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In my 'Classical Dynamics of Particles and Systems, THORNTON/MARION, 5th Edition' book of classical mechanics it is given the following general solution for a damped oscillation solving $\ddot{x}+2\beta\dot{x}+w_o^2x=0$:

$$x(t)=e^{-\beta t}[A_{1}\exp(\sqrt{\beta^2-w_o^2}t)+A_{2}\exp(-\sqrt{\beta^2-w_o^2}t)]$$

as $A_1$ and $A_2$ some arbitrary constants, $\beta$ the damping parameter and $w_0$ the the characteristic angular frequency.

So, if I want to find the $x(t)$ expression for the critical damped case, I have to consider that $$w_0^2=\beta^2$$

so my critical damped motion is described by:

$$x(t)=(A_1+A_2)(e^{-\beta t})$$

But, it we solve the specific differential equation for the critical damped case, the result is (as equal roots exists):

$$x(t)=(A+B t)e^{-\beta t}$$

So, my question is: What am I missing?

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The first solution is not correct since it implies a strange connection $\dot x (0)=-\beta x(0)$. Check the general solution for its region of validity.

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  • $\begingroup$ You are absolutely right. This book is not making a case for equal roots, so is not that general, thank you very much. $\endgroup$ – Marc C Nov 8 '14 at 12:49
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I think the solution: x(t)=(A1+A2)(e^−βt) is not a correct solution because there is no cause to write constant in form of (A1+A2) instead of A. I also solve the equation for Critical damping and get x(t)=(A+Bt)e^−βt as solution.

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