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For resonance to occur, is it true that the force lags behind the motion by $\pi/2$? I saw some notes written that the motion lags behind the force by $\pi/2$ which makes no sense to me. As I watched many videos and I worked out the motion, it always happens before the force pushes. E.g. if the force is $F=\cos\omega t$ and $x = A\cos(\omega t -\pi/2)$, is it true that force lags behind the motion?

If it's $\pi/2$ , does there happen anything special?

Also, I realised that if there is no damping, we can only get 180 or 0 phase difference, which is quite counterintuitive to me. Can anyone give any example to help me feel better?

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marked as duplicate by Emilio Pisanty, Waffle's Crazy Peanut, Chris White, Alan Rominger, Brandon Enright May 31 '13 at 1:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Possible duplicate: – Qmechanic Apr 13 '13 at 17:44
up vote 1 down vote accepted

Imagine that the oscillator is a swing and you are the force pushing it. The phase shift is nothing more than the statement that you have to act differently than the swing. Obviously, you shouldn't push in the exact opposite direction (which rules out a phase shift of $\pi$).

enter image description here Imagine the red line being the amplitude of the swing, and the green line is your push strength.

What the optimal phase shift of $\pi/2$ (which is equivalent to switching $\sin$ with $\cos$) tells you is that you change your pushing direction every time the swing is at its maximum amplitude. So, instead of pushing the strongest when the swing amplitude is the biggest, you push the strongest when the amplitude is 0 and don't push at all when the amplitude is at its maximum.

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