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Two loudspeakers, Aand B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172Hz. You are 8.00​ mfrom role="math" localid="1655809995061" A. What is the closest you can be to Band be at a point of destructive interference?

Short Answer

Expert verified

The closet distance from B is 1m.

Step by step solution

01

Given Data

Frequency of sound emitted: f=172 Hz

The distance of observer from A:l=8 m

02

Condition of destructive interference

Condition of destructive interference is Δl=n2λ , (n=1,3,5,7,9,...)whereΔl is the path difference andλ is the wavelength of the sound emitted from the speakers.

03

Use the condition

Given thatf=172 Hz and we know that the speed of sound in air isv=344 m/s .

So,

λ=vf=344 m/s172 Hz=2 m

The two speakers are 8mapart. Let the distance from B is x. So, we can write Δl=8x.

Now,

8x=n2λ8x=n2×28x=nx=8n

From the given values of n, for n=7, we get the closest distance.

Hence, the closest distance from B is 1m.

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Most popular questions from this chapter

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