Boundary conditions of a forced oscillator (string)

[cromata]

-If we have string of length L that has fixed ends, then we can easily find frequencies with which this string can oscillate:
We just need to solve wave equation: ∂²y/∂x²=1/c²*∂²∂t² (c is determined by strings properties (linear density and tension), with Dirichlet boundary conditions (y(0,t)=0, y(L,t)=0) Of course to determine how the string is oscillating we also need to know initial shape/speed of string (but that only tells us amplitude of each harmonic)

-But what happens when some force is acting on the string? Let's say that some force F(t) is acting at some distance x_o from one end of the string? How do we find solution to this problem?
Can it be treated like some sort of boundary condition or should that force be added to wave equation or something else?

 

[Dr.D]

The enduring (lasting) solution is the particular solution; the free vibration will die away sooner or later due to unavoidable damping. The particular solution will oscillate at the frequency of the excitation.If this is a stick-slip situation, such as a violin string, then it is going to be a bit messy, depending on the frequency at which slipping is starting.

 

[sophiecentaur]

-But what happens when some force is acting on the string?

The problem has instantly got a lot harder. You would need to specify what causes this force. If you are hanging a mass on the string then the natural frequency of oscillation would change. If you use a spring, the force will vary with displacement so the frequency would change. If you have a rocket engine, applying a constant force then I cannot see how the frequency would change.

 

[cromata]

The enduring (lasting) solution is the particular solution; the free vibration will die away sooner or later due to unavoidable damping. The particular solution will oscillate at the frequency of the excitation

I know that this is the case when there is forced discrete oscillating system (like masses connected with springs), and it can easily be shown for discrete systems that enduring solution is particular solution. But I wasn't sure that same thing happens with continuous system.

 

[Dr.D]

The same processes are at work in the continuuous system as were in the discrete system. Air drag and internal hysteresis still serve to induce damping, so the homogeneous solution will die away, leaving only the particular solution.