91Ó°ÊÓ

Write the expression for the net force in a viscous medium.

$\mathit{F}\mathbf{=}\mathit{T}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{f}}{\mathbf{∂}{\mathbf{z}}^{\mathbf{2}}}\mathbf{∆}\mathit{z}\mathbf{-}\mathbf{∆}{\mathit{F}}_{\mathbf{d}\mathbf{r}\mathbf{a}\mathbf{g}}$ .....(1)

(a)

Substitute $∆F_{drag}=-y\frac{∂f}{∂t}∆z$in equation (1).

$F=T\frac{{∂}^{2}f}{∂z^2}∆z-y\frac{∂f}{∂t}∆z$ …… (2)

Write the expression for the net transverse force on the segment of string between z and $z+∆z$.

$F=\mathrm{μ}∆z\frac{{∂}^{2}f}{∂z^2}$

Substitute $F=\mathrm{μ}∆z\frac{{∂}^{2}f}{∂z^2}$in equation (2).\

$$\begin{gathered} \mathrm{μ}∆z\frac{{∂}^{2}f}{∂z^2}=T\frac{{∂}^{2}f}{∂z^2}∆z-y\frac{{∂}^{2}f}{∂t}∆z \\ T\frac{{∂}^{2}f}{∂z^2}=\mathrm{μ}\frac{{∂}^{2}f}{∂z^2}+y\frac{∂f}{∂t} \end{gathered}$$ .......(3)

Therefore, the modified wave equation describing the motion of the string is

$T\frac{{∂}^{2}f}{∂z^2}=\mathrm{μ}\frac{{∂}^{2}f}{∂z^2}+y\frac{∂f}{∂t}$

(b)

Write the standard form of the solution of the differential equation (3).

$\tilde{f}\left(z,t\right)=e^{-i\mathrm{Ó\neg }t}\widetilde{F\left(z\right)}$ …… (4)

Take the partial derivative of the equation (4) with respect to z.

$$\begin{gathered} \frac{∂f}{∂z}=\frac{∂}{∂z}\left(e^{-i\mathrm{Ó\neg }t}\tilde{F}\left(z\right)\right) \\ \frac{∂f}{∂z}=e^{-i\mathrm{Ó\neg }t}\frac{dF}{dz} \end{gathered}$$

Again differentiate the above equation with respect to z.

$$\begin{gathered} \frac{{∂}^{2}f}{∂z^2}=\frac{∂}{∂z}\left(e^{-i\mathrm{Ó\neg }t}\frac{dF}{dz}\right) \\ \frac{{∂}^{2}f}{∂z^2}=e^{-i\mathrm{Ó\neg }t}\frac{d^{2}F}{d{z}^2} \end{gathered}$$

Now, take the partial derivative of the equation (4) with respect to t.

$$\begin{gathered} \frac{∂f}{∂t}=\frac{∂}{∂t}\left(e^{i\mathrm{Ó\neg }t}\widetilde{F\left(z\right)}\right) \\ \frac{∂f}{∂t}=\tilde{F}\left(z\right)\left(-1\mathrm{Ó\neg }\right)e^{-i\mathrm{Ó\neg }t} \end{gathered}$$

Again differentiate the above equation with respect to t.

$$\begin{gathered} \frac{{∂}^{2}f}{∂t^2}=\frac{∂}{∂z}\left(\tilde{F}\left(z\right)\left(-i\mathrm{Ó\neg }\right)e^{-i\mathrm{Ó\neg }t}\right) \\ \frac{{∂}^{2}f}{∂t^2}=\left(-{\mathrm{Ó\neg }}^2\right)\tilde{F}\left(z\right)e^{-i\mathrm{Ó\neg }t} \end{gathered}$$

Substitute $\frac{{∂}^{2}f}{∂t^2}=e^{-i\mathrm{Ó\neg }t}\frac{d^{2}F}{d{z}^2}$and $\frac{∂f}{∂t}=\tilde{F}\left(z\right)\left(-i\mathrm{Ó\neg }\right)e^{-i\mathrm{Ó\neg }t}$in equation (3).

$$\begin{gathered} T\left(e^{-i\mathrm{Ó\neg }t}\frac{d^{2}F}{d{z}^2}\right)=\mathrm{μ}\left(e^{-i\mathrm{Ó\neg }t}\frac{d^{2}F}{d{z}^2}\right)+y\left(\tilde{F}\left(z\right)\left(-i\mathrm{Ó\neg }\right)e^{-i\mathrm{Ó\neg }t}\right) \\ T\left(\frac{d^2\tilde{F}}{d{z}^2}\right)=-\mathrm{μ}{\mathrm{Ó\neg }}^2\tilde{F}\left(z\right)-i\mathrm{Ó\neg }y\tilde{F}\left(z\right) \\ \left(\frac{d^2\tilde{F}}{d{z}^2}\right)=-\frac{\left(\mathrm{μ}{\mathrm{Ó\neg }}^2+i\mathrm{Ó\neg }y\right)}{T}\tilde{F}\left(z\right) \\ \left(\frac{d^2\tilde{F}}{d{z}^2}\right)=-\frac{\mathrm{Ó\neg }}{T}\left(\mathrm{μ}\mathrm{Ó\neg }+iy\right)\tilde{F}\left(z\right) \end{gathered}$$

Let${\tilde{k}}^2=\frac{\mathrm{Ó\neg }}{T}\left(\mathrm{μ}\mathrm{Ó\neg }+iy\right)$in the above equation.

$\left(\frac{d^2\tilde{F}}{d{z}^2}\right)=-{\tilde{k}}^2\tilde{F}\left(z\right)$

Write the solution of the differential equation.

$\tilde{F}\left(z\right)=\tilde{A}{e}^{i\tilde{k}z}+\tilde{B}{e}^{-i\tilde{k}z}$

Substitute $\tilde{k}=k+ik$in the above equation.

$$\begin{gathered} \tilde{F}\left(z\right)=\tilde{A}{e}^{i\left(k+ik\right)z}+\tilde{B}{e}^{-i\left(k+ik\right)z} \\ \tilde{F}\left(z\right)=\tilde{A}{e}^{ikz}{e}^{-kz}+\tilde{B}{e}^{-ikz}{e}^{kz} \end{gathered}$$

Since, k is greater than zero, the second term increases exponentially with increasing z, so B becomes zero.

$\tilde{F}\left(z\right)=\tilde{A}{e}^{ikz}{e}^{-kz}+0$

Substitute the value in equation (4).

$$\begin{gathered} \tilde{f}\left(z,t\right)=e^{-i\mathrm{Ó\neg }t}\tilde{A}{e}^{ikz}{e}^{-kz} \\ \tilde{f}\left(z,t\right)=\tilde{A}{e}^{-kz}{e}^{i\left(kz-\mathrm{Ó\neg }t\right)} \end{gathered}$$

Therefore, the solution of the equation is $\tilde{f}\left(z,t\right)=\tilde{A}{e}^{-kz}{e}^{i\left(kz-\mathrm{Ó\neg }t\right)}$.

(c)

As the amplitude$\frac{1}{e}$of its original value, write the penetration distance.

$$\begin{gathered} \frac{1}{e}=e^{-k{z}_0} \\ {z}_0=\frac{1}{k} \end{gathered}$$ …… (5)

Solve the complex $\tilde{k}$value.

$$\begin{gathered} \tilde{k}=k+ik \\ {\tilde{k}}^2=k^2+i^2{k}^2+2kki \\ {\tilde{k}}^2=k^2-k^2+2ikk \end{gathered}$$

Since, it is assumed as:

$$\begin{gathered} {\tilde{k}}^2=\frac{\mathrm{Ó\neg }}{T}\left(\mathrm{μ}\mathrm{Ó\neg }+i\mathrm{γ}\right) \\ {k}^2-k^2+2ikk=\frac{\mathrm{Ó\neg }}{T}\left(\mathrm{μ}\mathrm{Ó\neg }+i\mathrm{γ}\right)......\left(6\right) \end{gathered}$$

Compare the imaginary terms on both sides of equation (6)

$$\begin{gathered} 2kk=\frac{\mathrm{Ó\neg }}{T}\mathrm{γ} \\ k=\frac{\mathrm{Ó\neg }}{2kT} \end{gathered}$$

Compare the real terms on both sides of equation (6).

$k^2-k^2=\frac{\mathrm{μ}{\mathrm{Ó\neg }}^2}{T}$

Substitute $k=\frac{\mathrm{Ó\neg }\mathrm{γ}}{2kT}$in the above expression.

$$\begin{gathered} k^2-{\left(\frac{\mathrm{Ó\neg }\mathrm{γ}}{2kT}\right)}^2=\frac{\mathrm{μ}{\mathrm{Ó\neg }}^2}{T} \\ {k}^2-\frac{1}{k^2}{\left(\frac{\mathrm{Ó\neg }\mathrm{γ}}{2T}\right)}^2=\left(\frac{\mathrm{μ}{\mathrm{Ó\neg }}^2}{T}\right) \\ {k}^4-{\left(\frac{\mathrm{Ó\neg }\mathrm{γ}}{2T}\right)}^2=\left(\frac{\mathrm{μ}{\mathrm{Ó\neg }}^2}{T}\right)k^2 \\ \left(k^2\right)-\left(\frac{\mathrm{μ}{\mathrm{Ó\neg }}^2}{T}\right)k^2-{\left(\frac{\mathrm{Ó\neg }\mathrm{γ}}{2T}\right)}^2=0 \end{gathered}$$

Solve the above equation by using a quadratic formula.

localid="1657697166890" role="math" $$\begin{gathered} k^2=\frac{-\left(-\left({\displaystyle \frac{\mathrm{μ}\mathrm{Ó\neg }}{T}}\right)\right)Â\pm \sqrt{\left(-\left({\displaystyle \frac{\mathrm{μ}\mathrm{Ó\neg }}{T}}\right)-4\left(1\right)\left(-{\left({\displaystyle \frac{\mathrm{Ó\neg }\mathrm{γ}}{2T}}\right)}^2\right)\right)}}{2\left(1\right)} \\ {k}^2=\frac{\frac{\mathrm{μ}\mathrm{Ó\neg }}{T}+\sqrt{{\left({\displaystyle \frac{\mathrm{μ}\mathrm{Ó\neg }}{T}}\right)}^2+4{\left({\displaystyle \frac{\mathrm{Ó\neg }\mathrm{γ}}{2T}}\right)}^2}}{2} \\ {k}^2=\left(\frac{\mathrm{μ}{\mathrm{Ó\neg }}^2}{2T}\right)\left[1Â\pm \sqrt{1+{\left(\frac{\mathrm{γ}}{\mathrm{μ}\mathrm{Ó\neg }}\right)}^2}\right] \\ k=\frac{\mathrm{γ}}{\sqrt{2T\mathrm{μ}}}{\left(1+\sqrt{1+{\left(\frac{\mathrm{γ}}{\mathrm{μ}\mathrm{Ó\neg }}\right)}^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Substitute $\frac{\mathrm{γ}}{\sqrt{2T\mathrm{μ}}}{\left(1+\sqrt{1+{\left(\frac{\mathrm{γ}}{\mathrm{μ}\mathrm{Ó\neg }}\right)}^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$for k in equation (5).

role="math" localid="1657697517366" $$\begin{gathered} z_0=\frac{1}{\frac{\mathrm{γ}}{\sqrt{2T\mathrm{μ}}}{\left(1+\sqrt{1+{\left(\frac{\mathrm{γ}}{\mathrm{μ}\mathrm{Ó\neg }}\right)}^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ {z}_0=\frac{\sqrt{2T\mathrm{μ}}}{y}\sqrt{1+\sqrt{1+{\left(\frac{\mathrm{γ}}{\mathrm{μ}\mathrm{Ó\neg }}\right)}^2}} \end{gathered}$$

Therefore, the penetration distance is role="math" localid="1657697421671" $z_0=\frac{\sqrt{2T\mathrm{μ}}}{y}\sqrt{1+\sqrt{1+{\left(\frac{\mathrm{γ}}{\mathrm{μ}\mathrm{Ó\neg }}\right)}^2}}$.

(d)

As the incident wave amplitude is$A_1$ with the ${\mathrm{δ}}_1=0$, the reflected wave amplitude will be,

$$\begin{gathered} {\tilde{A}}_R=\left(\frac{k_1-k-ik}{k_1+k+ik}\right){\tilde{A}}_1 \\ \frac{{\tilde{A}}_R}{\widetilde{A_1}}=\left(\frac{k_1-k-ik}{k_1+k+ik}\right) \\ {\left(\frac{{\tilde{A}}_R}{\widetilde{A_1}}\right)}^2=\left(\frac{k_1-k-ik}{k_1+k+ik}\right)\left(\frac{k_1-k-ik}{k_1+k+ik}\right) \end{gathered}$$

On further solving,

role="math" localid="1657699259164" $$\begin{gathered} {\left|\frac{A_R}{A_I}\right|}^2=\frac{(k_1{)}^2-k^2-k^2-2ik{k}_1}{(k_1+k{)}^2+k^2} \\ {A}_R=\sqrt{\frac{(k_1-k{)}^2+k^2}{(k_1+k{)}^2+k^2}}{A}_l \end{gathered}$$

Calculate the phase of the reflected wave.

role="math" localid="1657699204843" $$\begin{gathered} tan{\mathrm{δ}}_R=\frac{lm\left({\displaystyle \frac{{\tilde{A}}_R}{A_I}}\right)}{lm\left(\frac{{\tilde{A}}_R}{A_I}\right)} \\ {\mathrm{δ}}_R=ta{n}^{-1}\left(\frac{-2k{k}_1}{{\displaystyle \frac{(k_1+k_2{)}^2+k^2}{{\displaystyle \frac{(k_1{)}^2-k^2-k^2}{(k_1+k_2{)}^2+k^2}}}}}\right) \\ {\mathrm{δ}}_R=ta{n}^{-1}\left(\frac{-2k{k}_1}{(k_1{)}^2-k^2-k^2}\right) \end{gathered}$$

Therefore, the reflected wave’s amplitude and phase is $A_R=\sqrt{\frac{(k_1-k{)}^2+k^2}{(k_1+k{)}^2+k^2}}{A}_l$and${\mathrm{δ}}_R=ta{n}^{-1}\left(\frac{-2k{k}_1}{(k_1{)}^2-k^2-k^2}\right)$ , respectively.