enter image description here

enter image description here

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:

enter image description here

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, $x(t)$

enter image description here

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:

enter image description here

Using Boundary Conditions I'm getting


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.


2 Answers 2


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 $t=0$, $x(t=0)= X_0$.


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

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


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.

enter image description here

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
  • $\begingroup$ "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. $\endgroup$ Commented Dec 26, 2021 at 7:32
  • $\begingroup$ 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 $\phi_0=\frac{\pi}{2}$ $\endgroup$ Commented Dec 26, 2021 at 7:35
  • $\begingroup$ 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? $\endgroup$ Commented 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:


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