Question

The sound level measured in a room by a person watching a movie on a home theater system varies from 50 dB

 during a quiet part to 90 dB

 during a loud part. Approximately how many times louder is the latter sound?

Answer 1

The sound intensity level $\beta$ of a sound wave is defined by the equation

$\beta = (10 \,dB) \log{\cfrac{I}{I_0}}$

Then, since we have two sounds ($\beta_1=50\,dB;\beta_2=90\,dB$) we substitute to find the intensity of the loud part I₂:

$\beta_2-\beta_1 = (10 \,dB) \Bigg( \log{\cfrac{I_2}{I_0}}-\log{\cfrac{I_1}{I_0}} \Bigg) \\ \beta_2-\beta_1 = (10 \,dB) \Big( \log{{I_2}}-\log{{I_0}}-\log{{I_1}}+\log{{I_0}} \Big) \\ \beta_2-\beta_1 = (10 \,dB) \big( \log{{I_2}}-\log{{I_1}} \big)$

$\cfrac{{I_2}}{{I_1}} = \Large{10}^{\frac{{\beta_2-\beta_1 }}{{(10 \,dB)}}}=\Large{10}^{\frac{{(90-50)dB }}{{(10 \,dB)}}}=\Large{10}^{4}$

$\Large \implies I_2=10^4\cdot I_1$

In conclusion, I₂ (the intensity of the loud part) is approximately 10⁴ times bigger than I₁ (the intensity of the quiet part).

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.