Question

Part A A microphone is located on the line connecting two speakers hat are 0 932 m apart and oscillating in phase. The microphone is 2 83 m from the midpoint of the two speakers What are the lowest two trequencies that produce an interflerence maximum at the microphone's location? Enter your answers numerically separated by a comma

Answer 1

a)
`f=368.025\ \textup{Hz}`

b)
`f_2=736.051\ \textup{Hz}`

Step-by-step explanation:

Given:

The distance between two speakers (d) = 0.932 m

The distance of the microphone from the midpoint = 2.83 m

Thus, distance of microphone from the nearest speaker (L) = 2.83 - (0.932/2) = 2.364 m

also, the distance of the microphone from the farther speaker (L') = 2.83 + (0.932/2) = 3.296 m

Now,

The path difference is calculated as

L' - L = d = 0.932 m

Now,for a maxima to be produced at the microphone, the waves must constructively interfere.

for this to happen the path difference should be integral multiple of the wavelength.

thus,

`\textup d = n\lambda`

hence, the largest wavelength will be for n = 1,

therefore,

0.932 = 1 × λ

or

λ = 0.932 m

now, the velocity of sound is given as c = 343 m/s

thus, the frequency will be

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

on substituting the values, we get

`f=(343)/(0.932)=368.025\ \textup{Hz}`

now, the 2nd largest wavelength will be for n = 2

0.932 = 2 × λ

or

λ = 0.466

thus, the frequency will be

`f_2=(343)/(0.466)=736.051\ \textup{Hz}`

hence, these are the lowest first two frequencies.