# Effect of initial shift on undamped forced oscillation

I have general equation for undamped forced oscillations (no friction) which is:

$$x(t)\ =\ \frac{v_0}{\omega_n}\cdot \text {sin}\left( \omega_n \cdot t \right)+\left(x_0\ -\ \frac {f_0}{\omega_n^2-\omega^2}\right)\cdot \text{cos}\left( \omega_n \cdot t \right)+\frac {f_0}{\omega_n^2-\omega^2}\cdot\text{cos}\left(\omega \cdot t\right)$$

I just wonder about,what type of motion should occur when initial conditions are both $$0$$ (i.e $$v_0=0\ and \ x_0=0$$). My intuitive expectation is that as there is no 'natural' oscillations at beginning,vibration has to be depending only on last term.

But in formulation we can see that it will not depend only on applied force and its frequency since it will still have second term as: $$-\left( \frac {f_0}{\omega_n^2-\omega^2}\right)\cdot \text{cos}\left( \omega_n \cdot t \right)$$ which will be summed with oscillation coming from last term yielding complicating oscillation with two different frequencies.

Why is that happening? Is that force during acting produces also that 'natural oscillation' which after joins together with last term?

and interestingly, to produce clear vibration caused just by last term without any other 'natural' vibrations (first two terms) we need to start movement with 0 velocity and SHIFTED position $$\left [x_0=\left( \frac {f_0}{\omega_n^2-\omega^2}\right)\cdot \text{cos}\left( \omega_n \cdot t \right)\right]$$. its so counter intuitive.