Chapter 15, Problem 69P

(a)

To determine

To show:

The coefficient of the reflection and transmission is zero and $+1$ respectively when ${\text{μ}}_{{}_2}={\text{μ}}_{{}_1}$ .

Explanation of Solution

Given:

  ${\text{μ}}_{{}_2}={\text{μ}}_{{}_1}$

Formula used:

Consider:

Mass per unit length for the first string is given as:

  ${\text{μ}}_1=\frac{{\text{m}}_1}{{\text{l}}_1}$

Mass per unit length for the second string is given as:

  ${\text{μ}}_2=\frac{{\text{m}}_2}{{\text{l}}_2}$

The coefficient of the reflection is given as:

  $\text{r =}\frac{{\text{v}}_2{\text{- v}}_1}{{\text{v}}_2{\text{+ v}}_1}$

The coefficient of the transmission is given as:

  $\text{τ =}\frac{{\text{2v}}_2}{{\text{v}}_2{\text{+ v}}_1}$

Calculation:

The speed of the pulse on the first string is calculated as:

  ${\text{v}}_1\text{=}\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_1}}$...... (1)

The speed of the pulse on the second string is calculated as:

  ${\text{v}}_2\text{=}\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_2}}$...... (2)

Divide the equation (1) by equation (2) as:

  $\frac{{\text{v}}_1}{{\text{v}}_2}\text{=}\left[\frac{\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_1}}}{\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_2}}}\right]$

  $\frac{{\text{v}}_1}{{\text{v}}_2}\text{=}\sqrt{\frac{{\text{μ}}_2}{{\text{μ}}_1}}$...... (3)

Now, the coefficient of the reflection is calculated as:

  $\begin{array}{l}\text{r =}\frac{{\text{v}}_2{\text{- v}}_1}{{\text{v}}_2{\text{+ v}}_1}\\ \text{r =}\frac{\left(\text{1-}\frac{{\text{v}}_1}{{\text{v}}_2}\right)}{\left(\text{1+}\frac{{\text{v}}_1}{{\text{v}}_2}\right)}\\ \text{r =}\frac{\left(\text{1-}\sqrt{\frac{{\text{μ}}_2}{{\text{μ}}_1}}\right)}{\left(\text{1+}\sqrt{\frac{{\text{μ}}_2}{{\text{μ}}_1}}\right)}\end{array}$

Now put the values of ${\text{μ}}_1$ and ${\text{μ}}_2$ by using the given condition:

  $\begin{array}{l}\text{r =}\frac{\left(\text{1-}\sqrt{\frac{{\text{μ}}_{\text{2}}}{{\text{μ}}_{{}_{\text{2}}}}}\right)}{\left(\text{1+}\sqrt{\frac{{\text{μ}}_{\text{2}}}{{\text{μ}}_{{}_{\text{2}}}}}\right)}\\ \text{r =}\frac{\text{1-}\sqrt{\text{1}}}{\text{1+}\sqrt{\text{1}}}\\ \text{r = 0}\end{array}$

Hence, the coefficient of reflection at ${\text{μ}}_{{}_2}={\text{μ}}_{{}_1}$ is zero $\left(\text{r = 0}\right)$ .

The coefficient of the transmission is given as:

  $\text{τ =}\frac{{\text{2v}}_2}{{\text{v}}_2{\text{+ v}}_1}$

Divide the numerator and denominator of the above equation by ${\text{v}}_2$ :

  $\begin{array}{l}\text{τ =}\frac{\text{2}\left({\text{v}}_2/{\text{v}}_2\right)}{\left({\text{v}}_2/{\text{v}}_2\right)\text{+ }\left({\text{v}}_1/{\text{v}}_2\right)}\\ \text{τ =}\frac{\text{2}}{\text{1+ }\left({\text{v}}_1/{\text{v}}_2\right)}\end{array}$

Now the put the value of the ratio of $\left({\text{v}}_1/{\text{v}}_2\right)$ from the equation (3):

  $\begin{array}{l}\text{τ =}\frac{\text{2}}{\text{1+ }\left({\text{v}}_1/{\text{v}}_2\right)}\\ \text{τ =}\frac{\text{2}}{\text{1+ }\sqrt{\frac{{\text{μ}}_2}{{\text{μ}}_1}}}\end{array}$

Now put the values of ${\text{μ}}_{\text{t}}$ and ${\text{μ}}_{\text{s}}$ in above equation:

  $\begin{array}{l}\text{τ =}\frac{\text{2}}{\text{1+ }\sqrt{\frac{{\text{μ}}_2}{{\text{μ}}_1}}}\\ \text{τ =}\frac{\text{2}}{\text{1+ }\sqrt{\frac{{\text{μ}}_2}{{\text{μ}}_2}}}\\ \text{τ =}\frac{\text{2}}{\text{1+ }\sqrt{1}}\\ \text{τ = }\frac{2}{2}\\ \text{τ =}+1\end{array}$

The coefficient of the transmission at ${\text{μ}}_{{}_2}={\text{μ}}_{{}_1}$ is $\text{τ =}+1$ .

(b)

To determine

To show:

The coefficient of the reflection and transmission is $-1$ and $0$ respectively when ${\text{μ}}_{{}_2}\gg {\text{μ}}_{{}_1}$ .

Explanation of Solution

Given:

  ${\text{μ}}_{{}_2}\gg {\text{μ}}_{{}_1}$

Formula used:

Consider:

Mass per unit length for the first string is given as:

  ${\text{μ}}_1=\frac{{\text{m}}_1}{{\text{l}}_1}$

Mass per unit length for the second string is given as:

  ${\text{μ}}_2=\frac{{\text{m}}_2}{{\text{l}}_2}$

The coefficient of the reflection is given as:

  $\text{r =}\frac{{\text{v}}_2{\text{- v}}_1}{{\text{v}}_2{\text{+ v}}_1}$

The coefficient of the transmission is given as:

  $\text{τ =}\frac{{\text{2v}}_2}{{\text{v}}_2{\text{+ v}}_1}$

Calculation:

The speed of the pulse on the first string is calculated as:

  ${\text{v}}_1\text{=}\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_1}}$...... (1)

The speed of the pulse on the second string is calculated as:

  ${\text{v}}_2\text{=}\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_2}}$...... (2)

Divide the equation (1) by equation (2) as:

  $\frac{{\text{v}}_1}{{\text{v}}_2}\text{=}\left[\frac{\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_1}}}{\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_2}}}\right]$

  $\frac{{\text{v}}_1}{{\text{v}}_2}\text{=}\sqrt{\frac{{\text{μ}}_2}{{\text{μ}}_1}}$...... (3)

But as per the given condition: if ${\text{μ}}_{{}_2}\gg {\text{μ}}_{{}_1}$ then ${\text{v}}_{{}_{\text{1}}}\gg {\text{v}}_{{}_{\text{2}}}$ or ${\text{v}}_2\approx 0$

Now, the coefficient of the reflection is calculated as:

  $\begin{array}{l}\text{r =}\frac{{\text{v}}_2{\text{- v}}_1}{{\text{v}}_2{\text{+ v}}_1}\\ \text{r}\approx \frac{{\text{-v}}_1}{{\text{v}}_1}\\ \text{r}=-1\end{array}$

The coefficient of the transmission is given as:

  $\text{τ =}\frac{{\text{2v}}_2}{{\text{v}}_2{\text{+ v}}_1}$

Divide the numerator and denominator of the above equation by ${\text{v}}_2$ :

  $\begin{array}{l}\text{τ =}\frac{\text{2}\left({\text{v}}_2/{\text{v}}_2\right)}{\left({\text{v}}_2/{\text{v}}_2\right)\text{+ }\left({\text{v}}_1/{\text{v}}_2\right)}\\ \text{τ =}\frac{\text{2}}{\text{1+ }\left({\text{v}}_1/{\text{v}}_2\right)}\\ \text{τ = }0\end{array}$

The coefficient of the reflection and transmission at ${\text{μ}}_{{}_2}\gg {\text{μ}}_{{}_1}$ is $-1$ and $0$ respectively due ${\text{v}}_{{}_{\text{1}}}\gg {\text{v}}_{{}_{\text{2}}}$ .

(c)

To determine

To show:

The coefficient of the reflection and transmission is $1$ and $2$ respectively when ${\text{μ}}_{{}_2}\ll {\text{μ}}_{{}_1}$ .

Explanation of Solution

Given:

  ${\text{μ}}_{{}_2}\ll {\text{μ}}_{{}_1}$

Formula used:

Consider:

Mass per unit length for the first string is given as:

  ${\text{μ}}_1=\frac{{\text{m}}_1}{{\text{l}}_1}$

Mass per unit length for the second string is given as:

  ${\text{μ}}_2=\frac{{\text{m}}_2}{{\text{l}}_2}$

The coefficient of the reflection is given as:

  $\text{r =}\frac{{\text{v}}_2{\text{- v}}_1}{{\text{v}}_2{\text{+ v}}_1}$

The coefficient of the transmission is given as:

  $\text{τ =}\frac{{\text{2v}}_2}{{\text{v}}_2{\text{+ v}}_1}$

Calculation:

The speed of the pulse on the first string is calculated as:

  ${\text{v}}_1\text{=}\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_1}}$...... (1)

The speed of the pulse on the second string is calculated as:

  ${\text{v}}_2\text{=}\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_2}}$...... (2)

Divide the eq. (1) by eq. (2) as:

  $\frac{{\text{v}}_1}{{\text{v}}_2}\text{=}\left[\frac{\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_1}}}{\sqrt{\frac{{\text{F}}_{\text{T}}}{{\text{μ}}_2}}}\right]$

  $\frac{{\text{v}}_1}{{\text{v}}_2}\text{=}\sqrt{\frac{{\text{μ}}_2}{{\text{μ}}_1}}$...... (3)

But as per the given condition: if ${\text{μ}}_{{}_2}\ll {\text{μ}}_{{}_1}$ then ${\text{v}}_{{}_{\text{2}}}\gg {\text{v}}_{{}_{\text{1}}}$ or $\left({\text{v}}_1/{\text{v}}_2\right)\approx 0$

Now, the coefficient of the reflection is calculated as:

  $\begin{array}{l}\text{r =}\frac{{\text{v}}_{\text{2}}{\text{- v}}_{\text{1}}}{{\text{v}}_{\text{2}}{\text{+ v}}_{\text{1}}}\\ \text{r =}\frac{\left({\text{v}}_{\text{2}}/{\text{v}}_{\text{2}}\right)\text{- }\left({\text{v}}_{\text{1}}/{\text{v}}_{\text{2}}\right)}{\left({\text{v}}_{\text{2}}/{\text{v}}_{\text{2}}\right)\text{+ }\left({\text{v}}_{\text{1}}/{\text{v}}_{\text{2}}\right)}\\ \text{r =}\frac{\text{1- }\left({\text{v}}_{\text{1}}/{\text{v}}_{\text{2}}\right)}{\text{1+ }\left({\text{v}}_{\text{1}}/{\text{v}}_{\text{2}}\right)}\\ \text{r =}\frac{\text{1- 0}}{\text{1+ 0}}\\ \text{r =1}\end{array}$

The coefficient of the transmission is given as:

  $\text{τ =}\frac{{\text{2v}}_2}{{\text{v}}_2{\text{+ v}}_1}$

Divide the numerator and denominator of the above equation by ${\text{v}}_2$ :

  $\begin{array}{l}\text{τ =}\frac{\text{2}\left({\text{v}}_2/{\text{v}}_2\right)}{\left({\text{v}}_2/{\text{v}}_2\right)\text{+ }\left({\text{v}}_1/{\text{v}}_2\right)}\\ \text{τ =}\frac{\text{2}}{\text{1+ }\left({\text{v}}_1/{\text{v}}_2\right)}\\ \text{τ =}\frac{\text{2}}{\text{1+0}}\\ \text{τ = }2\end{array}$

The coefficient of the reflection and transmission at ${\text{μ}}_{{}_2}\ll {\text{μ}}_{{}_1}$ is $1$ and $2$ respectively due to ${\text{v}}_{{}_{\text{2}}}\gg {\text{v}}_{{}_{\text{1}}}$ .