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2 years ago

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Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 8.0

0 m. A person is standing 12.0 m away from the wall, equidistant from the loudspeakers. When the person moves 3.00 m parallel to the wall, she experiences destructive interference for the second time. What is the frequency of the sound

1 answer:

zloy xaker [14]2 years ago

4 0

Answer: $278\ Hz$

Explanation:

Given

Distance between two speakers is 8 m

Man is standing 12 away from the wall

When the person moves 3 parallel to the wall

the parallel distances from the speaker become 4+3, 4-3

Now, the difference of distances from the speaker is

$\Delta d=\sqrt{12^2+(4+3)^2}-\sqrt{12^2-(4-3)^2}\\\Delta d=1.85\ m$

Condition for destructive interference is

$\Delta d=(2n-1)\dfrac{\lambda }{2}=(2n-1)\dfrac{\nu }{2f}\\\\\Rightarrow f=(2n-1)\dfrac{v}{2\Delta d}$

for second destructive interference; n=2

$\Rightarrow f=(2\times 2-1)\dfrac{343}{2\times 1.85}=278.10\approx 278\ Hz$

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