Chapter 16, Problem 51AP

(a)

To determine

The power $P(x)$ carried by this wave at a point.

Answer to Problem 51AP

The power $P(x)$ carried by this wave at a point is $\frac{1}{2k}\mu {\omega }^3{A}_0^2{e}^{-2bx}$.

Explanation of Solution

The wave function for string is given as,

  $y\left(x,t\right)=\left(A_0{e}^{-bx}\right)\sin \left(kx-\omega t\right)$

Formula to calculate the velocity of wave for small segment of string is,

  $v=\frac{dy}{dt}$

Here, $dy$ is the distance travel by wave in small segment and $dt$ is the small time interval.

Substitute $\left(A_0{e}^{-bx}\right)\sin \left(kx-\omega t\right)$ for $y$.

  $\begin{array}{c}v=\frac{d\left(A_0{e}^{-bx}\right)\sin \left(kx-\omega t\right)}{dt}\\ v=-\omega \left(A_0{e}^{-bx}\right)\cos \left(kx-\omega t\right)\end{array}$

Assume the mass presents in the small segment $dx$ of string is,

  $dm=\mu dx$

Here, $\mu$ is the linear mass density and $dx$ is the small length of the string.

Formula to calculate the kinetic energy for small segment is,

    $dK=\frac{1}{2}dm{v}^2$        (1)

Here, $dK$ is the kinetic energy for small segment and $dm$ is the mass of small segment of string.

Substitute $\mu dx$ for $m$ and $-\omega \left(A_0{e}^{-bx}\right)\cos \left(kx-\omega t\right)$ for $v$ in equation (1).

    $\begin{array}{c}dK=\frac{1}{2}\mu dx{\left(-\omega \left(A_0{e}^{-bx}\right)\cos \left(kx-\omega t\right)\right)}^2\\ =\frac{1}{2}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2{\cos }^2\left(kx-\omega t\right)dx\end{array}$        (2)

Integrate the equation (2) over all the string elements in the wavelength of the waves for total kinetic energy.

    $\begin{array}{c}{\displaystyle \int dK}={\displaystyle \underset{0}{\overset{\lambda }{\int }}\frac{1}{2}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2{\cos }^2\left(kx-\omega t\right)dx}\\ K=\frac{1}{4}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\lambda \end{array}$

Formula to calculate the potential energy for string in small segment is,

    $dp=\frac{1}{2}dm\omega y^2$        (3)

Here, $dp$ is the potential energy present in the small segment of string.

Substitute $\mu dx$ for $m$ and $\left(A_0{e}^{-bx}\right)\sin \left(kx-\omega t\right)$ for $y$ in equation (3).

    $\begin{array}{c}dp=\frac{1}{2}\mu dx\omega {\left(\left(A_0{e}^{-bx}\right)\sin \left(kx-\omega t\right)\right)}^2\\ =\frac{1}{2}\mu dx\omega {\left(A_0{e}^{-bx}\right)}^2{\sin }^2\left(kx-\omega t\right)\end{array}$        (4)

Integrate the equation (4) over all the string elements in the wavelength of the waves for total potential energy.

    $\begin{array}{c}{\displaystyle \int dp}={\displaystyle \underset{0}{\overset{\lambda }{\int }}\frac{1}{2}\mu \omega {\left(A_0{e}^{-bx}\right)}^2{\sin }^2\left(kx-\omega t\right)}dx\\ p=\frac{1}{4}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\lambda \end{array}$

Formula to calculate the power $P(x)$ is,

    $P(x)=K+p$        (5)

Substitute $\frac{1}{4}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\lambda$ for $p$ and $\frac{1}{4}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\lambda$ for $K$ in equation (5).

    $\begin{array}{c}P(x)=\frac{1}{4}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\lambda +\frac{1}{4}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\lambda \\ =\frac{1}{2}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\lambda \end{array}$

Formula to calculate the wave length is,

    $\lambda =\frac{\omega }{k}$

Here, $k$ is the wave number and $\lambda$ is the wave number.

Substitute $\frac{\omega }{k}$ for $\lambda$ in equation (5) to get $P(x)$.

    $\begin{array}{c}P(x)=\frac{1}{2}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\frac{\omega }{k}\\ =\frac{1}{2k}\mu {\omega }^3{A}_0^2{e}^{-2bx}\end{array}$

Conclusion:

Therefore, the power $P(x)$ carried by this wave at a point is $\frac{1}{2k}\mu {\omega }^3{A}_0^2{e}^{-2bx}$.

(b)

To determine

The power $P(0)$ carried by this wave at the origin.

Answer to Problem 51AP

The power $P(0)$ carried by this wave at the origin is $\frac{1}{2k}\mu {\omega }^3{A}_0{}^2$.

Explanation of Solution

The wave function for string is given as,

    $y\left(x,t\right)=\left(A_0{e}^{-bx}\right)\sin \left(kx-\omega t\right)$

Formula to calculate the power for wave in the string is,

    $P\left(x\right)=\frac{1}{2k}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2\lambda$        (6)

Substitute $0$ for $x$ in equation (6).

    $\begin{array}{c}P\left(0\right)=\frac{1}{2k}\mu {\omega }^2{\left(A_0{e}^{-b\times 0}\right)}^2\lambda \\ P\left(0\right)=\frac{1}{2k}\mu {\omega }^2{A}_0{}^2\lambda \end{array}$        (7)

Formula to calculate the wave length is,

    $\lambda =\frac{\omega }{k}$

Substitute $\frac{\omega }{k}$ for $\lambda$ in equation (7) to get $P(0)$.

    $\begin{array}{c}P\left(0\right)=\frac{1}{2k}\mu {\omega }^2{A}_0{}^2\frac{\omega }{k}\\ =\frac{1}{2k}\mu {\omega }^3{A}_0{}^2\end{array}$

Conclusion:

Therefore, the power $P(0)$ carried by this wave at the origin is $\frac{1}{2k}\mu {\omega }^3{A}_0{}^2$.

(c)

To determine

The ratio of $\frac{P\left(x\right)}{P\left(0\right)}$.

Answer to Problem 51AP

The ratio of $\frac{P\left(x\right)}{P\left(0\right)}$ is $e^{-2bx}$.

Explanation of Solution

The wave function for string is given as,

    $y\left(x,t\right)=\left(A_0{e}^{-bx}\right)\sin \left(kx-\omega t\right)$

The ratio of power given as,

    $\frac{P\left(x\right)}{P\left(0\right)}$

Substitute $\frac{1}{2k}\mu {\omega }^2{A}_0{}^2$ for $P(0)$ and $\frac{1}{2k}\mu {\omega }^2{\left(A_0{e}^{-bx}\right)}^2$ for $P(x)$.

    $\begin{array}{c}\frac{P\left(x\right)}{P\left(0\right)}=\frac{\frac{1}{2k}\mu {\omega }^3{\left(A_0{e}^{-bx}\right)}^2}{\frac{1}{2k}\mu {\omega }^3{A}_0{}^2}\\ =e^{-2bx}\end{array}$

Conclusion:

Therefore, the ratio of $\frac{P\left(x\right)}{P\left(0\right)}$ is $e^{-2bx}$.