Respuesta :
Answer:
distance from speaker is 17.87 m
Explanation:
given data
distance from loudspeaker = 2.5 m
distance between loudspeaker = 3.0 m
room temperature = 20c
wavelength f = 686Hz
to find out
what distances from the speaker
solution
we know sound velocity c = 331.5 + 0.6 × 20c = 343.5
so wavelength of sound λ = c / f
wavelength = 343.5 / 686 = 0.5 m
when the difference in distance of speaker destructive interference will be
d = λ/2 × (2n-1)
for n = 1, 2 3 4 ..
d = 0.5/2 × (2n-1)
d = 0.250 , 0.75 , 1.25 , 1.750............ for n = 1, 2 3 .............
so
for d = 0.250
side of triangle by hypotenuse of triangle are
[tex]\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1)[/tex] = 0.250
0.5 x1 = 7.6875
x1 = 15.375 m
for d = 0.75
side of triangle by hypotenuse of triangle are
[tex]\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2)[/tex] = 0.75
1.5 x2 = 4.6875
x2 = 3.125 m
for d = 1.250
side of triangle by hypotenuse of triangle are
[tex]\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3)[/tex] = 1.250
2.5 x2 = 1.1875
x3 = 0.475 m
for d = 1.750
x4 will be negative so we stop here
so the distance from speaker here is given below
distance = 2.5 + x
here x = 0.475 , 3.125 and 15.375 so
distance 1 = 2.5 + 0.475 = 2.975 m
distance 2 = 2.5 + 3.125 = 5.625 m
distance 3 = 2.5 + 15.375 = 17.875 m
final distance from speaker is 17.87 m
The minimum distances from the speaker, where the observer hear minimum sound intensity, is 3.8 metres.
What is the intensity of the sound?
Intensity of the sound is the intensity flowing per unit area which is perpendicular to the direction of sound wave travelling in.
The intensity of the sound wave is inversely proportional to the distance of the object and source of sound wave.
The room temperature is 20 c. The speed of the light at the 20 ⁰C is equal to the 343.5 m/s.
The frequency of the sound for both the loudspeaker is 686 Hz. As, the wavelength of the wave is the ratio of speed of light of the frequency of wave. Thus, the wavelength of the loudspeaker sound wave is,
[tex]\lambda=\dfrac{343.5}{686}\\\lambda=0.5\rm m[/tex]
The distance between the person and the first loudspeakers is 2.5 m and the distance between the first and second loudspeakers is 3.0 m.
As the observer, begin to walk directly away from the speaker. Thus the distance between them increases. The path difference from the initial distance can be given as,
[tex]\Delta L=\sqrt{3^2+2.5^2}-2.5\\\Delta L=1.4\rm m[/tex]
The minimum sound intensity can be given as,
[tex]\Delta L=\dfrac{2N+1}{2}\times \lambda\\\Delta L=\dfrac{2N+1}{2}\times (0.5)[/tex]
Put the value of N equal to 1 the value of path difference we get as 0.75 in the above equation.
Let the minimum distances from the speaker, where the observer hear a minimum sound intensity is L. Therefore,
[tex]\sqrt{0.25^2+L^2}-L=0.75\\L=3.8[/tex]
Hence, the minimum distances from the speaker, where the observer hear minimum sound intensity, is 3.8 metres.
Learn more about the sound intensity here;
https://brainly.com/question/14261338
