Chapter 17, Problem 29P

(a)

To determine

The wavelength of the initial note.

Answer to Problem 29P

The wavelength of the initial note is $\underset{\_}{2.34\text{m}}$.

Explanation of Solution

Write the expression for wavelength.

  $\lambda =\frac{\upsilon }{f}$                                                                                                                        (I)

Here, $\upsilon$ is the speed, $f$ is the frequency, and $\lambda$ is the wavelength.

Conclusion:

Substitute, $343\text{m}/\text{s}$ for $\upsilon$, and $146.8{\text{s}}^{-1}$ for $f$ in equation (I).

  $\begin{array}{c}\lambda =\frac{343\text{m}/\text{s}}{146.8{\text{s}}^{-1}}\\ =2.34\text{m}\end{array}$

Therefore, the wavelength of the initial note is $\underset{\_}{2.34\text{m}}$.

(b)

To determine

The wavelength of the final note.

Answer to Problem 29P

The wavelength of the final note is $\underset{\_}{0.390\text{m}}$.

Explanation of Solution

Use equation (I) to obtain the wavelength of the final high note.

Conclusion:

Substitute, $343\text{m}/\text{s}$ for $\upsilon$, and $880{\text{s}}^{-1}$ for $f$ in equation (I).

  $\begin{array}{c}\lambda =\frac{343\text{m}/\text{s}}{880{\text{s}}^{-1}}\\ =0.390\text{m}\end{array}$

Therefore, the wavelength of the final note is $\underset{\_}{0.390\text{m}}$.

(c)

To determine

The pressure amplitude of the initial note.

Answer to Problem 29P

The pressure amplitude of the initial note is $\underset{\_}{0.161\text{Pa}}$.

Explanation of Solution

Write the expression for the intensity of the wave.

  $I=\frac{\Delta P_{\text{max}}^2}{2\rho \upsilon }$                                                                                                                 (II)

Here, $\Delta P_{\text{max}}$ is the variation in pressure amplitude, $\rho$ is the density of the medium, and $\upsilon$ is the speed.

Rewrite the equation (II) to obtain an expression for $\Delta P_{\text{max}}$.

  $\Delta P_{\text{max}}=\sqrt{2\rho \upsilon I}$                                                                                                      (III)

Write the expression for sound level.

  $\beta =\left(10\text{dB}\right)\log \left(\frac{I}{{10}^{-12}\text{W}/{\text{m}}^2}\right)$                                                                               (IV)

Here, $\beta$ is the sound level, and $I$ is the intensity.

Conclusion:

Substitute, $75\text{dB}$ for $\beta$ in equation (IV), and rearrange to obtain a value for $I$

  $\begin{array}{c}75\text{dB}=\left(10\text{dB}\right)\log \left(\frac{I}{{10}^{-12}\text{W}/{\text{m}}^2}\right)\\ I={10}^{75\text{dB}/10}\left({10}^{-12}\text{W}/{\text{m}}^2\right)\\ =3.16\times {10}^{-5}\text{W}/{\text{m}}^2\end{array}$

Substitute, $1.20\text{kg}/{\text{m}}^3$ for $\rho$, $343\text{m}/\text{s}$ for $\upsilon$, and $3.16\times {10}^{-5}\text{W}/{\text{m}}^2$ for $I$ in equation (III).

  $\begin{array}{c}\Delta P_{\text{max}}=\sqrt{2\left(1.20\text{kg}/{\text{m}}^3\right)\left(343\text{m}/\text{s}\right)\left(3.16\times {10}^{-5}\text{W}/{\text{m}}^2\right)}\\ =0.161\text{Pa}\end{array}$

Therefore, the pressure amplitude of the initial note is $\underset{\_}{0.161\text{Pa}}$.

(d)

To determine

The pressure amplitude of the final note.

Answer to Problem 29P

The pressure amplitude of the final note is $\underset{\_}{0.161\text{Pa}}$

Explanation of Solution

The pressure amplitude is depends on the intensity, density, and speed, since all these quantities are same in case of initial and final note, the pressure amplitude of initial and final note are same.

Conclusion:

Substitute, $75\text{dB}$ for $\beta$ in equation (IV), and rearrange to obtain a value for $I$

  $\begin{array}{c}75\text{dB}=\left(10\text{dB}\right)\log \left(\frac{I}{{10}^{-12}\text{W}/{\text{m}}^2}\right)\\ I={10}^{75\text{dB}/10}\left({10}^{-12}\text{W}/{\text{m}}^2\right)\\ =3.16\times {10}^{-5}\text{W}/{\text{m}}^2\end{array}$

Substitute, $1.20\text{kg}/{\text{m}}^3$ for $\rho$, $343\text{m}/\text{s}$ for $\upsilon$, and $3.16\times {10}^{-5}\text{W}/{\text{m}}^2$ for $I$ in equation (III).

  $\begin{array}{c}\Delta P_{\text{max}}=\sqrt{2\left(1.20\text{kg}/{\text{m}}^3\right)\left(343\text{m}/\text{s}\right)\left(3.16\times {10}^{-5}\text{W}/{\text{m}}^2\right)}\\ =0.161\text{Pa}\end{array}$

Therefore, the pressure amplitude of the final note is $\underset{\_}{0.161\text{Pa}}$.

(e)

To determine

The displacement amplitude of the initial note.

Answer to Problem 29P

The displacement amplitude of the initial note is $\underset{\_}{4.25\times {10}^{-7}\text{m}}$.

Explanation of Solution

Write the expression for the intensity in terms of displacement amplitude.

  $I=\frac{1}{2}\rho \upsilon {\left(\omega s_{\text{max}}\right)}^2$                                                                                                    (V)

Here, $\omega$ is the angular frequency, $s_{\text{max}}$ is the displacement amplitude.

Substitute, $2\pi f$ for $\omega$ in equation (V), and rearrange to obtain an expression for $s_{\text{max}}$.

  $\begin{array}{c}I=\frac{1}{2}\rho \upsilon {\left(2\pi f{s}_{\text{max}}\right)}^2\\ s_{\text{max}}=\sqrt{\frac{I}{2{\pi }^2\rho \upsilon f^2}}\\ =\frac{1}{f}\sqrt{\frac{I}{2{\pi }^2\rho \upsilon }}\end{array}$                                                                                            (VI)

Conclusion:

Substitute, $1.20\text{kg}/{\text{m}}^3$ for $\rho$, $343\text{m}/\text{s}$ for $\upsilon$, $146.8{\text{s}}^{-1}$ for $f$, and $3.16\times {10}^{-5}\text{W}/{\text{m}}^2$ for $I$ in equation (VI).

  $\begin{array}{c}{s}_{\text{max}}=\frac{1}{146.8{\text{s}}^{-1}}\sqrt{\frac{3.16\times {10}^{-5}\text{W}/{\text{m}}^2}{2{\pi }^2\left(1.20\text{kg}/{\text{m}}^3\right)\left(343\text{m}/\text{s}\right)}}\\ =4.25\times {10}^{-7}\text{m}\end{array}$

Therefore, displacement amplitude of the initial note is $\underset{\_}{4.25\times {10}^{-7}\text{m}}$.

(f)

To determine

The displacement amplitude of the final note.

Answer to Problem 29P

The displacement amplitude of the final note is $\underset{\_}{7.09\times {10}^{-8}\text{m}}$.

Explanation of Solution

Use equation (VI) to obtain the value of the displacement amplitude of the final note.

Conclusion:

Substitute, $1.20\text{kg}/{\text{m}}^3$ for $\rho$, $343\text{m}/\text{s}$ for $\upsilon$, $880{\text{s}}^{-1}$ for $f$, and $3.16\times {10}^{-5}\text{W}/{\text{m}}^2$ for $I$ in equation (VI).

  $\begin{array}{c}{s}_{\text{max}}=\frac{1}{880{\text{s}}^{-1}}\sqrt{\frac{3.16\times {10}^{-5}\text{W}/{\text{m}}^2}{2{\pi }^2\left(1.20\text{kg}/{\text{m}}^3\right)\left(343\text{m}/\text{s}\right)}}\\ =7.09\times {10}^{-8}\text{m}\end{array}$

Therefore, displacement amplitude of the final note is $\underset{\_}{7.09\times {10}^{-8}\text{m}}$.