Two speakers emit sound in phase at a frequency that you can control. The speake

Question

Two speakers emit sound in phase at a frequency that you can control. The speakers and the receiver are arranged as shown in the diagram. They also emit sound equally in all directions and there are no echoes. For this problem assume that the speed sound is 340.0 m/s and that the density of air is 1.17[kg/m^3]. Find the lowest frequency for which there is destructive interference at the receiver. F = Suppose speaker two is turned off and speaker #1 emits total sound power of 0.100[w]. Find the intensity and the amplitude of the sound pressure wave at the receiver. I = p_max = Now assume speaker #2 is turned back on. The speakers are still emitting sound in phase, with the frequency you specified for destructive interference in part a, and each speaker emits 0.100[W] of sound power. Because the speakers are different distances away, the amplitude of the waves won't be equal at the receiver and there will still be a little sound at the receiver. Find the intensity of the sound at the receiver taking into account the difference in distance. I =

Explanation / Answer

Ans:- (a). Speed of the sound = 340 m/s

        1 Speaker distance = 50 m

        2 Speaker distance = 50 / cos 22degree = 50 / 0.927 = 53.94 m

     f = v / distance                          (frequency f , speed of sound v)

      1f = 340 / 50     = 6.8 Hz

      2f = 340 / 53.94 = 6.3 Hz

so the lowest frequency is speaker 2.

(b). sound power LW = 10 log( W / W0)

       the reference of sound power use is W0 = 10-12 Watt

      Lw = 10 log( 0.100 /10-12) = 110 watt

sound intensity = sound power / (4 pi R2)

sound intensity     = 110 / (4 *3.14 x 502)

sound intensity   = 110 / 31400 = 3.5 x 10-3 watt/m2

sound intensity level = L= 10 logritum ( I / I0 )

reference intensity defined as I0 = 10-12 W/m2.

sound wave is proportional to the square of the amplitude of the sound wave p2max/ p02 = I / I0. p0 is the negligible as compare to pmax

p2max = I / I0. = 3.82 X 10-4 / 10-12 = 3.82 x 108

pmax =3.82 x 108 = 1.95 x 104

(c) . Now Speaker 2 is turned back on

sound power LW = 10 log( W / W0)

       the reference of sound power use is W0 = 10-12 Watt

      Lw = 10 log( 0.100 /10-12) = 110 watt

sound intensity = sound power / (4 pi R2)

sound intensity     = 110 / (4 *3.14 x 53.942)

sound intensity   = 110 / 31400 = 3.0 x 10-3 watt/m2

destructive interference intensity I

the variable n ensures that we only add whole numbers of wavelengths.

I =I1 – I2                

I = 3.5 x 10-3 - 3.0 x 10-3

I = 0.5 X 10-3 so the difference of intensity.

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