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Consider a spring mass dashpot system subjected to a harmonic excitation force of

enter image description here

$$F(t)=F_0 sin(\omega t)$$

The response of a system assuming sufficient time has been passed and the complementary part of the solution has become zero will be

$$x(t)=Xsin(\omega t-\phi)$$

where $$tan\phi=\frac{2\zeta r}{1-r^2}$$ $$r=\frac{\omega}{\omega_n}$$

when $r=1, \phi=90^0$ which means at $t=0, x(0)=-X$

I'm not able to make sense of these results physically. The results say that block at t=0 will be at -X when the excitation force is 0. But I started (t=0) the movement of block at x=0, at which time the force was zero. Results are contradicting my actual conditions. Results say at t=0 x=-X but I started moving the block when x=0

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I think the confusion is due to the fact that you have not understood the significance of "The response of a system assuming sufficient time has been passed and the complementary part of the solution has become zero".

The response of a simple harmonic oscillator (like the one in the image), can be distinguished (one of the many different categorisations), to transient and steady state.

The transient state is the part which is associated with a $e^{-a\, t}$ term (where a is a positive number), and eventually "becomes zero after a long time has passed".

The steady state is the portion or the response that follows the excitation.

The following graph, shows a force excitation of 20N @ 24.9 rad/s with red, and the partial (steady state), and homogeneous (transient) response. As you can see the homogeneous dies out (although initially is of equal but opposite significance), and eventually dies out (after about 0.7 of a s). From that point on the total response coincides with the partial solution.

enter image description here

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  • $\begingroup$ So will I take t=0 when my complementary function part of the solution has died or in other words when steady state of vibration has been just reached. At this time (when steady state of vibration has just reached) x= -X ? $\endgroup$
    – Harshit Rajput
    Oct 30, 2021 at 10:12
  • $\begingroup$ I am not certain what "So will I take t=0" means exactly to you. The "absolute" (note the inverted commas) time t=0 is when the force starts to get applied. The integration constants are calculated so that for time t=0 the velocity and starting position of the total solution are in agreement with the boundary conditions (so I interpret those starting values for the partial and homogeneous solutions above as mathematical artifacts to satisfy the conditions). If you set an arbitrary time t=0 at a later time, then the partial solution might still have non zero values which can be confusing. $\endgroup$
    – NMech ♦
    Oct 30, 2021 at 13:33
  • $\begingroup$ I looked at the graphs you shared in the answer, and thought about what could be happening with the system, carefully, from the absolute t=0. So, I start applying the force at t=0, the block moves and is in the transient state till t=0.7. At t=0.7, the force again will be zero and the displacement of the block will be x=-0.005 (x=-X). If I take t=0.7 as a new reference t=0, then the results in the question make sense (correct me if wrong) - Force is 0 at this new t=0 and x=-X (-0.005). this is in accordance with the equations in the question $\endgroup$
    – Harshit Rajput
    Oct 30, 2021 at 17:17
  • $\begingroup$ You can do that -- although I would not recommend it in general-- but you need to sync it properly. The thing is that the homogeneous oscillates with the natural frequency while the partial with excitation frequency so it's not always achievable. $\endgroup$
    – NMech ♦
    Oct 30, 2021 at 18:53
  • $\begingroup$ Yes that makes sense. Anyways, I was able to understand the physical meaning of the results. Thank you for that. $\endgroup$
    – Harshit Rajput
    Oct 31, 2021 at 6:27

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