Question

The sound level at a point is increased by 30dB. By what factor is the pressure amplitude increased?

• A. 16 times
• B. 10 times
• C. 8 times
• D. 32 times

Answer 1

The correct option is D

32 times

An explanation of the correct option

D: 32 times

Given: $L_2-L_1=30dB$

The sound level is measured in decibels (dB).

$L=10{\log }_{10}\left(\frac{i}{i_0}\right)$, where$iand{i}_0$ are the intensities of the sound wave and $L$is the sound level.

Now,

$$\begin{gathered} L_2-L_1=30dB \\ \Rightarrow 10{\log }_{10}\left(\frac{i_2}{i_0}\right)-10{\log }_{10}\left(\frac{i_1}{i_0}\right)=30 \\ \Rightarrow 10{\log }_{10}\left(\frac{i_2}{i_1}\right)=30 \\ \Rightarrow {\log }_{10}\left(\frac{i_2}{i_1}\right)=3 \\ \Rightarrow \frac{i_2}{i_1}={10}^3 \end{gathered}$$

The intensity of sound is proportional to the square of pressure amplitude. (Since $i=p^{2}wherei=intensity,p=pressureamplitude$)

,$$\begin{gathered} \Rightarrow \frac{p_2}{p_1}=\sqrt{\frac{i_2}{i_1}} \\ =\sqrt{1000} \\ \approx 32 \end{gathered}$$

Thus, if the sound level is increased by 30 dB, pressure amplitude is increased by 32 times.