Solving a pendulum ODE with initial velocity v0

applestrudle · Sep 16, 2015

..when it is released from rest with velocity (v₀, 0)

Typeset equations from a physics text: complex-number form d²r/dt² + 2iΩ dr/dt + ω₀²r = 0 (1.6.4), with Ω = ω cos θ and ω₀² = g/l, and general solution r = Ae^(-i(Ω-ω₁)t) + Be^(-i(Ω+ω₁)t) (1.6.5).

I can get 1.6.5 but I can't get this:

Typeset equation block: where omega1 squared equals omega0 squared plus Omega squared; if released from origin with velocity (v0, 0), A = -B = v0/(2i omega1), so that r = (v0/omega1) e^(-i Omega t) sin(omega1 t), equation (1.6.6).

Geofleur · Sep 16, 2015

First, note that ## \tilde{r}(0) = A + B = 0 ##, because the pendulum starts off at the origin. This equation gives ## A = -B ##. Next, calculate the time derivative of ## \tilde{r} ##, evaluate it at time zero, and set it equal to ## v_0 + i(0) = v_0##:

## i[(\omega_1-\Omega)A-B(\Omega+\omega_1)] = v_0 ##.

But the only way this equation could be true is if

## (\omega_1 - \Omega)A - B(\Omega+\omega_1)=-i v_0 ##.

Do you see it? The rest follows by using ## A = -B ##.

Solving a pendulum ODE with initial velocity v0

Discussion

Similar threads