Question

(a) Formulate an appropriate boundary condition, to replace Eq. 9.27, for the case of two strings under tension T joined by a knot of mass m.

(b) Find the amplitude and phase of the reflected and transmitted waves for the case where the knot has a mass m and the second string is massless.

Short answer

(a) The boundary conditions are ${\overline{)\frac{\partial f}{\partial z}}}_{0^{+}}-{\overline{)\frac{\partial f}{\partial z}}}_{0^{-}}={\overline{)\frac{m}{T}\frac{{\partial }^{2}f}{\partial t^2}}}_{0}and{\overline{)\frac{\partial f}{\partial z}}}_0={\overline{)\frac{\partial f}{\partial z}}}_{0^{+}}$

(b) The amplitude and phase of the reflected wave is $A_R=A_I$and ${\delta }_R={\delta }_I+{\tan }^{-1}\left(\frac{2\beta }{1-{\beta }^2}\right)$respectively, and the amplitude and phase of the transmitted wave is $A_T=\frac{2}{\sqrt{1+{\beta }^2}}{A}_{1}and{\delta }_T={\delta }_l+{\tan }^{-1}\beta$respectively.

Step-by-step solution

Expression for the derivative of f.

Consider the knot is of negligible mass, write the expression for the derivative of f.

${\overline{)\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{z}}}}_{{\mathbf{0}}^{\mathbf{-}}}\mathbf{}\mathbf{=}{\overline{)\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{z}}}}_{{\mathbf{0}}^{\mathbf{+}}}$

As the two strings under tension T are joined by a knot of mass m, write an appropriate equation for the unbalanced forces.

$$\begin{gathered} T\sin {\theta }_{+}-T\sin {\theta }_{-}=m{\overline{)\frac{{\partial }^{2}f}{\partial t^2}}}_0 \\ T(\sin {\theta }_{+}-T\sin {\theta }_{-})=m{\overline{)\frac{{\partial }^{2}f}{\partial t^2}}}_0 \end{gathered}$$ …… (1)

Determine the boundary conditions:

(a)

Since, it is known that:

$\sin {\theta }_{+}{\overline{)=\frac{\partial \mathrm{f}}{\partial \mathrm{z}}}}_{0^{+}}$

$\sin {\theta }_{+}{\overline{)=\frac{\partial \mathrm{f}}{\partial \mathrm{z}}}}_{0^{-}}$

Substitute the values of$\sin {\theta }_{+}and\sin {\theta }_{-}$ in equation (1).

$$\begin{gathered} T{\overline{)\frac{\partial f}{\partial z}}}_{0^{+}}-{\overline{)\frac{\partial f}{\partial z}}}_{0^{-}}={\overline{)m\frac{{\partial }^{2}f}{\partial t^2}}}_0 \\ {\overline{)\frac{\partial f}{\partial z}}}_{0^{+}}-{\overline{)\frac{\partial f}{\partial z}}}_{0^{-}}={\overline{)m\frac{{\partial }^{2}f}{\partial t^2}}}_0 \end{gathered}$$

Hence, the boundary condition will be,

$$\begin{gathered} {\overline{)\frac{\partial f}{\partial z}}}_{0^{+}}-{\overline{)\frac{\partial f}{\partial z}}}_{0^{-}}={\overline{)m\frac{{\partial }^{2}f}{\partial t^2}}}_0 \\ {\overline{)\frac{\partial f}{\partial z}}}_{0^{+}}-{\overline{)\frac{\partial f}{\partial z}}}_{0^{-}} \end{gathered}$$

Therefore, the boundary conditions are ${\overline{)\frac{\partial f}{\partial z}}}_{0^{+}}-{\overline{)\frac{\partial f}{\partial z}}}_{0^{-}}={\overline{)m\frac{{\partial }^{2}f}{\partial t^2}}}_{0}and{\overline{)\frac{\partial f}{\partial z}}}_{0^{+}}-{\overline{)\frac{\partial f}{\partial z}}}_{0^{-}}$

Determine the amplitude and phase of the reflected and transmitted wave:

(b)

Write the disturbance on the string for a sinusoidal incident wave.

$f^{-}\left(z,t\right)=\left\{\begin{array}{l}{A}_1{e}^{i\left(k_{l}z-\varpi t\right)}+A_R{e}^{i\left(k_{l}z-\varpi t\right)}z<0\\ A_1{e}^{i\left(k_{l}z-\varpi t\right)\mathbf{}\mathbf{z}\mathbf{>}\mathbf{}\mathbf{0}}\end{array}\right.$

Write the expression for the outgoing amplitudes andA in terms ofA incoming one .

$$\begin{gathered} A_l+A_R=A_T \\ {k}_l\left(A_l-A_R\right)=k_2{A}_T \end{gathered}$$ …… (2)

From the second boundary condition,

$$\begin{gathered} T\left[i{k}_2{A}_T-i{k}_1\left(A_l-A_R\right)\right]=m{\varpi }^2{A}_T \\ i{k}_2{A}_T-i{k}_1\left(A_l-A_R\right)=\frac{m{\varpi }^2{A}_T}{T} \\ {k}_1\left(A_l-A_R\right)=k_2{A}_T-\frac{m{\varpi }^2{A}_T}{T} \\ {k}_1\left(A_l-A_R\right)=\left(k_2-\frac{im{\varpi }^2}{T}\right)A_T \end{gathered}$$ …… (3)

Multiply equation (2) with and add the obtained equation to equation (3).

$$\begin{gathered} k_1{A}_l+k_1{A}_R=k_1{A}_T \\ {k}_1{A}_l+k_1{A}_R+k_1\left(A_l-A_R\right)=\left(k_2-\frac{im{\varpi }^2}{T}\right)A_T+k_1{A}_T \\ 2k_1{A}_l+k_2{A}_T-\frac{im{\varpi }^2}{T}{A}_T+k_1{A}_T \\ {A}_T=\left(\frac{2k_1}{k_1+k_2-\frac{im{\varpi }^2}{T}}\right)A_l \end{gathered}$$

…… (4)

Multiply equation (2) with and add the obtained equation to equation (3).

$$\begin{gathered} A_R=A_T-A_l \\ {A}_R=\left(\frac{2k_1}{k_1+k_2-{\displaystyle \frac{im{\varpi }^2}{T}}}\right)A_l-A_l \\ {A}_R=\left(\frac{k_1+k_2-\frac{im{\varpi }^2}{T}}{k_1+k_2-{\displaystyle \frac{im{\varpi }^2}{T}}}\right)A_l \end{gathered}$$ …… (4)

Divide the above equation by .

$A_R=\left(\frac{1-{\displaystyle \frac{k_2}{k_1}}-\frac{im{\varpi }^2}{T}}{1+\frac{k_2}{k_1}-{\displaystyle \frac{im{\varpi }^2}{T}}}\right)A_l$Since, and .

Therefore, the amplitude and phase of the reflected wave is and respectively, and the amplitude and phase of the transmitted wave is and respectively.