Question

A person wears a hearing aid that uniformly increases the intensity level of all audible frequencies of sound by 30.0 dB. The hearing aid picks up sound having a frequency of 100 Hz and an intensity of 1.3e-10 W/m2. What is the intensity delivered to the eardrum?

Answer 1

Answer:

The intensity delivered to the ear drum is I' = 653.131x10^-11 w/m^2

Explanation:

The standard intensity Io = 10^{-12} w/m^2

The intensity of $1.63 \times 10^{-10}$ corresponds to an intensity level of

$10 log[\frac{I}{Io}] = 10 log[\frac{(1.3\times 10^{-10}}{10^{-12}}] = 11.13 dB$

This is increased by the hearing aid by a factor of 30 dB

It now becomes 30 + 11.13 = 41.13 dB

This corresponds to an intensity of I'

10 log(I'/Io) = 41.13 dB

log (I'/Io) = 4.11

I'/Io + 10^4.11

I' = 10^{-12} x 12882.49 w/m^2

The intensity delivered to the ear drum is

I' = 653.131x10^-11 w/m^2

Answer 2

The intensity level at the eardrum is 1.3 × 10⁻⁷ W/m²

Sound intensity level

The sound intensity level is given by L = 10㏒(I/I₀) where

• I = intensity and
• I₀ = reference intensity

Now given that

• I = intensity at eardrum,
• I₀ = intensity of hearing aid = 1.3 10⁻¹⁰ W/m² and
• L = 30 dB (since the hearing aid that uniformly increases the intensity level of all audible frequencies of sound by 30.0 dB and thus the difference in their sound levels)

Since  L = 10㏒(I/I₀)

Intensity at eardrum

Making I subject of the formula, we have

$I = I_{0} (10^{\frac{L}{10} } )$

Substituting the values of the variables into the equation, we have

$I = I_{0} (10^{\frac{L}{10} } )$

$I = 1.3 X 10^{-10}W/m^{2} (10^{\frac{30}{10} } )\\I = 1.3 X 10^{-10}W/m^{2} (10^{3} )\\I = 1.3 X 10^{-7}W/m^{2}$

So, the intensity at the eardrum is 1.3 × 10⁻⁷ W/m²

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