Question

A sound level of 50 decibels is a) 2.5 times as intense as a sound of 20 decibels. b) 6.25 times as intense as a sound of 20 decibels. c) 10 times as intense as a sound of 20 decibels. d) 100 times as intense as a sound of 20 decibels. e) 1000 times as intense as a sound of 20 decibels.

Short answer

Answer: The intensity of a 50-decibel sound is 1000 times as intense as a sound of 20 decibels.

Step-by-step solution

Write the given values and formula

We know that: - A 50-decibel sound has a decibel level of 50 dB - A 20-decibel sound has a decibel level of 20 dB The formula to convert decibel levels to intensities is: $dB=10\times lo{g}_{10}(\frac{I}{I_0})$

Find the ratio of intensities#

First, let's consider the 50-decibel sound. Let $I_1$ be its intensity. Using the formula, we have: $50=10\times lo{g}_{10}(\frac{I_1}{I_0})$ Now consider the 20-decibel sound. Let $I_2$ be its intensity. Using the formula, we have: $20=10\times lo{g}_{10}(\frac{I_2}{I_0})$ We need to find the ratio $\frac{I_1}{I_2}$.

Solve the equations for $I_1$ and $I_2$

First, divide both sides of the equations by 10: $5=lo{g}_{10}(\frac{I_1}{I_0})$ $2=lo{g}_{10}(\frac{I_2}{I_0})$ Now, eliminate the logarithms using the property $a=lo{g}_{10}(b)\Rightarrow b={10}^a$: $I_1=I_0\times {10}^5$ $I_2=I_0\times {10}^2$

Calculate the intensity ratio

The intensity ratio between the two sounds can be found by dividing $I_1$ by $I_2$: $\frac{I_1}{I_2}=\frac{I_0\times {10}^5}{I_0\times {10}^2}$ Simplify the expression by canceling the $I_0$ terms: $\frac{I_1}{I_2}=\frac{{10}^5}{{10}^2}$ Now, use the property $\frac{a^b}{a^c}=a^{b-c}$ for exponents: $\frac{I_1}{I_2}={10}^{(5-2)}$ $\frac{I_1}{I_2}={10}^3$ Thus, the intensity of a 50-decibel sound is 1000 times as intense as a sound of 20 decibels.

Select the correct answer

The intensity ratio $\frac{I_1}{I_2}$ is 1000, which corresponds to option (e). Hence, the correct answer is: e) 1000 times as intense as a sound of 20 decibels.