A 60 year old person has a threshold of hearing of 93.0 dB for a sound with frequency f=10,000 Hz. By what factor must the intensity of a sound wave of that frequency, audible to a typical young adult, (sound level=43.0 dB) be increased so that it is heard by the older person.

Respuesta :

Answer:

12.06 times

Explanation:

The ration of intensity of sound of 93 dB and 43 dB is given as follows

I₁ / I₂ = [tex](10)^\frac{9.3}{4.3}[/tex]

I₁ / I₂ = 145.44

The intensity of sound heard by old person is 145.44 times the sound heard by young adult.

We know that intensity of sound is proportional to square of frequency  , ie

I ∝ n²

In order to increase the intensity of later sound 145.44 times , we shall have to increase its frequency √145.44  times or 12.06 times

So 12.06 times is the answer.