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.

Short answer

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

Step-by-step solution

Given data

The increase in sound level intensity is 13 dB.

Concept

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

(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.