Answered

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.Part A
What is the first resonant frequency?
Part B
What is the wavelength at second resonance?
Part C
How does the first resonant frequency change if the ear canal is full of water?Part D
At a listener's ear, the intensity of a conversation is 4×10−6W/m24×10−6W/m2 and a typical adult's ear has a surface area of 210×10−3m2210×10−3m2. What is the power of the sound waves on the ear?
Part E
Long-term exposure to loud noises can damage hearing. If a loud machine produces sounds with an intensity level of 100dB100dB, what would the intensity level be if the intensity were reduced by a factor of 4?



Answer :

The first resonance frequency is 3400 Hz and the wavelength at second frequency is 3.33 cm. After the ear canal is full of water the first resonance frequency increases. The power of sound waves on the year is  840 X 10⁻⁹ W and if the intensity of sound is reduced by a factor of 4, then the intensity level will be 135 dB.

(a) The ear, being open at one end and closed at the other, is equivalent to a closed pipe. For a closed pipe, the first resonant frequency occurs when one quarter of the wavelength is equal to the length of the pipe.

= λ/4 = l

= λ = 4l

Now, Speed is given by

= v = fλ

where f = frequency

= f = v/λ

At the first resonant frequency,

= f = v/λ

= f = v/4l

= f = 340 / (4 X 2.5 X 10⁻²) m

= f = 3400 Hz

(b) At second resonance,

= (3/4)λ = l

= λ = (4/3) X l

= λ = (4/3) X 2.5 X 10⁻²

= λ = 3.33 cm

(c) Since, sound travels faster in a liquid than in gas, the first resonance frequency increases when the ear canal is filled with water as speed increases. This is because, speed is directly proportional to intensity.

(d) Intensity = i = 4 X 10⁻⁶ W/m²
Surface area = a = 210 X 10 ⁻³ m²

Power of the sound waves = p =

= p = i X a

= p = 4 X 10⁻⁶ X 210 X 10 ⁻³

= p = 840 X 10⁻⁹ W

(e) Intensity level = l = 100 dB

If the intensity is reduced by 4, then using the relationship between decibel and sound intensity is :

= β = 10 X log(I₁ / I₂)

= β = 10 log (0.25/10⁻¹²) dB

=  β = 135 dB

To know more about Sound Intensity:

https://brainly.com/question/14980416

#SPJ4

Other Questions