Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity is minimum when speaker 2 is 10 cm behind speaker 1. The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude 2a, when it is 30 cm in front of speaker 1. What is What isThe amplitude of the sound (as a multiple of a) if the speakers are placed side by side?

Answers

Answer 1

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

Answer 2

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

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Two Loudspeakers Emit Sound Waves Of The Same Frequency Along The X-axis. The Amplitude Of Each Wave

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