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Chapter 2: Q38E (page 540)

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 172鈥�Hz. You are 8.00鈥嬧赌�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鈥塇锄

The distance of observer from A:l=8鈥尘

02

Condition of destructive interference

Condition of destructive interference is 螖濒=n2位 , (n=1,3,5,7,9,...)where螖濒 is the path difference and位 is the wavelength of the sound emitted from the speakers.

03

Use the condition

Given thatf=172鈥塇锄 and we know that the speed of sound in air isv=344鈥尘/s .

So,

位=vf=344鈥尘/s172鈥塇锄=2鈥尘

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

Now,

8鈭�x=n2位8鈭�x=n2脳28鈭�x=nx=8鈭�n

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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