Q21E - University Physics with Modern Physics

Question

A baby’s mouth is 30 cm from her father’s ear and 1.50 m from her mother’s ear. What is the difference between the sound intensity levels heard by the father and by the mother? 

Step-by-Step Solution

Answer

The difference in the sound intensity heard by the father and by the mother is 14 dB

STEP 1 Concept of the difference in the sound intensity level

The sound intensity level is given by ${\mathrm{\beta }}_2-{\mathrm{\beta }}_1=\left(10\mathrm{dB}\right)\log \left(\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}\right)$ Where, ${\mathrm{\beta }}_2-{\mathrm{\beta }}_1$ is the difference in the sound intensity level, ${\mathrm{l}}_2$ is the intensity of the sound heard by the father, ${\mathrm{l}}_1$ is the intensity of the sound heard by the mother. 

STEP 2 Calculate the ratio of the sound intensity heard by the father to the sound heard by the mother with respect distance

The ratio of the sound intensity heard by the father to the sound heard by the mother 

with respect distance is, $\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}=\frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}$ Where, ${\mathrm{r}}_1$ is the distance of the mother from the baby, ${\mathrm{r}}_2$ is the distance of the father from the baby.

Substitute 30 cm for ${\mathrm{r}}_2$ and 1.5 m for ${\mathrm{r}}_1$ in the above equation to find $\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}$

$$\begin{gathered} \frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}=\frac{\left(1.5\mathrm{m}\right)\left({\displaystyle \frac{100{\mathrm{cm}}^2}{1\mathrm{m}}}\right)}{\left(30{\mathrm{cm}}^2\right)} \\ =25 \end{gathered}$$

Thus, ratio of the sound intensity heard by the father to the sound heard by the mother with respect distance is 25

STEP 3 To find the difference in the sound intensity level

Substitute 25 for $\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}$ to find ${\mathrm{\beta }}_2-{\mathrm{\beta }}_1$

$$\begin{gathered} {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=\left(10\mathrm{dB}\right)\log \left(25\right) \\ =\left(10\mathrm{dB}\right)1.4 \\ =14\mathrm{dB} \end{gathered}$$

Therefore, the difference in the sound intensity heard by the father and by the mother is 14dB