Respuesta :
Answer:
Explanation:
To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.
For the wavelength;
Recall that, the separation between two successive max. and min. intensity points are [tex]\dfrac{\lambda}{2}[/tex]
Thus; for both speakers; the wavelength of the sound is:
[tex]\dfrac{\lambda}{2} = (10+30) cm[/tex]
[tex]\dfrac{\lambda}{2} = (40) cm[/tex]
λ = 80 cm
The relation between the path difference(Δx) and the phase difference(Δ∅) is:
[tex]\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o[/tex]
where;
Δx = 10 cm
λ = 80 cm
Δ∅ = π rad
∴
[tex]\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o[/tex]
[tex]\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o[/tex]
[tex]\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o[/tex]
[tex]\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o[/tex]
[tex]\Delta \phi_o = \pi -\dfrac{ \pi}{4}[/tex]
[tex]\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}[/tex]
[tex]\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad[/tex]
Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm
Thus, we have:
[tex]\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o[/tex]
[tex]\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad[/tex]
[tex]\Delta \phi = \dfrac{3 \pi}{4} \ rad[/tex]
∴
The amplitude of the sound wave if the two speakers are placed side-by-side is:
[tex]A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)[/tex]
[tex]A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)[/tex]
[tex]A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)[/tex]
A = 0.765a
The expressions for sound interference allows to find the amplitude for the wave when the two speakers are together is:
- The amplitud is: A = 0.765a
Given parameters
- Minimum interference speaker 2 behind speaker 1 is: Δr₁ = 10 cm
- Maximum interference haughty 2 in front of speaker 1 Δr₂ = 30 cm
To find
- The amplitude if the speakers are side by side.
The interference phenomenon occurs when two coherent waves have paths of different lengths to reach a point, we have two extreme cases:
- Constructive. When the waves arrive in phase, the expression is:
Δr = [tex]2n \ \frac{\lambda}{2}[/tex]
- Destructive. When the waves arrive with a phase difference of 180º, the expression is
Δr = [tex](2n+1) \ \frac{\lambda}{2}[/tex]
Where Δr₁ and Δr₂ are the path difference, λ is the wavelength and n is an integer.
Let's start by looking for the wavelength that the speakers emit, see attached.
Destructive Interference Δr₂ = (2n + 1) la / 2
Constructive interference Δr₁ = 2n lam / s
Let's solve the system.
Δr₂ - Δr₁ = [tex]\frac{\lambda}{2}[/tex]
Let's calculate
30 - (-10) = [tex]\frac{\lambda}{2}[/tex]
λ = 80 cm
Now we can use the general relation for the path change and the phase.
[tex]\frac{\delta r}{\lambda } = \frac{\phi - \phi_o}{2\pi }[/tex]
[tex]\phi = \frac{\Delta r \ 2\pi }{\lambda } + \phi_o[/tex]fi = Dr 2pi / lam + fio
Where [tex]\phi[/tex] is the possible initial phase difference between the speakers.
Let's find the initial phase difference emitted by the two speakers, let's use destructive interference, for which the phase difference is:
[tex]\phi = \pi \ rad[/tex]
Let's calculate
[tex]\pi = \frac{2\pi }{80} \ 10 + \phi_o \\\phi_o = \pi - \frac{\pi}{4}[/tex]
[tex]\phi_o[/tex] = ¾ [tex]\pi[/tex] rad
This value is kept constant, let's find the phase angle for when the speakers are together, so the path difference is zero Δr = 0
[tex]0= \frac{\phi - \phi_o}{2\pi }\\\phi = \phi_o[/tex]
[tex]\phi[/tex] = ¾ pi
The amplitude of the sound wave is
A = 2a cos [tex]\frac{\phi}{2}[/tex]
A = 2a cos ⅜ π
A = 0.765 a
In conclusion using the expression for sound interference we can find the amplitude for the wave when the two speakers are together is:
A = 0.765a
Learn more here: brainly.com/question/14006922
