Respuesta :
Answer:
Explanation:
Given
Distance between two loud speakers [tex]d=1.6\ m[/tex]
Distance of person from one speaker [tex]x_1=3\ m[/tex]
Distance of person from second speaker [tex]x_2=3.5\ m[/tex]
Path difference between the waves is given by
[tex]x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}[/tex]
for destructive interference m=0 I.e.
[tex]x_2-x_1=\frac{\lambda }{2}[/tex]
[tex]3.5-3=\frac{\lambda }{2}[/tex]
[tex]\lambda =0.5\times 2[/tex]
[tex]\lambda =1\ m[/tex]
frequency is given by
[tex]f=\frac{v}{\lambda }[/tex]
where [tex]v=velocity\ of\ sound\ (v=343\ m/s)[/tex]
[tex]f=\frac{343}{1}=343\ Hz[/tex]
For next frequency which will cause destructive interference is
i.e. [tex]m=1[/tex] and [tex]m=2[/tex]
[tex]3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda[/tex]
[tex]\lambda =\frac{1}{3}\ m[/tex]
frequency corresponding to this is
[tex]f_2=\frac{343}{\frac{1}{3}}=1029\ Hz[/tex]
for [tex]m=2[/tex]
[tex]3.5-3=\frac{5}{2}\cdot \lambda [/tex]
[tex]\lambda =\frac{1}{5}\ m[/tex]
Frequency corresponding to this wavelength
[tex]f_3=\frac{343}{\frac{1}{5}}[/tex]
[tex]f_3=1715\ Hz[/tex]
The expression for destructive interference allows to find the results for the frequencies of destructive interference are:
a) f = 343 Hz
b) f = 1029 Hz and f = 1715 Hz
The phenomenon of interference, two sources in phase emit sound, the difference in top between these sounds to an explicit point determines the perceived intensity, we have two types of interference:
- Constructive. When the path difference is an integer number of wavelengths and a maximum is heard.
- Destructive. When the path difference is a number of half wavelengthmeasurements and at this point no sound is heard.
Δr = (2n + 1) [tex]\frac{\lambda}{2} \ \ \ n= 0, 1, 2, ...[/tex]
where Δr is the path difference, λ the wavelength and n an integer.
They indicate the length from each speaker to the listener is r₁ = 3 m and r₂=3.5 m, see attached.
a) The path difference is
Δr = r₂-r₁
Δr = 3.5-3 = 0.5 m
The wavelength of the interference is
λ = [tex]\frac{2 \Delta r}{(2n+1)}[/tex]
let's find the wavelength, for the first destructive interference m = 0
λ = 2 Δr
λ = 2 0.5
λ = 1 m
The speed of the wave is related to its wavelength and its frequency.
v = λ f
f = [tex]\frac{v}{\lambda }[/tex]
the speed of sound is v = 343 m/s
Let's calculate the frequency.
f = 343/1 = 343 Hz
b) let's find the wavelengths for the following two destructive interferences.
m = 1
λ = [tex]\frac{ 2 \Delta r }{(2n+1)}[/tex]
λ = [tex]\frac{2 \ 0.5 }{( 2 \ 1 + 1)}[/tex]
λ = ⅓ = 0.333 m
Let's look for the frequency.
f = [tex]\frac{343}{0.33333}[/tex]
f = 1029 Hz
m = 2
λ = [tex]\frac{2 \ 0.5}{(2 \ 2 +1)}[/tex]2 0.5 / 2 2 +1
λ = 1/5 = 0.2 m
f = [tex]\frac{343 }{0.2}[/tex]
f = 1715 Hz
In conclusion using the expression for destructive interference we can find the results for the frequencies of destructive interference are:
a) f = 343 Hz
b) f = 1029 Hz and f = 1715 Hz
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