Q14E - University Physics with Modern Physics

Question

Question: (a) By what factor must the sound intensity be increased to raise the sound intensity level by 13.0 dB? (b) Explain why you don’t need to know the original sound intensity.

Step-by-Step Solution

Answer

(a) the factor for the sound intensity to be increased to raise the sound intensity level by 13.0 dB is 20.

Step 1: Given data

The increase in sound level intensity is 13 dB.

Step 2: Concept

Sound intensity can be measured as the number of wave particles falling on the surface that can be measured in 1 second.

Step 3: (a) Calculation of the factor

The sound intensity level was \({\beta _1}\) , and it got increased, and the value can be written as,\({\beta _2}\).

The formula for the initial level is,

\({\beta _1} = 10\log \left( {\frac{{{I_1}}}{{{I_0}}}} \right)\) 

Here \({I_0}\) is the original intensity, \({I_1}\)is the measured intensity.

 

The expression for final intensity level is,

\({\beta _2} = 10\log \left( {\frac{{{I_2}}}{{{I_0}}}} \right)\)

Here \({I_2}\)is the measured intensity.

 

The increment or change  is 13 dB; we can write it as,

\(\Delta \beta  = {\beta _2} - {\beta _1}\) 

Substitute the values in the above expression, and we get,

\(\begin{array}{c}13 = 10\log \left( {\frac{{{I_2}}}{{{I_0}}}} \right) - 10\log \left( {\frac{{{I_1}}}{{{I_0}}}} \right)\\13 = 10\log \left( {\frac{{\frac{{{I_2}}}{{{I_0}}}}}{{\frac{{{I_1}}}{{{I_0}}}}}} \right)\end{array}\)

Solving further as,

\(\begin{array}{c}13 = 10\log \left( {\frac{{{I_2}}}{{{I_1}}}} \right)\\1.3 = \log \left( {\frac{{{I_2}}}{{{I_1}}}} \right)\\\frac{{{I_2}}}{{{I_1}}} = 20\end{array}\) 

Thus, the factor for the sound intensity to be increased to raise the sound intensity level by 13.0 dB is 20.