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Two loudspeakers separated by 3.0 meters emit out-of-phase sound waves. Both speakers...
Two loudspeakers separated by 3.0 meters emit out-of-phase sound waves. Both speakers are playing a 686 Hz tone. If you are standing 4.0 meters directly in front of one of the loudspeakers. Do you hear maximum sound intensity, minimum sound intensity, or something in between? Justify.
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Let us suppose the loudspeaker A emits waves with initial phase zero, and loudspeaker B emits waves with initial phase `phi` . See the figure below. Thus they emit out of phase sounds.
`Y_A(x_1,t) = A*sin(2*pi*(x_1/lambda -t/T))` (1)
`Y_B(x_2,t) = A*sin(2*pi(x_2/lambda-t/T) +fi)` (2)
Because both loudspeakers emit the same tone `F =686 Hz` the period of both waves is `T =1/F =0.00146s =1.46 ms`
and the wavelength is `lambda = v*T =343*0.00146 =0.5 m`
(speed of sound in air at 20 degree Celsius is `v =343 m/s` )
The path difference between the two waves coming from A and B is
`P =(x_1-x_2)= sqrt(D^2+d^2) -D =sqrt(16+9) -4 =5-4 =1m`
Therefore the difference in paths is an integer multiple of lambda.
Now, looking at expressions (1) and (2), at the same moment in time there are three different cases:
a) if `phi =2*k*pi` with `k` integer (including zero) then a person at point C will hear a maximum sound intensity.
b) if `phi=(2*k+1)*pi/2` , with `k` integer (including zero) then a person at point C will hear a minimum of sound intensity
c) for other dephasing between the two loudspeakers the person at point C will hear an intermediate sound intensity between maxima and minima.
Answer: Unless special out of phase conditions of points a) and b) above, the person will hear something between maxima and minimum intensity.
Posted by valentin68 on October 16, 2013 at 3:03 PM (Answer #1)
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