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Suppose I have a string of mass density $\rho$ clumped between two points so that the tension is $T$.

The resonant standing waves of this vibrating string are those in which the restoring force on the elements of the string are proportional to their displacements $\psi$ from equilibrium. In short, this string obeys the wave equation $$\frac{\partial^2 \psi}{\partial t^2}+v^2 \frac{\partial^2 \psi}{\partial x^2}=0$$

Now, consider the inhomogeneous equation of this set up: $$\frac{\partial^2 \psi}{\partial t^2}+v^2 \frac{\partial^2 \psi}{\partial x^2}=f(x)$$

I am familiar with the methods of solution to the above equation using Green's methods and Sturm-Liouville theory of linear operators, so that I know how to solve this problem.

My question is about the meaning of the inhomogeneous term to the set up. Is there any meaning to the graph of $f(x)$ and to the form of solution of the string at later times?

What if I had a time-dependent function $g(t)$ as the inhomogeneous term, instead of a spatial-dependence function? Does $g(t)$ has a meaning only at $t=0$?

Any given insight would be helpful.

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  • $\begingroup$ I think this is an interesting and important question, and I'm sorry no one responded properly and I personally can be of no help. Maybe it will be better received on the Physics StackExchange. $\endgroup$ Sep 5, 2016 at 4:27

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Compare the form of these equations:$$\begin{align} \frac{\partial^2 x}{\partial t^2} + \frac{b}{m} \frac{\partial x}{\partial t} & = \frac{1}{m} F(t) \\ \frac{\partial^2 \psi}{\partial t^2} + \nu^2 \frac{\partial^2 \psi}{\partial x^2} & = f(x, t). \end{align}$$ For the first equation, it is pretty clear that the inhomogeneity is an external forcing function. In the exact form you have given, the external force is constant in time, if non-uniform in $x$. Although, in usual physics problems the sign of the spatial derivative term will differ from the time derivative term as: $$ \frac{\partial^2 \psi}{\partial t^2} - \nu^2 \frac{\partial^2 \psi}{\partial x^2} = f(x, t), $$ to produce a wave equation. Switching that sign is the same as looking for a stationary state in a 2-dimensional problem, but that doesn't affect the interpretation of $f$ as an external force.

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