Question

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

Answer 1

f = 632 Hz

Step-by-step 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)(\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 = (7\lambda)/(2)`

`\lambda = (2 * 1.90)/(7)`

`\lambda = 0.54 m`

now for frequency we know that

`f = (v)/(\lambda)`

`f = (343)/(0.54) = 632 Hz`