I've been working on a multi-step problem. Step one involved solving the initial value problem:
Consider the frictionless rod. The equation of motion becomes $$mr’’-mw^2r=-mg\sin(\omega t)$$ With $g=9.81m/s^2$ and a constant angular speed $\omega$. The rod is initially horizontal, and the initially horizontal, and the initial conditions for the bead are $r(0)=r_0$ and $r’(0)=v_0$.
I solved this problem. The answer was $$r(t)=\frac{2wv_0-9.81+2r_0}{4\omega^2}(e^{\omega t})+\left(r_0-\frac{2\omega v_0-9.81+2r_0}{4 \omega^2}\right)(e^{-\omega t})+\frac{9.81}{2\omega^2}\sin(\omega t)$$ answer to step A
Step B involves finding $v_0$.
Consider the initial position to be zero, i.e. $r_0=0$. Find the initial velocity, $v_0$, that results in a solution, r(t), which displays simple harmonic motion, i.e. a solution that does not tend toward infinity.
I've tried solving for $v_0$ by substituting every $r_0$ with $0$, and by taking the derivative of the solution and then substituting every $r_0$ with $0$. All I find is that $0=0$, or $v_0=v_0$.
Through experimenting with a graphing calculator, I've found that $v_0=\frac{g}{2\omega}$.
How can I find the answer through the differential equation?
If $\omega>0$ then the coefficient of $e^{\omega t}$ must be zero. $$\frac{2\omega v_0-g +2r_0}{4\omega ^2}=0$$ Plug in $r_0=0$ and mutiply the above equation by $4\omega^2$: $$2\omega v_0-g=0\\v_0=\frac{g}{2\omega}$$ Plugging in all back into your $r(t)$, the first two terms are zero, so $$r(t)=\frac{g}{2\omega^2}\sin(\omega t)$$