longitudinal vibration of continuous system (1D bar) 1

[Hoomanya]

Hi,
I have been looking at working out the natural frequency and mode shapes of a 1D bar. I have now understood the standard cases, for instance the fixed-free bar with BCs: u@(x=0) = 0; and du/dx@(x=L) = 0.
I need to tackle a case where the free end is replaced by du/dx@(x=L) = P(t), where P(t) is a pressure force and it comes in the form of a set of numbers.
So I have started by expressing u as:
u(x,t) = G(x)*F(t) = [B1*cos(w*x/c)+B2*sin(w*x/c)]*[A1*cos(w*t)+A2*sin(w*t)]

Applying u@(x=0) = 0 gives B1 = 0
Applying du/dx@(x=L) = P(t) gives u(x,t) = (w*L/c)*[B2*cos(w*L/c)]*[A1*cos(w*t)+A2*sin(w*t)] = P(t)

(Note: B1,B2,A1,A2 are constants, c = sqrt(E/rho))
This is the point I get stuck as the 2nd boundary condition does not help. I would appreciate any hints ASAP. Please let me know if this cannot be solved.
Thank you!

 

[amanuensis]

Congratulation for this work.

Sometimes you can use B1*exp(jkx) and B2*exp(-jkx) which is more physical instead of sin() and cos() but from mathematics it's completly identical.

The first case is a free response, the second case is a forced response.

The time is given by the external force P(t) = P1*exp(jwt). So I believe that you can forget the temporal part of the solution.

 

[Hoomanya]

Hi and thank you.
I will have a think about that and possible come back here with more questions on the same problem.

 

[amanuensis]

The vibration response is always the overlap of a free response, which is a transitional regime, and a forced response.
There are always the both.

Nevertheless, if you apply an impact (with a hammer for example), the vibration response is just composed of the free response (forced response only exists during the impact which lasts a very short time).

If a forced excitation is applied (with a shaker for example), you have first the both regimes (free and forced), but very quickly, the free regime decreases and becomes completely negligeable. Then the vibration response is only equal to the forced response.

This physical explanation should help you to writte the right équations.