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The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force on at $t=0$ and onwards, say, to push your particle in a positive direction, then, depending on the particle phase, the force will accelerate or decelerate the particle.

Generally you write down the external force in the same way: $$F_{ext}(t)=F_0\cdot \cos(\omega t +\Phi_0).$$ This expression stays in the driven oscillator equation, namely, in the right-had side. The resonance happens always, i.e., the external force will in the and supply energy to the particle, but this supplying can start immediately if the force direction and the particle velocity direction coincide. Otherwise the external force first slows down the particle and only then starts pumping its amplitude.

The particle velocity phase is shifted with respect to the particle coordinate $$v(t) = -A \sin(\omega t + \phi_0)=A\cos(\omega t + \phi_0 +\pi/2),\; t<0.\;$$ So when $\Phi_0 =\phi_0 +\pi/2\;$ the force is in phase with velocity (not with coordinate) - you have not only the same direction for the velocity and for the force, but also coincidence of instants when both the velocity and the force become zero (no time intervals with their opposite signs).

EDIT: The permanent phase shift of $\pi/2$ in a resonant case with friction (as described in user17581 answer) is a self-established thing and its meaning is simple - the external force in the end compensates exactly the friction force; the latter being proportional to velocity which is shifted by $\pi/2$ with respect to the coordinate time-dependence (so the oscillator oscillates as if it were free, without losses).

The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force on at $t=0$ and onwards, say, to push your particle in a positive direction, then, depending on the particle phase, the force will accelerate or decelerate the particle.

Generally you write down the external force in the same way: $$F_\text{ext}(t)=F_0\cdot \cos(\omega t +\Phi_0).$$ This expression stays in the driven oscillator equation, namely, in the right-hand side. The resonance happens always, i.e., the external force will supply energy to the particle, but this supplying can start immediately if the force direction and the particle velocity direction coincide. Otherwise the external force first slows down the particle and only then starts pumping its amplitude.

The particle velocity phase is shifted with respect to the particle coordinate $$v(t) = -A \sin(\omega t + \phi_0)=A\cos(\omega t + \phi_0 +\pi/2),\; t<0.\;$$ So, when $\Phi_0 =\phi_0 +\pi/2\;$ the force is in phase with velocity (not with coordinate) - you have not only the same direction for the velocity and for the force, but also coincidence of instants when both the velocity and the force become zero (no time intervals with their opposite signs).

EDIT: The permanent phase shift of $\pi/2$ in a resonant case with friction (as described in user17581 answer) is a self-established thing and its meaning is simple - the external force in the end compensates exactly the friction force; the latter being proportional to velocity which is shifted by $\pi/2$ with respect to the coordinate time-dependence (so the oscillator oscillates as if it were free, without losses).

The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force on at $t=0$ and onwards, say, to push your particle in a positive direction, then, depending on the particle phase, the force will accelerate or decelerate the particle.

Generally you write down the external force in the same way: $$F_{ext}(t)=F_0\cdot \cos(\omega t +\Phi_0).$$ This expression stays in the driven oscillator equation, namely, in the right-had side. The resonance happens always, i.e., the external force will in the and supply energy to the particle, but this supplying can start immediately if the force direction and the particle velocity direction coincide. Otherwise the external force first slows down the particle and only then starts pumping its amplitude.

The particle velocity phase is shifted with respect to the particle coordinate $$v(t) = -A \sin(\omega t + \phi_0)=A\cos(\omega t + \phi_0 +\pi/2),\; t<0.\;$$ So when $\Phi_0 =\phi_0 +\pi/2\;$ the force is in phase with velocity (not with coordinate) - you have not only the same direction for the velocity and for the force, but also coincidence of instants when both the velocity and the force become zero (no time intervals with their opposite signs).

The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force on at $t=0$ and onwards, say, to push your particle in a positive direction, then, depending on the particle phase, the force will accelerate or decelerate the particle.

Generally you write down the external force in the same way: $$F_{ext}(t)=F_0\cdot \cos(\omega t +\Phi_0).$$ This expression stays in the driven oscillator equation, namely, in the right-had side. The resonance happens always, i.e., the external force will in the and supply energy to the particle, but this supplying can start immediately if the force direction and the particle velocity direction coincide. Otherwise the external force first slows down the particle and only then starts pumping its amplitude.

The particle velocity phase is shifted with respect to the particle coordinate $$v(t) = -A \sin(\omega t + \phi_0)=A\cos(\omega t + \phi_0 +\pi/2),\; t<0.\;$$ So when $\Phi_0 =\phi_0 +\pi/2\;$ the force is in phase with velocity (not with coordinate) - you have not only the same direction for the velocity and for the force, but also coincidence of instants when both the velocity and the force become zero (no time intervals with their opposite signs).

EDIT: The permanent phase shift of $\pi/2$ in a resonant case with friction (as described in user17581 answer) is a self-established thing and its meaning is simple - the external force in the end compensates exactly the friction force; the latter being proportional to velocity which is shifted by $\pi/2$ with respect to the coordinate time-dependence (so the oscillator oscillates as if it were free, without losses).

The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force on at $t=0$ and onwards, say, to push your particle in a positive direction, then, depending on the particle phase, the force will accelerate or decelerate the particle.

Generally you write down the external force in the same way: $$F_{ext}(t)=F_0\cdot \cos(\omega t +\Phi_0).$$ This expression stays in the driven oscillator equation, namely, in the right-had side. The resonance happens always, i.e., the external force will in the and supply energy to the particle, but this supplying can start immediately if the force direction and the particle velocity direction coincide. Otherwise the external force first slows down the particle and only then starts pumping its amplitude.

The particle velocity phase is shifted with respect to the particle coordinate $$v(t) = -A \sin(\omega t + \phi_0)=A\cos(\omega t + \phi_0 +\pi/2),\; t<0.\;$$ So when $\Phi_0 =\phi_0 +\pi/2\;$ you have not only the same direction for the velocity and for the force, but also coincidence of instants when both the velocity and the force become zero (no time intervals with their opposite signs).

The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force on at $t=0$ and onwards, say, to push your particle in a positive direction, then, depending on the particle phase, the force will accelerate or decelerate the particle.

Generally you write down the external force in the same way: $$F_{ext}(t)=F_0\cdot \cos(\omega t +\Phi_0).$$ This expression stays in the driven oscillator equation, namely, in the right-had side. The resonance happens always, i.e., the external force will in the and supply energy to the particle, but this supplying can start immediately if the force direction and the particle velocity direction coincide. Otherwise the external force first slows down the particle and only then starts pumping its amplitude.

The particle velocity phase is shifted with respect to the particle coordinate $$v(t) = -A \sin(\omega t + \phi_0)=A\cos(\omega t + \phi_0 +\pi/2),\; t<0.\;$$ So when $\Phi_0 =\phi_0 +\pi/2\;$ the force is in phase with velocity (not with coordinate) - you have not only the same direction for the velocity and for the force, but also coincidence of instants when both the velocity and the force become zero (no time intervals with their opposite signs).