Question

The sound level at a point is increased by \[30dB\] . What factor is the pressure amplitude increased?

Answer 1

Hint: In order to answer this question, first we will write the formula of sound level in terms of intensities. And then we will take 2 sound levels and the corresponding two intensities, and solve in the sound level. And atlast we will show the relation between intensities and the pressure amplitude to find the factor increasement.

Complete step by step solution:
The sound level in dB is:
\[L = 10lo{g_{10}}(\dfrac{I}{{{I_0}}})\]
If ${L_1}\,and\,{L_2}$ are the sound levels and ${I_1}\,and\,{I_2}$ are corresponding intensities in the two cases then:
$
  {L_2} - {L_1} = 10[{\log _{10}}(\dfrac{{{I_2}}}{{{I_0}}}) - {\log _{10}}(\dfrac{{{I_1}}}{{{I_0}}})] \\
   \Rightarrow 30 = 10{\log _{10}}(\dfrac{{{I_2}}}{{{I_1}}}) \\
   \Rightarrow \dfrac{{{I_2}}}{{{I_1}}} = {10^3} \\
 $
As the intensity is proportional to the square of the pressure amplitude, thus we have:
$\therefore \dfrac{{\vartriangle {p_2}}}{{\vartriangle {p_1}}} = \sqrt {\dfrac{{{I_2}}}{{{I_1}}}} = \sqrt {1000} \approx 32$


Note: \[MPa\] is the unit of pressure used to quantify amplitude (MegaPascals). When the power controls on the machine are set to maximum levels, pulsed ultrasound scanners can produce peak pressure amplitudes of several million pascals in water.