91影视

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

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

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

$$\begin{gathered} T\sin {胃}_{+}-T\sin {胃}_{-}=m{\overline{)\frac{{鈭�}^{2}f}{鈭�t^2}}}_0 \\ T(\sin {胃}_{+}-T\sin {胃}_{-})=m{\overline{)\frac{{鈭�}^{2}f}{鈭�t^2}}}_0 \end{gathered}$$ 鈥︹�� (1)

(a)

Since, it is known that:

$\sin {胃}_{+}{\overline{)=\frac{鈭�\mathrm{f}}{鈭�\mathrm{z}}}}_{0^{+}}$

$\sin {胃}_{+}{\overline{)=\frac{鈭�\mathrm{f}}{鈭�\mathrm{z}}}}_{0^{-}}$

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

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

Hence, the boundary condition will be,

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

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

(b)

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

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

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

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

From the second boundary condition,

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

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

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

鈥︹�� (4)

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

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

Divide the above equation by .

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