# A problem from Damped free Vibrations

Problem:

I can understand the problem physically that the block is pulled by some amount X so that its displacement at t=0 is X. It is then released from that position so that its velocity at t=0 is zero. So by the physical nature of the problem I can deduce that the displacement vs time graph will look like:

However I'm having trouble understanding the problem from mathematical standpoint. I've learnt that the equation of motion for an underdamped system is given as, $$x(t)$$

where $$X_o$$ and $$\phi_o$$ can be determined from initial conditions. So my understanding tells me that if I apply the boundary conditions as given in the problem I must obtain $$X_o = X$$ and $$\phi_o = \frac{\pi}{2}$$.

So I proceeded as follows:

Using Boundary Conditions I'm getting

$$tan\phi_o=\frac{\sqrt{1-\zeta^2}}{\zeta}$$

which wont reduce further to $$\phi=\frac{\pi}{2}$$

where am I going wrong?

P.S. - I'm just having trouble with mathematically coming to the conclusion that $$X_o = X$$ and $$\phi_o = \frac{\pi}{2}$$. I will be able to work out for what the problem is actually asking - amplitude after n cycles, by myself.

UPDATE: after your comment I sat down and derived it myself (I couldn't make it out from the image), and I realised that your assumption that $$\phi_0= \frac{\pi}{2}$$ is not valid.

I prefer the following notation

$$x(t) = A e^{-\zeta \omega t } \sin\left(\sqrt{1-\zeta^2}\omega_n t +\phi_0\right)$$

and also substituting $$\omega_d = \sqrt{1-\zeta^2}\omega_n$$ (to keep equations shorter), this becomes: $$x(t) = A e^{-\zeta \omega_n t } \sin\left(\omega_d t +\phi_0\right)$$

## displacement BC

So for time $$t=0$$, $$x(t=0)= X_0$$.

therefore:

$$x(t=0) = A e^{0} \sin\left(\sqrt{1-\zeta^2}\cdot 0 +\phi_0\right)$$ $$x(t=0) = A \sin\left(\phi_0\right)$$ $$X_0 = A \sin\left(\phi_0\right)$$

## velocity BC

By differentiating:

$$\dot{x}(t) = A \left((-\zeta \omega_n) e^{-\zeta \omega_n t } \sin\left(\omega_d t +\phi_0\right) + \omega_d e^{-\zeta \omega_n t } \cos\left(\omega_d t +\phi_0\right) \right)$$

Collecting term $$e^{-\zeta \omega_n t }$$: $$\dot{x}(t) = A e^{-\zeta \omega_n t } \left((-\zeta \omega_n) \sin\left(\omega_d t +\phi_0\right) + \omega_d \cos\left(\omega_d t +\phi_0\right) \right)$$

substituting and simplifying:

$$\dot{x}(t=0) = A \cdot 1 \cdot \left((-\zeta \omega_n) \sin\left(\phi_0\right) + \omega_d \cos\left( \phi_0\right) \right)$$

$$0 = A \cdot 1 \cdot \left((-\zeta \omega_n) \sin\left(\phi_0\right) + \omega_d \cos\left( \phi_0\right) \right)$$

$$\zeta \omega_n \sin\left(\phi_0\right) = \omega_d \cos\left( \phi_0\right)$$

$$\frac{\sin\left(\phi_0\right)}{\cos\left( \phi_0\right)} = \frac{\omega_d}{\zeta \omega_n }$$

therefore $$\tan\phi_0= \frac{\sqrt{1-\zeta^2}}{\zeta }$$

## Interpretation

The two equations are $$\begin{cases} X_0 = A \sin\left(\phi_0\right)\\ \tan\phi_0= \frac{\sqrt{1-\zeta^2}}{\zeta } \end{cases}$$

This is the actual correct solution. The assumption that $$\phi_0=\frac{\pi}{2}$$ is valid only for the case of the undamped free response. Whenever there is a damping ratio there is an angle $$\phi_0$$.

This can be observed in the following diagrams (I;ll get around to make them) for a $$\zeta$$ close to zero and close to 1.

The inflection point (change of curvature changes for different values of zeta) thus indicating the the phase angle changes.

In the case of

• $$\zeta \rightarrow 0$$ the initial change in displacement is almost horizontal (i.e. the velocity)
• $$\zeta \rightarrow 1$$ the initial change in displacement is very steep
• "that means that the initial position X, is the maximum displacement .." Nmech, that makes sense, however, I want to arrive at the result from pure mathematical considerations. Using the second boundary condition of velocity, we get a contradictory result. Dec 26, 2021 at 7:32
• Could it be that the value of $\zeta$ is very small so that the $tan(\phi_0)$ result in my question (at the end), becomes large enough on the right hand side of eqn, so that i can assume $\phi_0=\frac{\pi}{2}$ Dec 26, 2021 at 7:35
• Thank you Nmech, understood now. So that means my displacement vs time graph is not right. It should be like this -drive.google.com/file/d/1G7TEQ_bcxoIPZpW2cDZc200iPc2l1fJ0/…, correct? Dec 26, 2021 at 11:34

I think your math is taking you to the right place. Looking at the this sketch from the problem statement, it does not look like $$\dot{x}(0) = 0$$ when $$\phi_0=\pi/2$$; for example see here. Perhaps that may be the source of confusion?

So here's my crack at it. I changed the notation slightly. $$\omega_0$$ and $$\omega_1$$ would be expressed in terms of $$\omega_n$$ and $$\zeta$$ . Also used $$A$$ for the scale factor, since I don't think $$X_0$$ is the same as $$x(0)$$.

Quick check with wolfram alpha, using A=1, $$\omega_0$$=1, $$\omega_1$$=3: value and slope: