Undamped pendulum with pushing force getting phase angle

Question

I managed to get the solution to the ODE of an undamped pendulum with a pushing force, ODE with pushing force is $$\frac{d^2x}{dt^2} + \omega_0x = sin\omega_1t $$ where $\omega_0 =$ natural angular frequency $\omega_1 = $ driving\input frequency solution I have is $$x(t) = Acos(\omega_0t+\phi)-\frac{sin\omega_1t}{(\omega_1^2-\omega_0^2)}$$ I made the assumption that the acceleration and velocity at t = 0 is zero, so when I get the velocity i.e $\frac{dx}{dt} = -Asin(\phi)\omega_0-\frac{1}{(\omega_1^2-\omega_0^2)} =0$ which means $\phi$ is $\arcsin(\frac{1}{A\omega_0(\omega_1^2-\omega_0^2)})$ only thing is that at t = 0 the pendulum is stationary at its equilibrium position so $\phi$ should really be zero but my calculations are showing otherwise where did I go wrong?

Answer 1

Your derivation of ${dx\over dt}$ is wrong as it must be$${dx\over dt}|_{t=0}=-A\sin \phi-{\omega_1\over \omega_1^2-\omega_0^2}=0$$Also the second condition gives you$$A\cos \phi=0$$from which you can obtain $\phi={\pi\over 2}$. Probably $\phi=0$ is the horizontal line and if so, actually $\phi={\pi\over 2}$ becomes the equilibrium position by intuition.