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3 years ago

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Two identical loudspeakers are some distance apart. A person stands 5.20 m from one speaker and 4.10 m from the other. What is t

he third lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 344 m/s.

1 answer:

Igoryamba3 years ago

5 0

Answer:

125.09 Hz

Explanation:

Given : A person stands 5.20 from one speaker and 4.10 m away from the other speaker.

Distance between the speakers is 5.20 - 4.10 =1.10 m

We know that

For destructive interferences

n . λ / 2

where n =1,3,5,7....

Therefore difference between the speakers is

5.10 - 4.20 = 1 X 0.5 λ

λ = 2.2

Given velocity of sound is V = 344 m/s

Therefore frequency, f = $\frac{v}{\lambda }$

= $\frac{344}{2.2 }$

= 156.36 Hz

Now, the third lowest frequency is given by

λ = (5.20-4.10) X 5 X 0.5

= 2.75 m

Therefore frequency, f = $\frac{v}{\lambda}$

= $\frac{344}{2.75}$

= 125.09 Hz

Therefore third lowest frequency is 125.09 Hz

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