# Two in-phase loudspeakers emit identical 1000 Hz sound waves along

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 1.20 times that of each speaker alone?
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temnimam2
Solution:
Consider the formula to findout the wavelength is,
$\lambda =\frac{v}{f}$
Substute for v and 1000Hz for f in equation

Consider the formula to find the amplitude of the wave formed due to two waves with same frequency
$A=2a\mathrm{cos}\left(\frac{\mathrm{△}\varphi }{2}\right)$
Here, $\mathrm{△}\varphi =2\pi \frac{\mathrm{△}x}{\lambda }$
Therefore,
$A=2a\mathrm{cos}\left(\frac{\left(2\pi \frac{\mathrm{△}x}{\lambda }\right)}{2}\right)$
Substitute $\left(1.20×a\right)$ for A and 0.343 m for $\lambda$ in equation
$1.20a=2a\mathrm{cos}\left(\frac{\left(2\pi \frac{\mathrm{△}x}{\lambda }\right)}{2}\right)$
$\mathrm{△}x=\frac{0.343m}{\pi }{\mathrm{cos}}^{-1}\left(\frac{1.20}{2}\right)$

Thus, the distance that one speaker be placed behind the other for the sound to have an amplitude 1.20 times that of each speaker alone is $\mathrm{△}x=5.79$

ramirezhereva
I could not find the answer, thanks!