Chapter 15, Problem 68P

To determine

To calculate:

The coefficient of the reflection and transmission at the point of the junction.

Answer to Problem 68P

The coefficient of the reflection and transmission at the point of the junction is $-0.20$ and $0.80$ respectively.

Explanation of Solution

Given:

Length of the string: ${\text{l}}_{\text{s}}\text{= 3}\text{.00 m}$

Mass of the string: ${\text{m}}_{\text{s}}\text{= 25}\text{.0 g}$

Length of the twine: ${\text{l}}_{\text{t}}\text{= 4}\text{.00 m}$

Mass of the twine: ${\text{m}}_t\text{= 75}\text{.0 g}$

Formula used:

The coefficient of the reflection is given as:

  $\text{r =}\frac{{\text{v}}_{\text{t}}{\text{- v}}_{\text{s}}}{{\text{v}}_{\text{t}}{\text{+ v}}_{\text{s}}}$

The coefficient of the transmission is given as:

  $\text{τ =}\frac{{\text{2v}}_{\text{t}}}{{\text{v}}_{\text{t}}{\text{+ v}}_{\text{s}}}$

Calculation:

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

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

The speed of the pulse on the twine is calculated as:

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

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

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

  $\frac{{\text{v}}_{\text{s}}}{{\text{v}}_{\text{t}}}\text{=}\sqrt{\frac{{\text{μ}}_{\text{t}}}{{\text{μ}}_{\text{s}}}}$...... (3)

${\text{μ}}_{\text{t}}$ for the string is given as:

  ${\text{μ}}_{\text{t}}\text{=}\frac{{\text{m}}_{\text{t}}}{{\text{l}}_{\text{t}}}$

${\text{μ}}_{\text{s}}$ for the string is given as:

  ${\text{μ}}_{\text{s}}\text{=}\frac{{\text{m}}_{\text{s}}}{{\text{l}}_{\text{s}}}$

Now, the coefficient of the reflection is calculated as:

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

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

  $\begin{array}{l}\text{r =}\frac{\left(\text{1-}\sqrt{\frac{\frac{{\text{m}}_{\text{t}}}{{\text{l}}_{\text{t}}}}{\frac{{\text{m}}_{\text{s}}}{{\text{l}}_{\text{s}}}}}\right)}{\left(\text{1+}\sqrt{\frac{\frac{{\text{m}}_{\text{t}}}{{\text{l}}_{\text{t}}}}{\frac{{\text{m}}_{\text{s}}}{{\text{l}}_{\text{s}}}}}\right)}\\ \text{r =}\frac{\left(\text{1-}\sqrt{\frac{{\text{m}}_{\text{t}}{\text{l}}_{\text{s}}}{{\text{m}}_{\text{s}}{\text{l}}_{\text{t}}}}\right)}{\left(\text{1+}\sqrt{\frac{{\text{m}}_{\text{t}}{\text{l}}_{\text{s}}}{{\text{m}}_{\text{s}}{\text{l}}_{\text{t}}}}\right)}\end{array}$

Put all the given values in above equation.:

  $\begin{array}{l}\text{r =}\frac{\left(\text{1-}\sqrt{\frac{{\text{m}}_{\text{t}}{\text{l}}_{\text{s}}}{{\text{m}}_{\text{s}}{\text{l}}_{\text{t}}}}\right)}{\left(\text{1+}\sqrt{\frac{{\text{m}}_{\text{t}}{\text{l}}_{\text{s}}}{{\text{m}}_{\text{s}}{\text{l}}_{\text{t}}}}\right)}\\ \text{r =}\frac{\left(\text{1-}\sqrt{\frac{\text{75}\times {\text{10}}^{-3}\times 3.00}{\text{25}\times {\text{10}}^{-3}\times 4.00}}\right)}{\left(\text{1+}\sqrt{\frac{\text{75}\times {\text{10}}^{-3}\times 3.00}{\text{25}\times {\text{10}}^{-3}\times 4.00}}\right)}\\ \text{r =}-0.20\end{array}$

The coefficient of the transmission is given as:

  $\text{τ =}\frac{{\text{2v}}_{\text{t}}}{{\text{v}}_{\text{t}}{\text{+ v}}_{\text{s}}}$

Divide the numerator and denominator of the above eq. by ${\text{v}}_{\text{t}}$ :

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

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

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

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

  $\begin{array}{l}\text{τ =}\frac{\text{2}}{\text{1+ }\sqrt{\frac{{\text{μ}}_{\text{t}}}{{\text{μ}}_{\text{s}}}}}\\ \text{τ =}\frac{\text{2}}{\text{1+ }\sqrt{\frac{\left(\frac{{\text{m}}_{\text{t}}}{{\text{l}}_{\text{t}}}\right)}{\left(\frac{{\text{m}}_{\text{s}}}{{\text{l}}_{\text{s}}}\right)}}}\\ \text{τ =}\frac{\text{2}}{\text{1+ }\sqrt{\frac{{\text{m}}_{\text{t}}{\text{l}}_{\text{s}}}{{\text{m}}_{\text{s}}{\text{l}}_{\text{t}}}}}\\ \text{τ}=\frac{\text{2}}{\left(\text{1+}\sqrt{\frac{{\text{75×10}}^{\text{-3}}\text{×3}\text{.00}}{{\text{25×10}}^{\text{-3}}\text{×4}\text{.00}}}\right)}\\ \text{τ}=0.80\end{array}$

Conclusion:

Hence, the coefficient of the reflection and transmission at the point of the junction is $\text{r =}-0.20$ and $\text{τ}=0.80$ respectively.