Question

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 8.00 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

Answer 1

Answer:
`278\ Hz`

Step-by-step 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=√(12^2+(4+3)^2)-√(12^2-(4-3)^2)\\\Delta d=1.85\ m`

Condition for destructive interference is

`\Delta d=(2n-1)(\lambda )/(2)=(2n-1)(\\u )/(2f)\\\\\Rightarrow f=(2n-1)(v)/(2\Delta d)`

for second destructive interference; n=2

`\Rightarrow f=(2* 2-1)(343)/(2* 1.85)=278.10\approx 278\ Hz`