Chapter 16, Problem 60CP

(a)

To determine

To draw: The force diagram for this element showing the force exerted on the left and the right surface.

Answer to Problem 60CP

The force diagram for this element showing the force exerted on the left and the right surface is

Explanation of Solution

Introduction: Force diagram contains all the forces acting on the body. It contains the direction of the each force acting on the body represents at its top and bottom end or left and right sides.

The force diagram for this element showing the force exerted on the left and the right surface is shown below.

Figure (1)

The force diagram of the element of gas in Figure (1) indicates the force exerted on the right and left surfaces due the pressure of the gas on the either side of the gas.

(b)

To determine

To show: The expression, $-\frac{\partial \left(\Delta P\right)}{\partial x}A\Delta x=\rho A\Delta x\frac{{\partial }^{2}s}{\partial t^2}$ by applying the Newton’s second law to the element.

Answer to Problem 60CP

The expression $-\frac{\partial \left(\Delta P\right)}{\partial x}A\Delta x=\rho A\Delta x\frac{{\partial }^{2}s}{\partial t^2}$ is proved by applying the Newton’s second law to the element.

Explanation of Solution

Let $P\left(x\right)$ represents absolute pressure as a function of $x$.

The net force to the right on the chunk of air in Figure (1) is,

$+P\left(x\right)A-P\left(x+\Delta x\right)A$

The force due to atmosphere is,

$F=-\Delta P\left(x+\Delta x\right)A+\Delta P\left(x\right)A$ (1)

Here,

$A$ is the area of the surface.

$\Delta P(x)$ is the atmospheric pressure on the surface.

$\Delta x$ is the smallest distance.

Differentiate the equation (1) with respect to $x$.

$\begin{array}{c}\frac{\partial F}{\partial x}=\left[-\Delta P\left(x+\Delta x\right)+\Delta P\left(x\right)\right]A\\ =\left[-\frac{\partial \Delta P}{\partial x}x-\frac{\partial \Delta P}{\partial x}\Delta x+\frac{\partial \Delta P}{\partial x}x\right]A\\ =-\frac{\partial \Delta P}{\partial x}\Delta xA\end{array}$

Formula to calculate the mass of the air is,

$\Delta m=\rho \Delta V$

Here,

$\rho$ is the density of the air.

$\Delta m$ is the mass of the air.

$\Delta V$ is the volume of the air.

Formula to calculate the acceleration is,

$a=\frac{{\partial }^{2}s}{\partial t^2}$

Here,

$s$ is the distance.

$t$ is the time.

From Newton’s second law, formula to calculate the Force is,

$\frac{\partial F}{\partial x}=\Delta ma$ (2)

Substitute $\frac{{\partial }^{2}s}{\partial t^2}$ for $a$, $\rho \Delta V$ for $\Delta m$ and $-\frac{\partial \Delta P}{\partial x}\Delta xA$ for $\frac{\partial F}{\partial x}$ in equation (2).

$-\frac{\partial \Delta P}{\partial x}\Delta xA=\rho \Delta V\frac{{\partial }^{2}s}{\partial t^2}$

Conclusion:

Therefore the expression, $-\frac{\partial \left(\Delta P\right)}{\partial x}A\Delta x=\rho A\Delta x\frac{{\partial }^{2}s}{\partial t^2}$ by applying the Newton’s second law to the element.

(c)

To determine

To show: The wave equation for sound is $\frac{B}{\rho }\frac{{\partial }^{2}s}{\partial x^2}=\frac{{\partial }^{2}s}{\partial t^2}$.

Answer to Problem 60CP

The following wave equation for sound is $\frac{B}{\rho }\frac{{\partial }^{2}s}{\partial x^2}=\frac{{\partial }^{2}s}{\partial t^2}$.

Explanation of Solution

Given info:  The value of the $\Delta P$ is $B\frac{\partial s}{\partial x}$.

From part (b), the given expression is,

$-\frac{\partial \left(\Delta P\right)}{\partial x}A\Delta x=\rho A\Delta x\frac{{\partial }^{2}s}{\partial t^2}$

Substitute $B\frac{\partial s}{\partial x}$ for $\Delta P$.

$\begin{array}{c}-\frac{\partial }{\partial x}\left(B\frac{\partial s}{\partial x}\right)A\Delta x=\rho A\Delta x\frac{{\partial }^{2}s}{\partial t^2}\\ \frac{B}{\rho }\frac{{\partial }^{2}s}{\partial x^2}=\frac{{\partial }^{2}s}{\partial t^2}\end{array}$

Thus, the wave equation for sound is $\frac{B}{\rho }\frac{{\partial }^{2}s}{\partial x^2}=\frac{{\partial }^{2}s}{\partial t^2}$.

Conclusion:

Therefore, the wave equation for sound is $\frac{B}{\rho }\frac{{\partial }^{2}s}{\partial x^2}=\frac{{\partial }^{2}s}{\partial t^2}$.

(d)

To determine

To show: The function $s\left(x,t\right)=s_{\max }\cos \left(kx-\omega t\right)$ satisfies the wave equation $\frac{\omega }{k}=v=\sqrt{\frac{B}{\rho }}$.

Answer to Problem 60CP

The function $s\left(x,t\right)=s_{\max }\cos \left(kx-\omega t\right)$ satisfies the wave equation $\frac{\omega }{k}=v=\sqrt{\frac{B}{\rho }}$.

Explanation of Solution

The given wave equation is,

$s\left(x,t\right)=s_{\max }\cos \left(kx-\omega t\right)$ (3)

Apply the trial solution in the above equation.

Double differentiate the equation (1) with respect to $x$.

$\begin{array}{c}\frac{\partial s}{\partial x}=-k{s}_{\max }\sin \left(kx-\omega t\right)\\ \frac{{\partial }^{2}s}{\partial x^2}=-k^2{s}_{\max }\cos \left(kx-\omega t\right)\end{array}$

Double differentiate the equation (1) with respect to $t$.

$\begin{array}{c}\frac{\partial s}{\partial t}=+\omega s_{\max }\sin \left(kx-\omega t\right)\\ \frac{{\partial }^{2}s}{\partial t^2}=-\omega s_{\max }\cos \left(kx-\omega t\right)\end{array}$

The wave equation for sound in part (c) is,

$\frac{B}{\rho }\frac{{\partial }^{2}s}{\partial x^2}=\frac{{\partial }^{2}s}{\partial t^2}$ (4)

Substitute $-\omega s_{\max }\cos \left(kx-\omega t\right)$ for $\frac{{\partial }^{2}s}{\partial t^2}$ and $-k^2{s}_{\max }\cos \left(kx-\omega t\right)$ for $\frac{{\partial }^{2}s}{\partial x^2}$ in equation (2).

$\begin{array}{c}\frac{B}{\rho }\left(-k^2{s}_{\max }\cos \left(kx-\omega t\right)\right)=-{\omega }^2{s}_{\max }\cos \left(kx-\omega t\right)\\ \frac{B}{\rho }{k}^2={\omega }^2\\ \frac{\omega }{k}=\sqrt{\frac{B}{\rho }}\\ \frac{\omega }{k}=\sqrt{\frac{B}{\rho }}\end{array}$

Thus, the function $s\left(x,t\right)=s_{\max }\cos \left(kx-\omega t\right)$ satisfies the wave equation $\frac{\omega }{k}=v=\sqrt{\frac{B}{\rho }}$.

Conclusion:

Therefore, the function $s\left(x,t\right)=s_{\max }\cos \left(kx-\omega t\right)$ satisfies the wave equation $\frac{\omega }{k}=v=\sqrt{\frac{B}{\rho }}$.