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3 years ago

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Two identical loudspeakers are some distance apart. A person stands 5.80 m from one speaker and 3.90 m from the other. What is t

he fourth lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 343 m/s.

1 answer:

NikAS [45]3 years ago

7 0

Answer:

f = 632 Hz

Explanation:

As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength

here we know that

$\Delta x = (2n + 1)\frac{\lambda}{2}$

given that path difference from two loud speakers is given as

$\Delta x = 5.80 m - 3.90 m$

$\Delta x = 1.90 m$

now we know that it will have fourth lowest frequency at which destructive interference will occurs

so here we have

$\Delta x = 1.90 = \frac{7\lambda}{2}$

$\lambda = \frac{2 \times 1.90}{7}$

$\lambda = 0.54 m$

now for frequency we know that

$f = \frac{v}{\lambda}$

$f = \frac{343}{0.54} = 632 Hz$

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