Suppose that the sound level of a conversation in initially at an angry$\mathbf{\text{70dB}}$ and then drops to a soothing$\mathbf{\text{50dB}}$. Assuming that the frequency of the sound is$\mathbf{\text{500Hz}}$, determine the (a)Initial (b) Final sound intensities and(c)Initial and(d)Final sound wave amplitude

• Initial sound level$70\text{DB}$
• Final sound level$5\text{0DB}$
• Velocity of sound wave,$v=34\text{3m/s}$
• Density of sound wave is$\rho =1.21{\text{\hspace{0.17em}\hspace{0.17em}kg/m}}^{\text{3}}$
• Standard reference intensity is$I_0={10}^{-12}{\text{W/m}}^{\text{2}}$
• Frequency of sound wave is$f=500\text{Hz}$

Using the formula for the sound level in terms of intensity of sound and threshold intensity, find the intensity of the sound for the given sound levels. Using the formula for intensity in terms of density, velocity, frequency and displacement amplitude, find the displacement amplitude.

The expression for the sound level is given by,

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

Here,$\beta$ issound level and l is the final intensity and$I_0$ is the initial intensity.

From the relation of sound level $\beta$and intensity l

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

Substitute${10}^{-12}{\text{\hspace{0.17em}W/m}}^{\text{2}}$ for $I_0$into the above equation,

role="math" localid="1661697802755" $\begin{array}{c}I={10}^{-12}\times {10}^{\left(\frac{\beta }{10}\right)}\\ ={{10}^{-12+}}^{\left(\frac{\beta }{10}\right)}{\text{W/m}}^{\text{2}}\end{array}$

High intensity is obtained at high$\beta$value, so the initial sound level is$70\text{DB}$.

Substitute$70\text{DB}$ for$\beta$ into the above equation,

$\begin{array}{c}I={{10}^{-12+}}^{\left(\frac{70}{10}\right)}{\text{W/m}}^{\text{2}}\\ ={{10}^{-12+}}^{\left(7\right)}{\text{W/m}}^{\text{2}}\\ ={10}^{-5}{\text{W/m}}^{\text{2}}\end{array}$

$\therefore I=10{\text{μW/m}}^{\text{2}}$

Hence, the final sound intensity is$10{\text{μW/m}}^{\text{2}}$ .

Now, for,$\beta =5\text{0DB}$

$I={10}^{-12}\times {10}^{\left(\frac{\beta }{10}\right)}$

Substitute $50\text{DB}$for $\beta$into the above equation,

$\begin{array}{c}I={10}^{-12}\times {10}^{\left(\frac{50}{10}\right)}\\ ={10}^{-12}\times {10}^{\left(5\right)}\\ ={10}^{-7}{\text{\hspace{0.17em}W/m}}^{\text{2}}\end{array}$

$\therefore I=0.10{\text{μW/m}}^{\text{2}}$

Hence, the final sound intensity is$0.10{\text{μW/m}}^{\text{2}}$ .

Intensity in terms of displacement amplitude is

$I=\frac{1}{2}\rho v{\omega }^2{s}_m^2$

Now, find the displacement amplitude for high intensity.

By rearranging the intensity equation,

$s_m^2=\frac{2I_{high}}{\rho v{\omega }^2}$

Substitute ${10}^{-5}{\text{W/m}}^{\text{2}}$for $I_{high}$,$1.21kg/m^3$ for$\rho$ , $343m/s$for$v$, $(2\pi \times 500)rad/sec$for$\omega$ into the above equation,

$\begin{array}{c}{s}_m^2=\frac{2\times {10}^{-5}}{1.21\times 343\times {(2\pi \times 500)}^2}\\ =4.89\times {10}^{-15}\\ =70\times 10^-9m\end{array}$

$\therefore s_m=7\text{0nm}$

Hence, the initial sound wave amplitude is $7\text{0nm}$.

Intensity in terms of displacement amplitude is,

$I=\frac{1}{2}\rho v{\omega }^2{s}_m^2$

Now, find displacement amplitude for low intensity.

By rearranging the intensity equation,

$s_m^2=\frac{2I_{low}}{\rho v{\omega }^2}$

Substitute ${10}^{-7}{\text{W/m}}^{\text{2}}$for$I_{high}$ , $1.21kg/m^3$for $\rho$, $343m/s$for $v$,$(2\pi \times 500)rad/sec$ for into the above equation,

$\begin{array}{c}{s}_m^2=\frac{2\times {10}^{-7}}{1.21\times 343\times {(2\pi \times 500)}^2}\\ =4.89\times {10}^{-17}\\ =70\times 10^-10m\end{array}$

$\therefore s_m=7\text{nm}$

Hence, the final sound wave amplitude is$7\text{nm}$