A person wears a hearing aid that uniformly increases the intensity level of all audible frequencies of sound?

Question by Alex M: A person wears a hearing aid that uniformly increases the intensity level of all audible frequencies of sound?
A person wears a hearing aid that uniformly increases the intensity level of all audible frequencies of sound by 24.2 dB. The hearing aid picks up sound having a frequency of 250 Hz at an intensity of 3.60 10-11 W/m2. What is the intensity delivered to the eardrum?

Best answer:

Answer by ebs
Given:
Sound intensity I = 3.60×10^-11 W/m².

Reference sound intensity Io = 10^−12 W/m² (Threshold of hearing).
Reference sound intensity level LIo = 0 dB-SIL (Threshold of hearing level).

Get sound intensity level LI in dB when entering sound intensity I = 3.60×10^-11 W/m².
Sound intensity level LI = 10×log (I / Io) dB = 10×log (3.60×10^-11 / 10^−12) dB = 11.34 dB.

The hearing aid adds to this sound intensity level of 11.34 dB a level of 24.2 dB.
That makes a total of 35.54 dB.

The intensity ratio = 10^(11.34/10) = 13.6.
The sound intensity at the eardrum is 3.60×10^-11×13.6 W/m² = 0.0000000004896 W/m² = 4.896×10^-10 W/m².

Cheers ebs

PS: My eardrums are moved only by the sound pressure deviations – that is a sound field quantity – and my hearing aid is using an amplifier that decreases the input audio voltage of the microphones. Why do you like the sound intensity as energy quantity?
Sound is ususally measured by a SPL meter (SPL = Sound Pressure Level).

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