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Consider a bead sliding on smooth straight wire. Wire is rotating in vertical plane with constant angular velocity. Gravitational force is vertically downward as usual.

Equation of motion of the bead is:

$$ m\ddot r = m\omega^2r - mg\sin(\omega t), $$

where symbols have their usual meanings.

Rewritten equation becomes:

$$ \frac{\Bbb d^2r}{\Bbb dt^2} = \omega^2r - g\sin(\omega t). $$

How to solve this differential equation (analytically)?

Assume that at time $t=0$, position of the bead is $r_0$ and its velocity is zero.

Rócherz
  • 3,976
atom
  • 173

2 Answers2

4

$r'' - \omega^2 r = g\sin\omega t$

The general solution solves

$r'' - \omega^2 r = 0$

$(\frac {d}{dt} + \omega)(\frac {d}{dt} - \omega)r = 0$

The solutions to the diff-eq

$r' - \omega r = 0$ and $r' + \omega r = 0$ form the general solution.

$r_g = Ae^{\omega t} + Be^{-\omega t} = A\cosh \omega t + B\sinh \omega t$

Next, we find the particular solution.

Taking a guess

$r_p = C\sin \omega t$ will do it...

$r_p'' - \omega^2r = (-2C)\omega^2 \sin\omega t = -g\sin\omega t$

$-2a\omega^2 = g$

$r = A\cosh \omega t + B\sinh \omega t + \frac {g}{2\omega^2} \sin\omega t$

Solve $A,B$ to suit your initial conditions.

$A = r_0$
$B\omega + \frac {g}{2\omega} = r'_0$

user317176
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The final solution is:

$$r(t) = r_0 \cosh (\omega t)-\dfrac{g}{2\omega^2}\sinh(\omega t) + \dfrac{g}{2\omega^2} \sin(\omega t)$$

You can get that by summing up the homogeneous ODE solution (first two summed terms in my solution), which come from:

$$\ddot{r}-\omega^2 r=0$$

Whose general solution is:

$$r(t) = A \cosh(\omega t) + B\sinh(\omega t)$$

Plus the non homogeneous particular solution (the last summed term in my solution). The latter can come from any non homogeneous solution method, I recommend you to look for "Undetermined coefficients for 2nd order non homogeneous differential equations".

omivela17
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