The equation of motion for the string is given by the wave equation:

$\rho \frac{\partial^2 u}{\partial t^2}=T \frac{\partial^2 u}{\partial x^2}-\alpha u$

where $u(x,t)$ is the displacement of the string from its equilibrium position at point $x$ and time $t$. The first term on the right-hand side represents the tension in the string, and the second term represents the elastic foundation. We can assume a solution of the form:

$u(x, t)=U(x) e^{i \omega t}$

where $\omega = \Omega + i\beta$ is the complex frequency and $U(x)$ is the complex amplitude of the displacement.

Substituting this solution into the wave equation and simplifying, we get:

$\frac{d^2 U}{d x^2}+k^2 U=0$

where $k^2 = \frac{\omega^2 \rho}{T} – i\frac{\alpha}{T}$. This is a second-order homogeneous differential equation with constant coefficients, which has solutions of the form:

$U(x)=A e^{i k x}+B e^{-i k x}$

where $A$ and $B$ are complex constants to be determined by the boundary conditions.

At $x=0$, the displacement of the string must be equal to the displacement of the mass, which is given by:

$u(0, t)=U(0) e^{i \omega t}=\frac{F}{M} e^{i \omega t}$

Therefore, $U(0) = \frac{F}{M}$.

At $x=\pm \infty$, the string must be taut and have no displacement. Therefore, we can assume that $U(\pm \infty) = 0$.

Using these boundary conditions, we can solve for $A$ and $B$:

$A=-\frac{F}{2 M} e^{-i k x} \quad$ and $\quad B=\frac{F}{2 M} e^{i k x}$

Therefore, the displacement of the string is given by:

$u(x, t)=\frac{F}{2 M}\left(e^{i(k x-\omega t)}-e^{i(-k x-\omega t)}\right)$

Simplifying, we get:

$u(x, t)=\frac{F}{M} \sin (k x) \sin (\omega t)$

where $k = \sqrt{\frac{\omega^2 \rho}{T} – i\frac{\alpha}{T}}$. This is the steady-state displacement response of the string for $-\infty < x < \infty$.

To determine the wavenumber(s) of the excited steady-state radiating wave(s), we start by assuming a solution of the form:

$\eta(x, t)=A e^{i(k x-\omega t)}$

where $A$ is the amplitude of the wave, $k$ is the wavenumber, and $\omega$ is the frequency. Substituting this solution into the given equation yields:

$-i \omega A+i c_0 k A-i \beta k^3 A=0$

Dividing through by $-iA$ and solving for $k$, we get:

$k^2=\frac{\omega}{c_0} \pm \sqrt{\left(\frac{\omega}{c_0}\right)^2-\frac{\beta}{c_0}}$

Now suppose that an external localized pressure disturbance traveling with constant speed $V$ acts on the free surface. This means that the phase velocity of the wave excited by the disturbance will be $V$, i.e. $\omega/k=V$. Substituting this into the above expression for $k^2$ yields:

$k^2=\frac{V^2}{c_0^2} \pm \sqrt{\left(\frac{V^2}{c_0^2}\right)^2-\frac{\beta}{c_0^3}}$

Depending on the sign of $\beta$, we have two cases to consider:

Case 1: $\beta>0$

In this case, there are two possible values of $k$ given by:

$k_1=\frac{V}{c_0}+\sqrt{\left(\frac{V}{c_0}\right)^2-\frac{\beta}{c_0^3}}, \quad k_2=\frac{V}{c_0}-\sqrt{\left(\frac{V}{c_0}\right)^2-\frac{\beta}{c_0^3}}$

The two corresponding frequencies are:

$\omega_1=c k_1=c_0\left(\frac{V}{c_0}+\sqrt{\left(\frac{V}{c_0}\right)^2-\frac{\beta}{c_0^3}}\right), \quad \omega_2=c_0 k_2=c_0\left(\frac{V}{c_0}-\sqrt{\left(\frac{V}{c_0}\right)^2-\frac{\beta}{c_0^3}}\right)$

where $c=\sqrt{c_0^2+\beta k^2}$ is the phase velocity of the wave.

Note that $k_1$ and $k_2$ have opposite signs, which means that the two waves have opposite directions of propagation. Furthermore, the two waves have different wavelengths and frequencies. The wave with wavenumber $k_1$ has a longer wavelength and a lower frequency than the wave with wavenumber $k_2$.

To sketch the position of these waves relative to the forcing, we note that the wave with wavenumber $k_1$ travels in the same direction as the external disturbance, while the wave with wavenumber $k_2$ travels in the opposite direction. The amplitude of the waves decreases with distance from the source, so the wave with wavenumber $k_1$ will be closer to the source than the wave with wavenumber $k_2$.

Case 2: $\beta<0$

In this case, there is only one possible value of $k$ given

To sketch qualitatively the time history of the response for $t>0$ at a fixed station $x=L>0$, we first need to find the solution to the given equation. We can use the method of characteristics to solve this equation.

Let $dx/dt=c_0$ and $d^2x/dt^2=0$, which gives $x=c_0t+x_0$ and $x_0=L$ for $t=0$. Then, we have

$\frac{d \eta}{d t}=-\beta \frac{\partial^3 \eta}{\partial x^3}$

We assume a solution of the form $\eta(x,t)=f(x-ct)$, where $c$ is the wave speed. Substituting this into the above equation and integrating three times with respect to $x$, we get

$\eta(x, t)=A e^{-k(x-c t)} \cos (k(x-c t)+\phi)$

where $k=\sqrt[3]{\beta/c_0}$, $A$ and $\phi$ are constants determined by the initial conditions.

Now, we can sketch the time history of the response for $t>0$ at a fixed station $x=L>0$. At $x=L$, the free surface elevation is given by

$\eta(L, t)=A e^{-k(L-c t)} \cos (k(L-c t)+\phi)$

As $t$ increases, the amplitude of the cosine function decays exponentially with a decay rate of $k(c t-L)$. Therefore, the free surface elevation decreases in amplitude and oscillates with a decreasing frequency as time progresses. The decay rate is faster for larger values of $k$ (i.e., for larger values of $\beta$ or smaller values of $c_0$). If $\beta<0$, then $k$ is imaginary, which means the wave amplitude decays exponentially without oscillation.

To sketch qualitatively a snapshot of the disturbance for $-\infty<x<\infty$ at time $t=T$, long after the initial excitation, we can use the same solution as above, but with $t=T$. Thus, we have

$\eta(x, T)=A e^{-k(x-c T)} \cos (k(x-c T)+\phi)$、

The cosine function represents a wave with wavelength $\lambda=2\pi/k$ and wave number $k$. The wave is propagating to the right if $c>c_0$, or to the left if $c<c_0$. The amplitude of the wave decreases exponentially as $x$ increases or decreases from the initial position of the disturbance. The decay rate is faster for larger values of $k$ (i.e., for larger values of $\beta$ or smaller values of $c_0$). If $\beta<0$, then $k$ is imaginary, which means the wave amplitude decays exponentially without oscillation. The phase velocity of the wave is given by $c/k$, which decreases as $k$ increases. If $\beta>0$, then the group velocity of the wave is given by $c-(2/3)k^2c_0$, which is smaller than the phase velocity. If $\beta<0$, then the group velocity is undefined because $k$ is imaginary.