Chapter 12, Problem 15P

(a)

To determine

The pressure amplitude of sound wave.

Answer to Problem 15P

The pressure amplitude is $28.7\text{N}/{\text{m}}^{\text{2}}$.

Explanation of Solution

The intensity level of air is $120.0\text{dB}$.

Write the equation for sound intensity levels in bels.

$\beta =\left(10\text{dB}\right)\log \frac{I}{I_0}$

Here, $\beta$ is the sound intensity level in bels, $I$ is the intensity of sound, and $I_0$ is threshold of hearing.

Rewrite the above relation in terms of $I$.

$\begin{array}{c}\frac{\beta }{10\text{dB}}=\log \frac{I}{I_0}\\ {10}^{\frac{\beta }{10\text{dB}}}=\frac{I}{I_0}\\ I=I_0{10}^{\frac{\beta }{10\text{dB}}}\end{array}$ (I)

Write the relation between $I$ and the pressure amplitude of sound.

$\frac{p_0^2}{2\rho v}=I$ (II)

Here, $p_0$ is the pressure amplitude, $\rho$ is the density of air, and $v$ is the speed of sound in air.

Conclusion:

Substitute $120.0\text{dB}$ for $\beta$, and ${10}^{-12}\text{W}/{\text{m}}^{\text{2}}$ for $I_0$ in equation (I) to find $I$.

$\begin{array}{c}I=\left({10}^{-12}\text{W}/{\text{m}}^{\text{2}}\right){10}^{\frac{120.0\text{dB}}{10\text{dB}}}\\ =1.0\text{W}/{\text{m}}^{\text{2}}\end{array}$

Substitute $1.20\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $343\text{m}/\text{s}$ for $v$, and $1.0\text{W}/{\text{m}}^{\text{2}}$ for $I$ in equation (II) to find $p_0$.

$\begin{array}{c}\frac{p_0^2}{2\left(1.20\text{kg}/{\text{m}}^{\text{3}}\right)\left(343\text{m}/\text{s}\right)}=1.0\text{W}/{\text{m}}^{\text{2}}\\ p_0=\sqrt{823.69}\text{N}/{\text{m}}^{\text{2}}\\ =28.7\text{N}/{\text{m}}^{\text{2}}\end{array}$

Therefore, the pressure amplitude is $28.7\text{N}/{\text{m}}^{\text{2}}$.

(b)

To determine

The force exerted by air on eardrum.

Answer to Problem 15P

The force is $1.58\text{mN}$.

Explanation of Solution

The intensity level of air is $120.0\text{dB}$ and the area of eardrum is $0.550\times {10}^{-4}{\text{m}}^{\text{2}}$.

Write the equation to find the force exerted by air on eardrum.

$F=p_{0}A$

Here, $F$ is the force exerted by air on eardrum and $A$ is the area of eardrum.

Conclusion:

Substitute $28.7\text{N}/{\text{m}}^{\text{2}}$ for $p_0$ and $0.550\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A$ in the above equation.

$\begin{array}{c}F=\left(28.7\text{N}/{\text{m}}^{\text{2}}\right)\left(0.550\times {10}^{-4}{\text{m}}^{\text{2}}\right)\\ =15.8\times {10}^{-4}\text{N}\left(\frac{{10}^{-3}\text{mN}}{1\text{N}}\right)\\ =1.58\text{mN}\end{array}$

Therefore, the force is $1.58\text{mN}$.