A problem from Damped free Vibrations

Question

Problem:

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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?

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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.

Answer 1

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

Answer 2

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: