91Ó°ÊÓ

• 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, $\mathrm{Δ}{p}_s=4.0 Pa$.

The expression for the pressure amplitude is given by,

$\mathrm{Δ}{P}_m=\left(v\mathrm{ÒÏ}\mathrm{Ó\neg }\right)s_m$

Here $\mathrm{Δ}{P}_m$ is the pressure amplitude, v is the speed of the sound, $\mathrm{ÒÏ}$is the density, $\mathrm{Ó\neg }$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,

$k=\frac{\mathrm{Ó\neg }}{v}$

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 $\mathrm{Δ}{p}_m=\text{8.0mpma}$

$\mathrm{Δ}{p}_m=0.00{\text{8N/m}}^{\text{2}}$

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

$$\begin{gathered} ∆p_m=\left(v\mathrm{ÒÏ}\mathrm{Ó\neg }\right)s_m \\ {s}_m=\frac{\mathrm{Δ}{p}_m}{\left(v\mathrm{ÒÏ}\mathrm{Ó\neg }\right)} \end{gathered}$$

Substitute $2\mathrm{Ï€}/Tfor\mathrm{Ó\neg }$ into the above equation,

$s_m=\frac{\mathrm{Δ}{p}_m}{\left(v\mathrm{ÒÏ}2\mathrm{Ï€}/T\right)}$

Substitute localid="1661322974677" ${\text{0.008N/m}}^{2}for\mathrm{Δ}{p}_m,343 m/sforv,0.002 sforTand1.21  kg/m^{3}for\mathrm{ÒÏ}$into the above equation,

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

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

The angular wave vector is,

$k=\frac{\mathrm{Ó\neg }}{v}$

Substitute $2\mathrm{Ï€}/Tfor\mathrm{Ó\neg }$into the above equation,

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

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

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

The angular frequency formula is,

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$

Substitute 0.002s for T into the above equation,

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2×3.14}{0.002} \\ =3.1×1{\text{0}}^{\text{3}}\text{rad/s} \end{gathered}$$

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

$$\begin{gathered} \mathrm{Δ}{p}_m=8.0\text{mPa} \\ =0.008 {\text{N/m}}^{\text{2}} \end{gathered}$$

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

$$\begin{gathered} \mathrm{Δ}{p}_m=\left(v\mathrm{ÒÏ}\mathrm{Ó\neg }\right)s_m^{\text{'}} \\ {s}_m=\frac{\mathrm{Δ}{p}_m}{\left(v\text{'}\mathrm{ÒÏ}\text{'}\mathrm{Ó\neg }\right)} \end{gathered}$$

Substitute $2\mathrm{Ï€}/Tfor\mathrm{Ó\neg }$ into the above equation,

$s_m=\frac{\mathrm{Δ}{p}_m}{\left(v\text{'}\mathrm{ÒÏ}\text{'}2\mathrm{Ï€}/T\right)}$

Substitute ${\text{0.008N/m}}^{2}for\mathrm{Δ}{p}_m,320 m/sforv,0.002 sforTand1.35  kg/m^{3}for\mathrm{ÒÏ}$ into the above equation,

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

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

The angular wave number is,

$k=\frac{\mathrm{Ó\neg }}{v\text{'}}$

Substitute $2\mathrm{Ï€}/Tfor\mathrm{Ó\neg }$ into the above equation,

$k=\frac{2\mathrm{Ï€}}{v\text{'}T}$

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

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

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