Suppose string 2 is embedded in a viscous medium (such as molasses), which imposes a drag force that is proportional to its (transverse) speed:

$\mathbf{∆}{\mathit{F}}_{\mathbf{d}\mathbf{r}\mathbf{a}\mathbf{g}}\mathbf{=}\mathbf{-}\mathit{Y}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{t}}\mathbf{∆}\mathit{z}$.

(a) Derive the modified wave equation describing the motion of the string.

(b) Solve this equation, assuming the string vibrates at the incident frequency. That is, look for solutions of the form $\stackrel{\mathbf{~}}{\mathbf{f}}\left(z,t\right)\mathbf{=}{\mathit{e}}^{\mathbf{i}\mathbf{\omega }\mathbf{t}}\stackrel{\mathbf{~}}{\mathbf{F}}\left(z\right)$.

(c) Show that the waves are attenuated (that is, their amplitude decreases with increasing z). Find the characteristic penetration distance, at which the amplitude is of its original value, in terms of $\mathit{{\rm Y}}\mathbf{,}\mathit{T}\mathbf{,}\mathit{\mu }$and $\mathit{\omega }$.

(d) If a wave of amplitude A , phase $\mathit{\delta }$,= 0 and frequency$\mathit{\omega }$ is incident from the left (string 1), find the reflected wave’s amplitude and phase.

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

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

(a)

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

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

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

$F=\mu ∆z\frac{{\partial }^{2}f}{\partial z^2}$

Substitute $F=\mu ∆z\frac{{\partial }^{2}f}{\partial z^2}$in equation (2).\

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

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

$T\frac{{\partial }^{2}f}{\partial z^2}=\mu \frac{{\partial }^{2}f}{\partial z^2}+y\frac{\partial f}{\partial t}$

(b)

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

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

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

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

Again differentiate the above equation with respect to z.

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

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

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

Again differentiate the above equation with respect to t.

$$\begin{gathered} \frac{{\partial }^{2}f}{\partial t^2}=\frac{\partial }{\partial z}\left(\tilde{F}\left(z\right)\left(-i\omega \right)e^{-i\omega t}\right) \\ \frac{{\partial }^{2}f}{\partial t^2}=\left(-{\omega }^2\right)\tilde{F}\left(z\right)e^{-i\omega t} \end{gathered}$$

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

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

Let${\tilde{k}}^2=\frac{\omega }{T}\left(\mu \omega +iy\right)$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).

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

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

(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:

$$\begin{gathered} {\tilde{k}}^2=\frac{\omega }{T}\left(\mu \omega +i\gamma \right) \\ {k}^2-k^2+2ikk=\frac{\omega }{T}\left(\mu \omega +i\gamma \right)......\left(6\right) \end{gathered}$$

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

$$\begin{gathered} 2kk=\frac{\omega }{T}\gamma \\ k=\frac{\omega }{2kT} \end{gathered}$$

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

$k^2-k^2=\frac{\mu {\omega }^2}{T}$

Substitute $k=\frac{\omega \gamma }{2kT}$in the above expression.

$$\begin{gathered} k^2-{\left(\frac{\omega \gamma }{2kT}\right)}^2=\frac{\mu {\omega }^2}{T} \\ {k}^2-\frac{1}{k^2}{\left(\frac{\omega \gamma }{2T}\right)}^2=\left(\frac{\mu {\omega }^2}{T}\right) \\ {k}^4-{\left(\frac{\omega \gamma }{2T}\right)}^2=\left(\frac{\mu {\omega }^2}{T}\right)k^2 \\ \left(k^2\right)-\left(\frac{\mu {\omega }^2}{T}\right)k^2-{\left(\frac{\omega \gamma }{2T}\right)}^2=0 \end{gathered}$$

Solve the above equation by using a quadratic formula.

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

Substitute $\frac{\gamma }{\sqrt{2T\mu }}{\left(1+\sqrt{1+{\left(\frac{\gamma }{\mu \omega }\right)}^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$for k in equation (5).

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

Therefore, the penetration distance is role="math" localid="1657697421671" $z_0=\frac{\sqrt{2T\mu }}{y}\sqrt{1+\sqrt{1+{\left(\frac{\gamma }{\mu \omega }\right)}^2}}$.

(d)

As the incident wave amplitude is$A_1$ with the ${\delta }_1=0$, 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" $$\begin{gathered} tan{\delta }_R=\frac{lm\left({\displaystyle \frac{{\tilde{A}}_R}{A_I}}\right)}{lm\left(\frac{{\tilde{A}}_R}{A_I}\right)} \\ {\delta }_R=ta{n}^{-1}\left(\frac{-2k{k}_1}{{\displaystyle \frac{(k_1+k_2{)}^2+k^2}{{\displaystyle \frac{(k_1{)}^2-k^2-k^2}{(k_1+k_2{)}^2+k^2}}}}}\right) \\ {\delta }_R=ta{n}^{-1}\left(\frac{-2k{k}_1}{(k_1{)}^2-k^2-k^2}\right) \end{gathered}$$

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${\delta }_R=ta{n}^{-1}\left(\frac{-2k{k}_1}{(k_1{)}^2-k^2-k^2}\right)$ , respectively.