Question: Figure shows the output from a pressure monitor mounted at a point along the path taken by a sound wave of single frequency traveling at343 m/sthrough air with a uniform density of 1.12 kg/m³ . The vertical axis scale is set by $\mathit{\Delta }{\mathbf{p}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{}{\mathbf{p}}_{\mathbf{a}}$.If the displacement function of the wave is $\mathbf{s}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{s}}_{\mathbf{m}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{kx}\mathbf{-}\mathbf{\omega t}\mathbf{)}\mathbf{,}$what are (a) s_m (b) k (c) $\omega$ ? The air is then cooled so that its density is 1.35 kg/m³ and the speed of a sound wave through it is 320 m/s. The sound source again emits the sound wave at the same frequency and same pressure amplitude. Find the following quantities (d) s_m (e) k (f) $\omega$?

• The velocity of sound 343 m/s and uniform density $1.21 {\text{kg/m}}^{\text{3}}$.
• The velocity of sound 320 m/s and uniform density $1.35 {\text{kg/m}}^{\text{3}}$.
• Pressure amplitude is, $\Delta p_s=4.0 Pa$.

The expression for the pressure amplitude is given by,

$\Delta P_m=\left(v\rho \omega \right)s_m$

Here $\Delta P_m$ is the pressure amplitude, v is the speed of the sound, $\rho$is the density, $\omega$is the angular frequency, $s_m$ is the displacement amplitude.

The expression for the angular frequency is given by,

w =2 pi /T

Here T is the time period.

The expression for the spring constant is given by,

$k=\frac{\omega }{v}$

Here k is the spring constant.

From the figure, the period is

$$\begin{gathered} T=2.0\text{ms} \\ =\text{0.002} \text{s} \end{gathered}$$

The pressure amplitude is $\Delta p_m=\text{8.0mpma}$

$\Delta p_m=0.00{\text{8N/m}}^{\text{2}}$

From the equation, calculate the displacement amplitude of the wave,

$$\begin{gathered} ∆p_m=\left(v\rho \omega \right)s_m \\ {s}_m=\frac{\Delta p_m}{\left(v\rho \omega \right)} \end{gathered}$$

Substitute $2\pi /Tfor\omega$ into the above equation,

$s_m=\frac{\Delta p_m}{\left(v\rho 2\pi /T\right)}$

Substitute localid="1661322974677" ${\text{0.008N/m}}^{2}for\Delta p_m,343 m/sforv,0.002 sforTand1.21 kg/m^{3}for\rho$into the above equation,

$$\begin{gathered} s_m=\frac{0.008}{\left(343\times 1.21\times \frac{2\pi }{0.002}\right)} \\ =6.1\times {10}^{-9}\text{m} \end{gathered}$$

Hence, the displacement amplitude of the wave is,$s_m=6.1\times {10}^{\text{- 9}}\text{m}$.

The angular wave vector is,

$k=\frac{\omega }{v}$

Substitute $2\pi /Tfor\omega$into the above equation,

Substitute 343 m/s for v , 0.002 s for T into the above equation,

$$\begin{gathered} k=\frac{2\times 3.14}{343\times 0.002} \\ =9.\text{2rad/m} \end{gathered}$$

Hence, the angular wave vector is 9.2 rad / m .

The angular frequency formula is,

$\omega =\frac{2\pi }{T}$

Substitute 0.002s for T into the above equation,

$$\begin{gathered} \omega =\frac{2\times 3.14}{0.002} \\ =3.1\times 1{\text{0}}^{\text{3}}\text{rad/s} \end{gathered}$$

Hence, the angular frequency is 3.1 X 10³rad/s.

From the figure given in the question, the time period is,

$$\begin{gathered} T=2.0\text{ms} \\ =\text{0.002} \text{s} \end{gathered}$$

The pressure amplitude is

$$\begin{gathered} \Delta p_m=8.0\text{mPa} \\ =0.008 {\text{N/m}}^{\text{2}} \end{gathered}$$

From the equation, calculate the displacement amplitude of the wave,

$$\begin{gathered} \Delta p_m=\left(v\rho \omega \right)s_m^{\text{'}} \\ {s}_m=\frac{\Delta p_m}{\left(v\text{'}\rho \text{'}\omega \right)} \end{gathered}$$

Substitute $2\pi /Tfor\omega$ into the above equation,

$s_m=\frac{\Delta p_m}{\left(v\text{'}\rho \text{'}2\pi /T\right)}$

Substitute ${\text{0.008N/m}}^{2}for\Delta p_m,320 m/sforv,0.002 sforTand1.35 kg/m^{3}for\rho$ into the above equation,

$$\begin{gathered} s_m=\frac{0.008}{\left(320\times 1.35\times \frac{2\pi }{0.002}\right)} \\ =5.9\times 1{\text{0}}^{\text{- 9}} \text{m} \end{gathered}$$

Hence, the displacement amplitude of the wave is $s_m=5.9\times {\text{10}}^{\text{- 9}}\text{m}$.

The angular wave number is,

$k=\frac{\omega }{v\text{'}}$

Substitute $2\pi /Tfor\omega$ into the above equation,

$k=\frac{2\pi }{v\text{'}T}$

Substitute 320 m/s for v , 0.002 s forT into the above equation,

$$\begin{gathered} k=\frac{2\times 3.14}{320\times 0.002} \\ =9.\text{8rad/m} \end{gathered}$$

Hence, the angular wave vector is k = 9.8 rad/m .