Respuesta :
The sound intensity level is defined as:
[tex]L=10\log _{10}(\frac{I}{I_0})dB[/tex]Where dB is decibel, I₀ is the reference sound intensity (commonly 1 pW/m²) and I is the intensity of the sound wave. The intensity of a sound is the same as the power delivered per unit area.
Isolate I from the equation:
[tex]I=I^{}_0\cdot10^{\frac{L}{10dB}}[/tex]To compare the intensities of two sounds, consider I₁ and I₂ and find the ratio between them:
[tex]\begin{gathered} I_1=I^{}_0\cdot10^{\frac{L_1_{}_{}}{10dB}} \\ I_2=I^{}_0\cdot10^{\frac{L_2}{10dB}} \end{gathered}[/tex]Then:
[tex]\begin{gathered} \frac{I_1}{I_2}=\frac{I^{}_0\cdot10^{\frac{L_1_{}_{}}{10dB}}}{I^{}_0\cdot10^{\frac{L2}{10dB}}} \\ =10^{\frac{L_1-L_2}{10dB}} \end{gathered}[/tex]Use the subindex 1 to represent the speaking voice and the subindex 2 to represent the whisper. Substitute L₁=60dB and L₂=30dB to find how many times greater is the power delivered per unit area by a normal speaking voice than by a whisper:
[tex]\begin{gathered} \frac{I_1}{I_2}=10^{\frac{60dB-30dB}{10dB}} \\ =10^{\frac{30dB}{10dB}} \\ =10^3 \\ =1000 \end{gathered}[/tex]Therefore, the power delivered per unit area by a normal speaking voice is 1000 times greater than the power delivered per unit area by a whisper.
