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Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from speaker A. Take the speed of sound in air to be 344 m/s.What is the closest you can be to speaker B and be at a point of perfectly destructive interference?

Respuesta :

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of  A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am  1 m away from B , the path difference will be

8 - 1 = 7 m  which is  odd multiple of 1 or λ /2

So path difference becomes odd multiple of  λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

asifejaz20

Answer:

1m.

Explanation:

Frequency = f = 172 Hz

Speed = V = 344 m/s

For the destructive interference,

Wavelength = λ = V/f = 344/172 = 2m

Path difference = d2 – d1 = n λ/2

Where “n” is an odd integer.

If you are 1m away from the loudspeaker B then the path difference will be,

Path difference = d2 - d1 = 8 – 1 = 7m  

Which is odd multiple of λ/2 and also condition for destructive interference. Hence, to hear the perfect destructive interference you have to stay 1m close to loudspeaker B.  

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