Question

Your ears are sensitive to differences in pitch, but they are not very sensitive to differences in intensity. You are not capable of detecting a difference in sound intensity level of less than 1 dB. By what factor does the sound intensity increase if the sound intensity level increases from 60 dB to 61 dB

Answer 1

Answer:

1.29

Explanation:

The level of intensity of sound (β) (in decibels, dB) is given by:

β  =  $10log_{10}(\frac{I}{I_0} )$, where I = sound intensity in watts per meter squared and $I_0$ = reference intensity.

Change in sound intensity level (Δβ) = $\beta _2 - \beta _1$ = 61 - 60 = 1

Δβ = $10log_{10}(\frac{I_2}{I_0}) - 10log_{10}(\frac{I_1}{I_0})$ = 1

       $10log_{10}(\frac{I_2}{I_0})/(\frac{I_1}{I_0})$ = 1

       $10log_{10}(\frac{I_2}{I_1})$   = 1

               $\frac{I_2}{I_1}$ = $10^{0.1}$ = 1.29

Hence, the sound intensity will increase by a factor of 1.29.