Oscillation 27-09-2020

If the position of the mass $m$ is described by polar coordinates $r,\theta$ , then (since the only force acting on that mass is radial), angular momentum is conserved, and so $r^2\dot{\theta}$ is constant. In this case, this tells us that $r^2\dot{\theta} = R^2\omega$ . If the tension in the string is $T$ , then we deduce that $m(\ddot{r} - r\dot{\theta}^2) \; = \; -T \hspace{2cm} M\ddot{r} \; = \; T - Mg$ and hence $(M + m)\ddot{r} - mR^4\omega^2 r^{-3} + Mg \; = \; (M + m)\ddot{r} - mr\dot{\theta}^2 + Mg \; = \; 0$ Since the solution $r \equiv R$ is possible, we deduce that $Mg = mR\omega^2$ . We want to consider small perturbations about the solution $r \equiv R$ . If we write $r = R + \rho$ , then we obtain $\begin{aligned} (M + m)\ddot{\rho} - mR\omega^2\big(1 + \tfrac{\rho}{R}\big)^{-3} + Mg & = \; 0 \\ (M + m)\ddot{\rho} + 3m\omega^2\rho & \approx \; 0 \end{aligned}$ and so $\rho$ performs small oscillations with angular frequency $\sqrt{3}\omega \sqrt{\frac{m}{M+m}}$ making $\alpha = \boxed{\sqrt{3}}$ .

Place an X-Y coordinate system such that the origin is the hole and the X-Y plane lies on the tabletop. At a general instant, let the length of the string on the tabletop be $r$ and the angle it makes with the X-axisThe be $theta$ . Let the total length of the string be $L$ . Then:

$x = r \cos{\theta}$ $y = r \sin{\theta}$

The Z coordinate of the mass $M$ is

$z = -(L-r)$

The kinetic energy of this system is:

$T = \frac{m}{2}(\dot{x}^2 + \dot{y}^2) + \frac{M}{2}\dot{z}^2$

Simplifying:

$T = \left(\frac{M+m}{2}\right)\dot{r}^2 +\frac{mr^2\dot{\theta}^2}{2}$

The potential energy of the system is:

$V = -Mg(L-r)$

Applying Lagrange's equations (Newton's laws are possible but tedious) gives the equations:

$r^2 \dot{\theta} = c$

Where $c$ is constant. The constant can be found by applying initial conditions:

$\implies r^2 \dot{\theta} =R^2 \omega$

And the other equation is:

$(M+m)\ddot{r} = mr\dot{\theta}^2 -Mg$

Replacing $\dot{\theta}$ gives:

$(M+m)\ddot{r} = \frac{mR^4\omega^2}{r^3}-Mg$

Now, it is said that the mass $M$ is displaced by a small distance and released. We are asked to analyse resulting oscillations. To analyse small oscillations, let us take:

$r = R - s$

Where $s <<R$

Replacing in the equation of motion:

$-(M+m)\ddot{s} = \frac{mR^4\omega^2}{(R-s)^3}-Mg$ $-(M+m)\ddot{s} =mR\omega^2\left(1 - \frac{s}{R}\right)^{-3} -Mg$

Doing a binomial approximation gives:

$-(M+m)\ddot{s} =mR\omega^2\left(1 + \frac{3s}{R}\right)-Mg$ $(M+m)\ddot{s} + 3m\omega^2s + mR\omega^2 -Mg=0$

At equilibrium, The weight of the Mass must balance the tension in the string. Working that out leads to:

$mR\omega^2 =Mg$

Therefore, the approximated equation of motion is:

$(M+m)\ddot{s} + 3m\omega^2s=0$

This implies that the frequency of small oscillations are:

$\omega_s^2 = \frac{3 m\omega^2}{M+m}$

Let the hanging mass be pulled downwards through a small distance $x$ , so small that all the values $x^2,x^3..$ can be safely neglected. Let the initial radius of the circular path be $r$ , angular velocity of the mass $m$ be $\omega$ , and the final angular velocity be $\omega_1$ . Then, angular momentum conservation equation yields

$m\omega r^2=m\omega_1(r-x)^2$

Force acting on the system is

$F=Mg-m\omega_1^2(r-x)$

Initially the system is in equilibrium, so,

$m\omega^2r=Mg$

So $F=Mg-\dfrac {m\omega^2r^4(r-x)}{(r-x) ^4}$

$\approx -3m\omega^2x$

So, $(m+M)\dfrac {d^2x}{dt^2}\approx -3m\omega^2x$

Therefore the angular frequency of the resulting oscillation is

$\omega \sqrt {\dfrac {3m}{m+M}}$

Hence $α=\sqrt 3\approx \boxed {1.732}$ .