos(0.85) / π
= 0.0606m
= 6.06cm