## Natural Vibration of Beam - PDE

I am just wondering the author is doing in this calculation step.

Given ##\displaystyle \rho A \frac {\partial^2 w}{\partial x^2} - \rho I \frac{\partial^4 w}{\partial t^2 \partial x^2} +\frac {\partial^2 }{\partial x^2}EI \frac {\partial^2 w}{\partial x^2}=q(x,t)##

where ##w(x,t)=W(x)e^{-i \omega t}##

##\omega## is the frequency of natural transverse motion and ##W(x)## is the mode shape of the transverse motion.

He substitutes the above into the PDE to get the following

##\displaystyle \frac {d^2 }{d x^2}EI \frac {d^2 W}{d x^2} - \lambda (\rho A W -\rho I \frac {d^2 W }{d x^2} ) =0## where ##\lambda=\omega^2##

However, I calculate the second derivative ##w''(x)=e^{-i\omega t} W''(x)## and ##w''(t)=- \lambda e^{-i\omega t} W(x)##

What is incorrect on my part? Ie, where did the exponentials go?

thanks
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 Blog Entries: 9 Recognitions: Homework Help Science Advisor If the non-homogeneity function q goes away, so that the PDE hat 0 in the RHS, you can show that the exponential gets factored in the LHS, so that the resulting ODE will no longer contain it.

 Quote by dextercioby If the non-homogeneity function q goes away, so that the PDE hat 0 in the RHS, you can show that the exponential gets factored in the LHS, so that the resulting ODE will no longer contain it.

1) How does the non homogeneity function just go away?

2) How is it shown that the exponential is absorbed in the LHS of PDE?

Thanks

## Natural Vibration of Beam - PDE

 Quote by bugatti79 1) How does the non homogeneity function just go away? 2) How is it shown that the exponential is absorbed in the LHS of PDE? Thanks
On second thoughts is it something along these lines...

If ##q(x,t)=0##then we can write the ODE in which each term on the LHS will have a ##e^{-i \omega t}## factor. Pull this out from each of the terms and thus we get

##e^{-i \omega t}[ODE]=0## but ##e^{-i \omega t}\ne 0## therefore

##[ODE]=0##...?

Recognitions:
Science Advisor
 Quote by bugatti79 1) How does the non homogeneity function just go away?
If "natural transverse motion" means the same as "free vibration", then q(x,t) can't depend on t by definition, otherwise you would have forced vibration not free vibration.

So this is the same idea as a vertical mass-on-a-spring, where the spring is stretched by the weight of the mass. You can find the equilibrium position as the statics problem
$$\frac{d^2}{dz^2}EI\frac{d^2}{dz^2}w_0(x) = q(x)$$
Then you measure w(x) from the equilibrum position. That makes the right hand side = 0.

You answered your own question 2).

 Quote by AlephZero If "natural transverse motion" means the same as "free vibration", then q(x,t) can't depend on t by definition, otherwise you would have forced vibration not free vibration. So this is the same idea as a vertical mass-on-a-spring, where the spring is stretched by the weight of the mass. You can find the equilibrium position as the statics problem $$\frac{d^2}{dz^2}EI\frac{d^2}{dz^2}w_0(x) = q(x)$$ Then you measure w(x) from the equilibrum position. That makes the right hand side = 0. You answered your own question 2).
Thanks, that makes good sense. Now I can proceed :-)
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