Q59P - University Physics with Modern Physics

Question

A soprano and a bass are singing a duet. While the soprano sings an A-sharp at 932 Hz, the bass sings an A-sharp but three octaves lower. In this concert hall, the density of air is 1.20 kg/m³ and its bulk modulus is $\mathbf{1}\mathbf{.}\mathbf{42}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }\mathbf{Pa}$. In order for their notes to have the same sound intensity level, what must be (a) the ratio of the pressure amplitude of the bass to that of the soprano and (b) the ratio of the displacement amplitude of the bass to that of the soprano? (c) What displacement amplitude (in m and in nm) does the soprano produce to sing her A-sharp at 72.0 dB?

Step-by-Step Solution

Answer

A) The ratio of pressure amplitude of the bass to that of the soprano is 1.00.B) The ratio of displacement amplitude of the bass to the soprano is 8.00.C) The displacement amplitude produced by the soprano to sing A-sharp at 72.0 dB is 47.3 nm

STEP 1 Calculate the ratio of pressure amplitude of the soprano from the relation between the intensity to pressure

Formula is ${\mathrm{l}}_{\mathrm{S}}=\frac{\mathrm{v}{\left({\mathrm{p}}^2\max \right)}_{\mathrm{s}}}{2\mathrm{B}}$ where, ${\left({\mathrm{p}}^2\max \right)}_{\mathrm{s}}$ is the pressure amplitude of the soprano. B is the bulk modulus of the air, ${\mathrm{l}}_{\mathrm{S}}$ is the intensity of the soprano. is the speed of sound in air.

The intensity of the bass is ${\mathrm{l}}_{\mathrm{b}}=\frac{1}{2}\left(4{\mathrm{\pi }}^2{\mathrm{fb}}^2\right){\mathrm{A}}_{{\mathrm{b}}^2}\sqrt{\mathrm{\rho B}}$ and the intensity of the soprano is ${\mathrm{l}}_{\mathrm{s}}=\frac{1}{2}\left(4{\mathrm{\pi }}^2{\mathrm{fs}}^2\right){\mathrm{A}}_{{\mathrm{s}}^2}\sqrt{\mathrm{\rho B}}$ where, ${\mathrm{l}}_{\mathrm{b}}$ is the intensity of the bass, B is the bulk modulus of the air. p is the density of the air. fb. is the frequency of the bass, fs is the frequency of soprano and is the intensity of soprano

Step 2 Calculate the ratio of pressure amplitude of the soprano

Rewrite the above relation in terms of ${\left({\mathrm{p}}^2\max \right)}_{\mathrm{s}}$ which is given as ${\left(\mathrm{pmax}\right)}_{\mathrm{s}}=\sqrt{\frac{2{\mathrm{Bl}}_{\mathrm{s}}}{\mathrm{v}}}$

Formula to calculate the pressure amplitude of the bass is ${\left(\mathrm{pmax}\right)}_{\mathrm{b}}=\sqrt{\frac{2{\mathrm{Bl}}_{\mathrm{b}}}{\mathrm{v}}}$ where, ${\left(\mathrm{pmax}\right)}_{\mathrm{b}}$ is the pressure amplitude of the bass ${\mathrm{l}}_{\mathrm{b}}$ is the intensity of the bass

It is given that the intensity of the both the bass and soprano is same. Divide equation (Il) by the equation (l) to find  ${\left(\mathrm{pmax}\right)}_{\mathrm{b}}/{\left(\mathrm{pmax}\right)}_{\mathrm{s}}$

$$\begin{gathered} \frac{{\left(\mathrm{pmax}\right)}_{\mathrm{b}}}{{\left(\mathrm{pmax}\right)}_{\mathrm{s}}}=\frac{\sqrt{{\displaystyle \frac{2{\mathrm{Bl}}_{\mathrm{s}}}{\mathrm{v}}}}}{\sqrt{{\displaystyle \frac{2{\mathrm{Bl}}_{\mathrm{s}}}{\mathrm{v}}}}} \\ =\sqrt{\frac{{\mathrm{l}}_{\mathrm{b}}}{{\mathrm{l}}_{\mathrm{s}}}} \end{gathered}$$

Substitute I for l_b and I for l_s in the above equation to find ${\left(\mathrm{pmax}\right)}_{\mathrm{b}}/{\left(\mathrm{pmax}\right)}_{\mathrm{s}}$

$$\begin{gathered} \frac{{\left(\mathrm{pmax}\right)}_{\mathrm{b}}}{{\left(\mathrm{pmax}\right)}_{\mathrm{s}}}=\frac{\sqrt{\mathrm{l}}}{\sqrt{\mathrm{l}}} \\ =1.00 \end{gathered}$$

Therefore, the ratio of pressure amplitude of the bass to that of the soprano is 1.00.

STEP 3 Calculate the ratio of displacement amplitude of the bass to the soprano

The intensity of the bass is  ${\mathrm{l}}_{\mathrm{b}}=\frac{1}{2}\left(4{\mathrm{\pi }}^2{\mathrm{fb}}^2\right){\mathrm{A}}_{{\mathrm{b}}^2}\sqrt{\mathrm{\rho B}}$  (Ill) and the intensity of the soprano is ${\mathrm{l}}_{\mathrm{s}}=\frac{1}{2}\left(4{\mathrm{\pi }}^2{\mathrm{fs}}^2\right){\mathrm{A}}_{{\mathrm{s}}^2}\sqrt{\mathrm{\rho B}}$  (IV) 

It is given that the intensity of the both the bass and soprano is same so that. 

Substitute equation (Ill) and (IV) in the equation (V) to find $\frac{{\mathrm{A}}_{\mathrm{b}}}{{\mathrm{A}}_{\mathrm{s}}}$

$$\begin{gathered} \frac{1}{2}\left(4{\mathrm{\pi }}^2{\mathrm{f}}_{\mathrm{b}}^2\right){\mathrm{A}}_{{\mathrm{b}}^2}\sqrt{\mathrm{\rho B}}=\frac{1}{2}\left(4{\mathrm{\pi }}^2{\mathrm{f}}_{\mathrm{s}}^2\right){\mathrm{A}}_{{\mathrm{s}}^2}\sqrt{\mathrm{\rho B}} \\ {{\mathrm{f}}_{\mathrm{b}}}^2{\mathrm{A}}_{{\mathrm{b}}^2}={{\mathrm{f}}_{\mathrm{s}}}^2{\mathrm{A}}_{{\mathrm{s}}^2} \\ \frac{{\mathrm{A}}_{{\mathrm{b}}^2}}{{\mathrm{A}}_{{\mathrm{s}}^2}}=\frac{{\mathrm{f}}_{\mathrm{s}}^2}{{{\mathrm{f}}_{\mathrm{b}}}^2} \\ \frac{{\mathrm{A}}_{\mathrm{b}}}{{\mathrm{A}}_{\mathrm{s}}}=\frac{{\mathrm{f}}_{\mathrm{s}}}{{\mathrm{f}}_{\mathrm{b}}} \end{gathered}$$

Formula to calculate the frequency of the bass is, Substitute 932 Hz for fs in the above equation to find fb

$$\begin{gathered} {\mathrm{f}}_{\mathrm{b}}=\frac{{\mathrm{f}}_{\mathrm{s}}}{8} \\ =116.5\mathrm{Hz} \end{gathered}$$

 

Thus, the frequency of the bass is 116.5Hz

Substitute 116.5 Hz for fb and 932 Hz for fs to find Ab/As

$$\begin{gathered} \frac{{\mathrm{A}}_{\mathrm{b}}}{{\mathrm{A}}_{\mathrm{s}}}=\frac{932\mathrm{Hz}}{116.5\mathrm{Hz}} \\ =8.00 \end{gathered}$$

Therefore, the ratio of displacement amplitude of the bass to the soprano is 8.00.

STEP 4 Calculate the displacement amplitude of the soprano

Formula is given by $\mathrm{A}=\frac{1}{2\mathrm{\pi f}}\sqrt{\frac{2\mathrm{l}}{\mathrm{\rho B}}}$ and the intensity of the level of sound is Where, $\mathrm{\beta }=10\log \left(\frac{\mathrm{l}}{{\mathrm{l}}_0}\right)$  is the intensity level of the sound, ${\mathrm{l}}_0$ is the intensity level of the reference sound.

Substitute 72.0 dB for $\mathrm{\beta }$ and $1\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2$ in the above equation to find I

$$\begin{gathered} 72.0\mathrm{dB}=\left(10\mathrm{dB}\right)\log \left(\frac{1}{1\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2}\right) \\ \log \left(\frac{1}{1\times {10}^{-12}W/m^2}\right)=\frac{72.0\mathrm{dB}}{\left(10\mathrm{dB}\right)} \\ I={10}^{72}\left(1\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2\right) \\ =1.585\times {10}^{-15}\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

 

Thus, the intensity of the soprano is $1.585\times {10}^{-15}\mathrm{W}/{\mathrm{m}}^2$

$$\begin{gathered} A=\frac{1}{2\pi \left(932\mathrm{Hz}\right)}\sqrt{\frac{2\left({1.5850}^{-15}\mathrm{W}/{\mathrm{m}}^2\right)}{\sqrt{\left(1.20\mathrm{kg}/{\mathrm{m}}^3\right)\left(1.42\times {10}^5\mathrm{Pa}\right)}}} \\ =4.73\times {10}^{-8}\mathrm{m} \\ =47.3\mathrm{nm} \end{gathered}$$

Therefore, the displacement amplitude produced by the soprano to sing A-sharp at 72.0 dB is 47.3 nm