91影视

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

The expression for the pressure amplitude is given by,

$螖P_m=\left(v蚁蝇\right)s_m$

Here $螖P_m$ is the pressure amplitude, v is the speed of the sound, $蚁$is the density, $蝇$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,

$k=\frac{蝇}{v}$

Here k is the spring constant.

From the figure, the period is

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

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

$螖p_m=0.00{\text{8N/m}}^{\text{2}}$

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

$$\begin{gathered} 鈭�p_m=\left(v蚁蝇\right)s_m \\ {s}_m=\frac{螖p_m}{\left(v蚁蝇\right)} \end{gathered}$$

Substitute $2蟺/Tfor蝇$ into the above equation,

$s_m=\frac{螖p_m}{\left(v蚁2蟺/T\right)}$

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

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

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

The angular wave vector is,

$k=\frac{蝇}{v}$

Substitute $2蟺/Tfor蝇$into the above equation,

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

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

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

The angular frequency formula is,

$蝇=\frac{2蟺}{T}$

Substitute 0.002s for T into the above equation,

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

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

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

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

The pressure amplitude is

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

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

$$\begin{gathered} 螖p_m=\left(v蚁蝇\right)s_m^{\text{'}} \\ {s}_m=\frac{螖p_m}{\left(v\text{'}蚁\text{'}蝇\right)} \end{gathered}$$

Substitute $2蟺/Tfor蝇$ into the above equation,

$s_m=\frac{螖p_m}{\left(v\text{'}蚁\text{'}2蟺/T\right)}$

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

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

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

The angular wave number is,

$k=\frac{蝇}{v\text{'}}$

Substitute $2蟺/Tfor蝇$ into the above equation,

$k=\frac{2蟺}{v\text{'}T}$

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

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

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