Question

3. What is the phase difference between driving force and velocity of the

forced oscillator?​

Answer 1

Explanation:

The velocity, of course, will be the derivative of the X(t) solution to the second-order differential equation derived either from Newton’s 2nd Law, F=m*d2x/dt^2, directly; or from canonical equations involving the Langrangian (L=T + V — where T is kinetic E and V is potential E). Newton suffices, easily here, though.

Your question is easier if the equation follows as m*d2x/dt^2 -b* dx/dt + c*x= F(app) — where b is zero (in the simple case) as a damping constant — and F(app) is the applied force. Note that c is often times related either to g (for a pendulum) or k (for a Hooke’s Law spring).

With b=0 we are in for resonance that has amplitude going to infinity (destruction of the system). This happens since, after some algebra, the solution winds up with w - that is Omega, as angular frequency - 2*Pi*f - in a subtraction with the w(app) or the frequency of the forced oscillation (like Sqrt[ (w^2-w(app)^2)]) in the denominator. When they match the infinities occur.

Adding b into a velocity term represents friction or a non-ideal spring, etc. This only quells the infinities, but leaves the max amplitude in the same spot. Taking the derivative of either solution, with or without b, leaves this term in the denominator, but does not affect phase.

Phase between drive and natural frequency of the oscillator can be determined by setting the derivative of the solution (including the w^2-w(app)^2 term) equal to zero and finding the closest order local maxima of the two sinusoids. This will involve a Tan^-2(…algebra terms… basically everything except F(app)/(w*m)).