Q14P - Fundamentals Of Physics

Question

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 at  343 m/s through 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) $\omega$ ?

Step-by-Step Solution

Answer

Answer

1. The displacement amplitude of the wave is,$s_m=6.1\times {10}^{\text{ - 9}}\text{m}$  .
2. The angular wave vector is$k=9.\text{2rad/m}$  .
3. The angular frequency is $\omega =3.1\times {10}^{\text{3}}\text{rad/s}$ .

After the air is cooled, the sound velocity is 320 m/s   and density is${\text{1.35kg/m}}^{\text{3}}$ with the same period and amplitude.

1. The displacement amplitude of the wave is$s_m=5.9\times {\text{10}}^{\text{ - 9}}\text{m}$  .
2. The angular wave vector is $k=9.\text{8rad/m}$ .
3. The angular frequency is  $\omega =3.1\times {10}^{\text{3}}\text{rad/s}$.

Step 1: Given

• 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, $\Delta p_s=4.0 Pa$ .

Step 2: Determining the concept

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.

 

Step 3: (a) Determining the displacement amplitude of the wave if sound velocity 343 m/s and 1.21 kg / m3 density

From the figure, the period is 

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

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

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

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   $2\pi /T for \omega$  into the above equation,

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

Substitute  ${\text{0.008N/m}}^2 for \Delta p_m , 343 m/s for v ,0.002 s for T and 1.21 kg/m^3 for \rho$ 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,$s_m=6.1\times {10}^{\text{ - 9}}\text{m}$  .

Step 4: (b) Determining the angular wave vector if sound velocity 343 m/s and density 1.21 kg /m3

 

The angular wave vector is,

 $k=\frac{\omega }{v}$

Substitute  $2\pi /T for \omega$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  .

Step 4: (c) Determining the angular frequency if sound velocity 343 m/s and density 1.21 kg/m 3

 

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 10³ rad/s.

Step 4: (d) Determine the displacement amplitude of the wave after the air is cooled

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  $2\pi /T for \omega$  into the above equation,

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

Substitute  ${\text{0.008N/m}}^2 for \Delta p_m , 320 m/s for v ,0.002 s for T and 1.35 kg/m^3 for \rho$ 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 $s_m=5.9\times {\text{10}}^{\text{ - 9}}\text{m}$.

Step 4: (e) Determine the angular wave vector after the air is cooled

The angular wave number is, 

 $k=\frac{\omega }{v\text{'}}$

Substitute   $2\pi /T for \omega$   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  .