Chapter 15, Problem 46P

(a)

To determine

To calculate:

${\text{C/A=2v}}_{\text{2}}\text{/}\left({\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}\right)$ and $\text{B/A=}\left({\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}\right)\text{/}\left({\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}\right)$

Answer to Problem 46P

It is proved that $\text{B/A}$ is equals to the ${\text{v}}_{\text{1}}{\text{-v}}_{\text{2}}{\text{/v}}_{\text{1}}{\text{+v}}_{\text{2}}$ and $\text{C/A}$ is equals to the ${\text{2v}}_{\text{2}}{\text{/v}}_{\text{1}}{\text{+v}}_{\text{2}}$ .

Explanation of Solution

Given:

The speed of the wave is ${\text{v}}_{\text{1}}$ when region $\text{x<0}$ .

The speed of the wave is ${\text{v}}_{\text{2}}$ when region $\text{x}\text{>0}$ .

Formula used:

Consideration: ${\text{lim}\text{B/A}|}_{{\text{v}}_{\text{2}}{\text{/v}}_{\text{1}}\to 0}$

This limit gives $\text{B/A=+1}$ , that is telling that the transmitted wave has standing wave with node at $\text{x=0}$ .

Calculation:

Both the wave function and its first spatial derivative are continuous at $\text{x=0}$ to establish equations relating $\text{A,B,C}$ and ${\text{k}}_{\text{1}}{\text{,k}}_{\text{2}}$ .

Let, the ${\text{y}}_{\text{1}}\left(\text{x,t}\right)$ represents the wave function in the region $\text{x<0}$ and ${\text{y}}_{\text{2}}\left(\text{x,t}\right)$ represents the wave function in the region $\text{x}\text{>0}$

Now, to expression for continuity of the two-wave function’s at $\text{x=0}$ are,

  $\text{y}\left(\text{0,t}\right){\text{=y}}_{\text{2}}\left(\text{0,t}\right)$

And

  $\text{Asin}\left[{\text{k}}_{\text{1}}\left(\text{0}\right)\text{-ωt}\right]\text{+B}\text{sin}\left[{\text{k}}_{\text{1}}\left(\text{0}\right)\text{+ωt}\right]\text{=Csin}\left[{\text{k}}_{\text{2}}\left(\text{0}\right)\text{-ωt}\right]$

Or

  $\text{Asin}\left(\text{-ωt}\right)\text{+B}\text{sin}\text{ωt=Csin}\left(\text{-ωt}\right)$

The sine function is odd as it is symmetric about the origin. So, $\text{Sin}\left(\text{-θ}\right)\text{=-sin}\text{θ}$ .

  $\text{-Asin}\text{ωt+B}\text{sin}\text{ωt=}\text{-Csin}\text{ωt}$ and $\text{A-B=C}$ ……(1)

Now, differentiate the wave functions with respect to $\text{x}$ to get,

  $\frac{\partial {\text{y}}_{\text{1}}}{\partial \text{x}}{\text{=Ak}}_{\text{1}}\text{cos}\left({\text{k}}_{\text{1}}\text{,x-ωt}\right){\text{+Bk}}_{\text{1}}\text{cos}\left({\text{k}}_{\text{1}}\text{,x+ωt}\right)$

And

  $\frac{\partial {\text{y}}_{\text{2}}}{\partial \text{x}}{\text{=Ck}}_{\text{2}}\text{cos}\left({\text{k}}_{\text{2}}\text{,x-ωt}\right)$

Now, express the continuity of the slopes of the two wave functions at $\text{x=0}$ ,

  ${\frac{\partial {\text{y}}_{\text{1}}}{\partial \text{x}}|}_{\text{x=0}}\text{=}{\frac{\partial {\text{y}}_{\text{2}}}{\partial \text{x}}|}_{\text{x=0}}$

And

  ${\text{Ak}}_{\text{1}}\text{cos}\left[{\text{k}}_{\text{1}}\left(\text{0}\right)\text{-ωt}\right]{\text{+Bk}}_{\text{1}}{\text{cosωt=Ck}}_{\text{2}}\text{cos}\left(\text{-ωt}\right)$

The cosine function is even as it is symmetric about the $\text{y}$ -axis. So, $\text{cos}\left(\text{-θ}\right)\text{=}\text{cos}\text{θ}$ .

  ${\text{Ak}}_{\text{1}}\text{cos}\text{ωt}{\text{+Bk}}_{\text{1}}\text{cos}{\text{ωt=Ck}}_{\text{2}}\text{cos}\text{ωt}$ and ${\text{k}}_{\text{1}}{\text{A+k}}_{\text{1}}{\text{B=k}}_{\text{2}}\text{C}$ …(2)

Multiply the equation $\left(\text{1}\right)$ by ${\text{k}}_{\text{1}}$ and add it to equation $\left(\text{2}\right)$ to get,

  ${\text{2k}}_{\text{1}}\text{A=}\left({\text{k}}_{\text{1}}{\text{+k}}_{\text{2}}\right)\text{C}$ .

Now to solving for,

  $\text{C=}\frac{{\text{2k}}_{\text{1}}}{{\text{k}}_{\text{1}}{\text{+k}}_{\text{2}}}\text{A=}\frac{\text{2}}{{\text{1+k}}_{\text{2}}{\text{/k}}_{\text{1}}}\text{A}$

Solve for $\text{C/A}$ and substitute ${\text{ω/v}}_{\text{1}}$ for ${\text{k}}_{\text{1}}$ and ${\text{ω/v}}_{\text{2}}$ for ${\text{k}}_{\text{2}}$ to get,

  $\begin{array}{l}\frac{\text{C}}{\text{A}}\text{=}\frac{\text{2}}{{\text{1+k}}_{\text{2}}{\text{/k}}_{\text{1}}}\\ \frac{\text{C}}{\text{A}}\text{=}\frac{\text{2}}{{\text{1+v}}_{\text{1}}{\text{/v}}_{\text{2}}}\\ \frac{\text{C}}{\text{A}}\text{=}\frac{{\text{2v}}_{\text{2}}}{{\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}}\end{array}$

Now, put the equation (1) to get,

  $\text{A-B=}\left(\frac{{\text{2v}}_{\text{2}}}{{\text{v}}_{\text{2}}{\text{+v}}_{\text{1}}}\right)\text{A}$

Solving for $\text{B/A}$ yields,

  $\frac{\text{B}}{\text{A}}\text{=}\frac{{\text{v}}_{\text{1}}{\text{-v}}_{\text{2}}}{{\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}}$

Conclusion:

Hence, it is proved $\frac{\text{B}}{\text{A}}\text{=}\frac{{\text{v}}_{\text{1}}{\text{-v}}_{\text{2}}}{{\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}}$ and $\frac{\text{C}}{\text{A}}\text{=}\frac{{\text{2v}}_{\text{2}}}{{\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}}$ .

(b)

To determine

To calculate: ${\text{B}}^{\text{2}}\text{+}\left({\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}\right){\text{C}}^{\text{2}}{\text{=A}}^{\text{2}}$

Answer to Problem 46P

The identified equation is,

  ${\text{B}}^{\text{2}}\text{+}\left(\frac{{\text{v}}_{\text{1}}}{{\text{v}}_{\text{2}}}\right){\text{C}}^{\text{2}}{\text{=A}}^{\text{2}}$

Explanation of Solution

Given:

The speed of the wave is ${\text{v}}_{\text{1}}$ when region $\text{x<0}$ .

The speed of the wave is ${\text{v}}_{\text{2}}$ when region $\text{x}\text{>0}$ .

Formula used:

Consideration: ${\text{lim}\text{B/A}|}_{{\text{v}}_{\text{2}}{\text{/v}}_{\text{1}}\to 0}$

This limit gives $\text{B/A=+1}$ , that is telling that the transmitted wave has standing wave with node at $\text{x=0}$ .

Calculation:

To show the ${\text{B}}^{\text{2}}\text{+}\left({\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}\right){\text{C}}^{\text{2}}{\text{=A}}^{\text{2}}$ , there is need to use results of the part (a) to get,

  $\text{B=-}\left[\left(\text{1-α}\right)\text{/}\left(\text{1+α}\right)\right]\text{A}$ and $\text{C=2A/}\left(\text{1+α}\right)$

In which,

  ${\text{α=v}}_{\text{1}}{\text{/v}}_{\text{2}}$

After substituting the values in equation,

  ${\text{B}}^{\text{2}}\text{+}\left({\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}\right){\text{C}}^{\text{2}}{\text{=A}}^{\text{2}}$

And then check to see if result equation is identified or not,

  $\begin{array}{l}{\text{B}}^{\text{2}}\text{+}\left({\text{v}}_{\text{1}}{\text{+v}}_{\text{2}}\right){\text{C}}^{\text{2}}{\text{=A}}^{\text{2}}\\ {\left(\frac{\text{1-α}}{\text{1+α}}\right)}^{\text{2}}{\text{A}}^{\text{2}}\text{+α}{\left(\frac{\text{2}}{\text{1+α}}\right)}^{\text{2}}{\text{A}}^{\text{2}}{\text{=A}}^{\text{2}}\\ {\left(\frac{\text{1-α}}{\text{1+α}}\right)}^{\text{2}}\text{+α}{\left(\frac{\text{2}}{\text{1+α}}\right)}^{\text{2}}\text{=1}\\ \frac{{\left(\text{1-α}\right)}^{\text{2}}\text{+4α}}{{\left(\text{1+α}\right)}^{\text{2}}}\text{=1}\\ \frac{{\text{1-2α+α}}^{\text{2}}\text{+4α}}{{\left(\text{1+α}\right)}^{\text{2}}}\text{=1}\\ \frac{{\text{1-2α+α}}^{\text{2}}}{{\left(\text{1+α}\right)}^{\text{2}}}\text{=1}\\ \frac{{\left(\text{1+α}\right)}^{\text{2}}}{{\left(\text{1+α}\right)}^{\text{2}}}\text{=1}\\ \text{1=1}\end{array}$

Therefore, identified equation is,

  ${\text{B}}^{\text{2}}\text{+}\left(\frac{{\text{v}}_{\text{1}}}{{\text{v}}_{\text{2}}}\right){\text{C}}^{\text{2}}{\text{=A}}^{\text{2}}$

Conclusion:

Therefore, identified equation is ${\text{B}}^{\text{2}}\text{+}\left(\frac{{\text{v}}_{\text{1}}}{{\text{v}}_{\text{2}}}\right){\text{C}}^{\text{2}}{\text{=A}}^{\text{2}}$