91Ó°ÊÓ

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 role="math" localid="1668079470446" ${\mathrm{l}}_{\mathrm{b}}=\frac{1}{2}\left(4{\mathrm{Ï€}}^2{\mathrm{fb}}^2\right){\mathrm{A}}_{{\mathrm{b}}^2}\sqrt{\mathrm{ÒÏ\mu þ}}$and the intensity of the soprano is ${\mathrm{l}}_{\mathrm{s}}=\frac{1}{2}\left(4{\mathrm{Ï€}}^2{\mathrm{fs}}^2\right){\mathrm{A}}_{{\mathrm{s}}^2}\sqrt{\mathrm{ÒÏ\mu þ}}$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

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 findrole="math" localid="1668080143577" ${\left(\mathrm{pmax}\right)}_{\mathrm{b}}/{\left(\mathrm{pmax}\right)}_{\mathrm{s}}$

role="math" localid="1668080153255" $$\begin{gathered} \frac{{\left(\mathrm{pmax}\right)}_{\mathrm{b}}}{{\left(\mathrm{pmax}\right)}_{\mathrm{s}}}=\frac{\sqrt{{\displaystyle \frac{2{\mathrm{Bl}}_{\mathrm{s}}}{\mathrm{v}}}}}{\sqrt{{\displaystyle \frac{2{\mathrm{Bl}}_{\mathrm{s}}}{\mathrm{v}}}}} \\ =\sqrt{\frac{{\mathrm{l}}_{\mathrm{b}}}{{\mathrm{l}}_{\mathrm{s}}}} \end{gathered}$$

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.

The intensity of the bass is ${\mathrm{l}}_{\mathrm{b}}=\frac{1}{2}\left(4{\mathrm{Ï€}}^2{\mathrm{fb}}^2\right){\mathrm{A}}_{{\mathrm{b}}^2}\sqrt{\mathrm{ÒÏ\mu þ}}$ (Ill)and the intensity of the soprano is ${\mathrm{l}}_{\mathrm{s}}=\frac{1}{2}\left(4{\mathrm{Ï€}}^2{\mathrm{fs}}^2\right){\mathrm{A}}_{{\mathrm{s}}^2}\sqrt{\mathrm{ÒÏ\mu þ}}$(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$\frac{{\mathrm{A}}_{\mathrm{b}}}{{\mathrm{A}}_{\mathrm{s}}}$

$$\begin{gathered} \frac{1}{2}\left(4{\mathrm{Ï€}}^2{\mathrm{f}}_{\mathrm{b}}^2\right){\mathrm{A}}_{{\mathrm{b}}^2}\sqrt{\mathrm{ÒÏ\mu þ}}=\frac{1}{2}\left(4{\mathrm{Ï€}}^2{\mathrm{f}}_{\mathrm{s}}^2\right){\mathrm{A}}_{{\mathrm{s}}^2}\sqrt{\mathrm{ÒÏ\mu þ}} \\ {{\mathrm{f}}_{\mathrm{b}}}^2{\mathrm{A}}_{{\mathrm{b}}^2}={{\mathrm{f}}_{\mathrm{s}}}^2{\mathrm{A}}_{{\mathrm{s}}^2} \\ \frac{{\mathrm{A}}_{{\mathrm{b}}^2}}{{\mathrm{A}}_{{\mathrm{s}}^2}}=\frac{{\mathrm{f}}_{\mathrm{s}}^2}{{{\mathrm{f}}_{\mathrm{b}}}^2} \\ \frac{{\mathrm{A}}_{\mathrm{b}}}{{\mathrm{A}}_{\mathrm{s}}}=\frac{{\mathrm{f}}_{\mathrm{s}}}{{\mathrm{f}}_{\mathrm{b}}} \end{gathered}$$

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.

Formula is given by$\mathrm{A}=\frac{1}{2\mathrm{Ï€´Ú}}\sqrt{\frac{2\mathrm{l}}{\mathrm{ÒÏ\mu þ}}}$ and the intensity of the level of sound is Where,$\mathrm{β}=10\log \left(\frac{\mathrm{l}}{{\mathrm{l}}_0}\right)$is the intensity level of the sound,${\mathrm{l}}_0$is the intensity level of the reference sound.

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

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

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

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

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