Chapter 17, Problem 17.29P

The most soaring vocal melody is in Johann Sebastian Bach's Mass in B Minor. In one section, the basses, tenors. altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of l 46.8 Hz and 880.0 Hz. Find the wave lengths of (a) the initial note and (b) the final note. Assume the chorus sings the melody with a uniform sound level of 75.0 dB. Find the pressure amplitudes of (c) the initial note and (d) the final note. Find the displacement amplitudes of (e) the initial note and (f) the final note.

To determine

The wavelength of the initial note

Answer to Problem 17.29P

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

Explanation of Solution

Given info: The initial assigned frequency is $146.8\text{Hz}$ and the final frequency is $880.0\text{Hz}$

The velocity of air in sound is $343\text{m}/\text{s}$.

Write the expression for the wavelength of the initial note.

$\lambda =\frac{v}{f_{\text{i}}}$

Here,

$v$ is the velocity.

$f_{\text{i}}$ is the initial note frequency.

Substitute $343\text{m}/\text{s}$ for $v$ and $146.8\text{Hz}$ for $f_{\text{i}}$ in above expression.

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

Conclusion:

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

To determine

The wavelength of the final note.

Answer to Problem 17.29P

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

Explanation of Solution

Given info: The final assigned frequency is $146.8\text{Hz}$ and the final frequency is $880.0\text{Hz}$.

The velocity of air in sound is $343\text{m}/\text{s}$.

Write the expression for the wavelength of the final note.

$\lambda =\frac{v}{f_{\text{f}}}$

Here,

$f_{\text{f}}$ is the final note frequency.

Substitute $343\text{m}/\text{s}$ for $v$ and $880.0\text{Hz}$ for $f_{\text{f}}$ in above expression.

$\begin{array}{c}\lambda =\frac{343\text{m}/\text{s}}{880.0\text{Hz}}\\ =0.3897\text{m}\\ \simeq \text{0}\text{.390}\text{m}\end{array}$

Conclusion:

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

To determine

The pressure amplitude of the initial note.

Answer to Problem 17.29P

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

Explanation of Solution

Given info: The final assigned frequency is $146.8\text{Hz}$ and the final frequency is $880.0\text{Hz}$.

The sound level of the melody is $75.0\text{dB}$.

Write the expression for the intensity level of the sound.

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

Here,

$I$ is the intensity of the sound.

$I_0$ is the reference intensity.

Substitute $1.00\times {10}^{-12}\text{W}/{\text{m}}^{\text{2}}$ for $I_o$ and $75.0\text{dB}$ for $I_o$ in above expression.

$\begin{array}{c}75.0\text{dB}=10\log \left(\frac{I}{1.00\times {10}^{-12}\text{W}/{\text{m}}^{\text{2}}}\right)\\ I=3.16\times {10}^{-5}\text{W}/{\text{m}}^{\text{2}}\end{array}$

Thus the intensity of the sound is $3.16\times {10}^{-5}\text{W}/{\text{m}}^{\text{2}}$.

Write the expression for the maximum change in pressure.

$\Delta P_{\max }=\sqrt{2\rho vI}$

Here,

$I$ is the intensity.

$\rho$ is the density.

$v$ is the velocity.

Substitute $1.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $343\text{m}/\text{s}$ for $v$ and $3.16\times {10}^{-5}\text{W}/{\text{m}}^{\text{2}}$ for $I$ in above expression.

$\begin{array}{c}\Delta P_{\max }=\sqrt{2\times 1.2\text{kg}/{\text{m}}^{\text{3}}\times 343\text{m}/\text{s}\times 3.16\times {10}^{-5}\text{W}/{\text{m}}^{\text{2}}}\\ =0.161\text{Pa}\end{array}$

Conclusion:

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

To determine

The pressure amplitude of the final note.

Answer to Problem 17.29P

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

Explanation of Solution

Since the level of the sound from melody is same for both the initial and final note so the intensity of the sound for final note also remains same due to which the pressure amplitude for the final note is same as the pressure amplitude for the initial note.

From Part (c) of the question, the speed of the final note is,

$\Delta P_{\max }=0.161\text{Pa}$.

Conclusion:

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

To determine

The displacement amplitude of the initial note.

Answer to Problem 17.29P

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

Explanation of Solution

Given info: The final assigned frequency is $146.8\text{Hz}$ and the final frequency is $880.0\text{Hz}$.

The sound level of the melody is $75.0\text{dB}$.

Write the expression for the displacement amplitude of the of the initial note.

${\left(s_{\max }\right)}_{\text{i}}=\frac{\Delta P_{\max }}{\rho v2\pi f_{\text{i}}}$

Substitute $0.161\text{Pa}$ for $\Delta P_{\max }$, $1.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $343\text{m}/\text{s}$ for $v$ and $146.8\text{Hz}$ for $f_{\text{i}}$ in above expression.

$\begin{array}{c}{\left(s_{\max }\right)}_{\text{i}}=\frac{0.161\text{Pa}}{1.2\text{kg}/{\text{m}}^{\text{3}}\times 343\text{m}/\text{s}\times 2\pi \times 146.8\text{Hz}}\\ =4.25\times {10}^{-7}\text{m}\end{array}$

Conclusion:

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

To determine

The displacement amplitude of the final note.

Answer to Problem 17.29P

The displacement amplitude of the final note is $7.07\times {10}^{-8}\text{m}$.

Explanation of Solution

Given info: The final assigned frequency is $146.8\text{Hz}$ and the final frequency is $880.0\text{Hz}$.

The sound level of the melody is $75.0\text{dB}$.

The expression for the displacement amplitude of the of the final note is,

${\left(s_{\max }\right)}_{\text{f}}=\frac{\Delta P_{\max }}{\rho v2\pi f_{\text{f}}}$

Substitute $0.161\text{Pa}$ for $\Delta P_{\max }$, $1.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $343\text{m}/\text{s}$ for $v$ and $880.0\text{Hz}$ for $f_{\text{f}}$ in above expression.

$\begin{array}{c}{\left(s_{\max }\right)}_{\text{f}}=\frac{0.161\text{Pa}}{1.2\text{kg}/{\text{m}}^{\text{3}}\times 343\text{m}/\text{s}\times 2\pi \times 880.0\text{Hz}}\\ =7.07\times {10}^{-8}\text{m}\end{array}$

Conclusion:

Therefore, the displacement amplitude of the final note is $7.07\times {10}^{-8}\text{m}$.