Question

The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end and open on the other. The length of the ear canal is partially responsible for our sensitivities to certain frequencies. Use 340m/s for the speed of sound when performing the following calculations.
(a) What is the first resonant frequency?

(b) What is the wavelength at second resonance?

(c) How does the first resonant frequency change if the ear canal is full of water? (increases, shortens, no change or decreases...?)

(d) At a listener's ear, the intensity of a conversation is 4×10−6W/m2 and a typical adult's ear has a surface area of 210×10−3m2. What is the power of the sound waves on the ear?

(e) Long-term exposure to loud noises can damage hearing. If a loud machine produces sounds with an intensity level of 110dB, what would the intensity level be if the intensity were reduced by a factor of 3?

Answer 1

Answer:

a)  f = 3400 Hz , b)  f5 = 17000 Hz , c) speed is much higher and the frequency is directly proportional to the speed so  the frequency increases

Explanation:

The resonance in a system with one end closed and the other open, we can find it because in the closed points we have nodes and in the open parts we have a maximum, so for this system we have the following resonances

 

    4L = λ                fundamental

    4L = 3 λ₃           third harmonic

    4L = 5 λ₅           fifth harmonic

   4L = (2n + 1)λ     general with n = 1, 2,3,

a) the speed of the wave is

              v = λ f

              f = v /λ

               

            λ₁ = 4L / 1

            λ₁ = 4 0.025 m

            λ₁ = 0.1 m

             f = 340 / 0.1

             f = 3400 Hz

b) the second resonance occurs for n = 5

           λ₅ = 4 0.025 / 5

           λ₅ = 0.02 m

           f5 = 340 / 0.02

           f5 = 17000 Hz

c) if the channel is full of water the speed of sound is v = 1436 m / s

     Since the speed is much higher and the frequency is directly proportional to the speed so  the frequency increases

d) The intensity is defined

            I = P / A

            P = I A

            P = 4.10-6 210.10-3

            P = 840 10-9 W

e) the intensity is β₁ = 110 if it is reduced by a factor of 3, Δβ = 110/3 = 36.67 f

The intensity of the system is now β₂ = 110 - 36.67 = 73.33  

We can calculate what the intensity reduction is

              β1 = 10 log (I1 / Io)

              β2 = 10 log (I2 / Io)

              β1 - β 2 = 10 [log (I1 / Io) - log (I2 / Io))

              β1 -β2 = 10 log (I1 / I2)

               I1 / I2 = 10 (β1-β2) / 10

Calculous

             I1-I2 = 10 (36.67 / 10)

             I1 / I2 = 4645

             I2 = 1/4645 I1