I'm having trouble trying to figure out what conditions should I apply to solve the wave equation for a driven string.
The string should follow the wave equation:
$$\dfrac{\partial^2 u(x,t)}{\partial x^2}=\dfrac{1}{v^2}\dfrac{\partial^2 u(x,t)}{\partial t^2}$$
where the velocity $v$ depends on the tension and linear density of the string.
This string has length L and is fixed on both ends. One of the ends is driven, forcing the string to oscillate. This end should barely move for some given frequencies of the driven force where we have standing waves on the string. There is nice experiment showing the harmonics of a driven string.
What I want to do is to solve the wave equation for a string in order to find a solution that gives me a function for the behavior of the string for different frequencies of the driven force, including the standing wave behavior.
I know how to solve the wave equation for common boundary conditions, such as the fixed ends conditions for a plucked string.
For this problem what initial conditions and what boundary conditions should I use to model the system? My first attempts were to apply a BC for the first time derivative on the moving end with the driven frequency, but the final solution wasn't correct. Am I missing information, should the equation be non-homogeneous, or maybe there is no simple solution at all?
Best Answer
The wave equation $$\frac{\partial^2 u}{\partial x^2} - \frac{1}{v^2}\frac{\partial^2 u}{\partial t^2} = 0$$ is to be solved for the transverse displacement $u(x,t)$ for the fixed-driven boundary conditions of a string \begin{align*} u(0,t)= 0\quad \text{and}\quad u(L,t) = u_0e^{j\omega_0 t}, \end{align*} where $u_0$ is the drive amplitude.
Employing separation of variables leads to the general solution \begin{align}\label{general}\tag{1} u(x,t) = (Ae^{-jkx} + Be^{jkx})(Ce^{j\omega t} + De^{-j\omega t})\,, \end{align} where $k = \omega/v$. Application of the fixed boundary condition $u(0,t)= 0$ necessitates $B= -A$, so equation (\ref{general}) becomes \begin{align}\label{general2}\tag{2} u(x,t) = (e^{-jkx} - e^{jkx})(C_1e^{j\omega t} + C_2e^{-j\omega t}) \end{align} where $A$ has now been factored into the constants $C$ and $D$ in equation (\ref{general}), giving two new constants $C_1$ and $C_2$ in equation (\ref{general2}). Meanwhile, application of the driven boundary condition $u(L,t) = u_0e^{j\omega_0 t}$ to equation (\ref{general2}) gives \begin{align} u(L,t) = u_0e^{j\omega_0t}&= (e^{-jkL} - e^{jkL})(C_1e^{j\omega t} + C_2e^{-j\omega t})\notag\\ &=\frac{j}{2}\sin(kL)(C_1e^{j\omega t} + C_2e^{-j\omega t})\tag{3}\label{particulars} \end{align} To satisfy equation (\ref{particulars}), the phase on both sides of the equation must match, implying that $C_2 = 0$ and $\omega = \omega_0$ (and thus $k = k_0 =\omega_0/v$): \begin{align*} u_0e^{j\omega_0t} =\frac{j}{2}\sin(k_0L)C_1e^{j\omega_0 t} \quad \implies \quad C_1 =\frac{2u_0}{j\sin(k_0L)}\,, \end{align*} so the general solution, equation (\ref{general2}), becomes $u(x,t) = \frac{2u_0}{j\sin(k_0L)}(e^{-jk_0x}-e^{jk_0x})e^{j\omega_0 t}$, or \begin{equation}\bbox[5px,border:2px solid purple] {u(x,t) = u_0 \frac{\sin(k_0x)}{\sin(k_0L)}e^{j\omega_0 t} }\label{solutioners}\tag{4} \end{equation} Note that equation (\ref{solutioners}) matches the form of the solution to the 1D acoustic wave equation for a source at one end of a tube that is closed at the other end.
To visualize this solution to the wave equation, equation (\ref{solutioners}) is rewritten in terms of three dimensionless parameters, the reciprocal of the number of wavelengths in the string $\lambda_0/L$, distance normalized by the length of the string $x/L$, and dimensionless time $\omega_0t$: \begin{align}\label{rewrite it}\tag{5} \frac{u(x,t)}{u_0} = \frac{\sin[(x/L)(2\pi L/\lambda_0)]}{\sin(2\pi L/\lambda_0)}e^{j\omega_0 t}\,. \end{align} Below are animations of equation (\ref{rewrite it}) for various values of $\lambda_0/L$. Note that the string has an infinite response for $\lambda_0/L = 1/n$, where $n = 1,2,\dots$, because at these frequencies correspond to the eigenfrequencies of the fixed-fixed string.
Finally, note that the d'Alembert solution solves the initial-value problem and is not of use to the boundary-value problem at hand.