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 the 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?

Short answer

The factor by which the sound intensity is increased is 1.25

Step-by-step solution

Write the given information

The sound intensity increased from 60 dB to 61 dB
Initial intensity level β_i= 60 dB

Final intensity level β_f= 61 dB

To determine the factor by which the sound intensity increased

The intensity of the ears before the intensity level is increased

$I=I_o{10}^{{\beta }_{i}dB/10}$

Substitute the values

$$\begin{gathered} I=\left({10}^{-12}W/m^2\right){10}^{60/10} \\ I={10}^{-6}W/m^2 \end{gathered}$$

Thus, initially, the intensity of the sound is 10^-6 W/m²

The intensity of the ears after the intensity level is increased is given as

$I=I_o{10}^{{\beta }_{f}dB/10}$
Substitute the values

$$\begin{gathered} I=\left({10}^{-12}W/m^2\right){10}^{61/10} \\ I=1.25\times {10}^{-6}W/m^2 \end{gathered}$$
Thus, the final intensity of the sound source is 1.25 x10^-6 W/m²

Compare both the values of intensity of sound

The factor by which the sound intensity is increased is 1.25