Two-mass pendulum

Relevant wiki: Pendulums

If $\theta$ is the angle between the upper string and the vertical, and $\phi$ is the angle between the lower string and the vertical, then the kinetic energy of the system is $T \; = \; \tfrac12m\ell^2\big[13\dot\theta^2 + 12\dot\phi^2 + 24\cos(\theta-\phi)\dot\theta\dot\phi \big]$ and the gravitational potential energy of the system is $V \; = \; -13mg\ell \cos\theta - 12mg\ell \cos\phi$ We are interested in small oscillations, and so we look for small order approximations. Thus $T \; \approx. \; \tfrac12m\ell^2 \dot{\mathbf{x}}^T \left(\begin{array}{cc} 13 & 12 \\ 12 & 12 \end{array}\right) \dot{\mathbf{x}} \hspace{2cm} V \; \approx \; \tfrac12mg\ell \mathbf{x}^T \left(\begin{array}{cc} 13 & 0 \\ 0 & 12 \end{array}\right) \mathbf{x} + c$ where $\mathbf{x} = \binom{\theta}{\phi}$ and $c$ is a constant. Thus Lagrange's equations of motion read $m\ell^2 \left(\begin{array}{cc} 13 & 12 \\ 12 & 12 \end{array}\right)\ddot{\mathbf{x}} \; = \; -mg\ell\left(\begin{array}{cc} 13 & 0 \\ 0 & 12 \end{array}\right) \mathbf{x}$ Looking for normal modes of oscillation, we look for solutions of the form $\mathbf{x} = e^{i\lambda t}\mathbf{u}$ for a constant $\lambda$ and a constant vector $\mathbf{u}$ , Thus $\begin{aligned} -m\ell^2\lambda^2 \left(\begin{array}{cc} 13 & 12 \\ 12 & 12 \end{array}\right)\ddot{\mathbf{u}}& = \; -mg\ell\left(\begin{array}{cc} 13 & 0 \\ 0 & 12 \end{array}\right) \mathbf{u} \\ \left[ \left(\begin{array}{cc} 13 & 0 \\ 0 & 12 \end{array} \right) - \tfrac{\ell \lambda^2}{g}\left(\begin{array}{cc}13 & 12 \\ 12 & 12 \end{array}\right)\right] \mathbf{e} & = \; \mathbf{0} \\ \left[ \left(\begin{array}{cc} 13 & -12 \\ -13 & 13 \end{array}\right) - \tfrac{\ell \lambda^2}{g} I\right]\mathbf{e} & = \; \mathbf{0} \end{aligned}$ Solving this eigensystem, we see that possible values of $\lambda$ are given by $\tfrac{\ell \lambda^2}{g} \; = \; 13 \pm 2\sqrt{39}$ with corresponding vectors $\mathbf{e} \; =\; \left(\begin{array}{c} \mp2\sqrt{\tfrac{3}{13}} \\ 1 \end{array}\right)$ The first of these has $\mathbf{e} = \binom{-0.961}{1} \approx \binom{-1}{1}$ , which corresponds to the motion where the second bob is roughly stationary. Thus we deduce that $\frac{T'}{T} \; = \; \frac{1}{\sqrt{13+2\sqrt{39}}} \; = \; 0.198068 \; \approx \; \boxed{0.20}$

Relevant wiki: Pendulums

We will use the same approximations about small amplitude oscillations that are used to solve the simple pendulum problem.

When the masses are at rest the tension in the lower rope is $Mg$ and the tension in the upper rope is $(M+m)g$ . For small amplitude oscillations the tensions will be the same.

Consider the situation when the mass $m$ has moved in the horizontal direction by distance $x$ . At this point in time the upper rope makes an angle $\phi$ to the vertical, where $\sin \phi =x/l$ . (Of course, in reality, the mass moves on an arc of length $\phi l$ , but for small displacements the vertical component of the movement is negligible.) The horizontal component of the force acting on the mass $m$ will be $F_1=-(x/l) (M+m)g$ , where the negative sign indicates that the force is opposite to the displacement. For the lower rope we can assume that the large mass is not moving sideways and therefore the lower rope will also make an angle $\phi$ to the vertical. Consequently, the horizontal component of the force due to the lower rope on $m$ will be $F_2=-(x/l) Mg$ . Here we neglected the vertical movement of the large mass, which is justified by the small displacement approximation.

The equation of motion is $ma=F_1+F_2= - (x/l) (m+2M)g$ , where $a$ is the acceleration. This results in harmonic oscillations with a period of $T=2 \pi \sqrt{\frac{m}{m+2M} \frac{l}{g}} = \sqrt{\frac {m}{m+2M}} T_0 = \frac{1}{5} T_0$ . We get $a=0.20$ .

Relevant wiki: Pendulums

For a simple pendulum, expressions for the energy yield (with $\sin \dot\theta \approx \dot\theta,\cos\theta \approx 1 - \tfrac12\theta^2$ ): $K \approx \tfrac12 m\ell^2{\dot\theta}^2;\ \ \ \ \ U \approx -\tfrac12 m g \ell\theta^2.$ Generally, if $K \approx \tfrac12M{\dot x}^2$ and $U \approx \tfrac12Cx^2$ , then the period of oscillation is $T = 2\pi\sqrt{M/C}$ . For the simple pendulum we get $M = m\ell^2$ and $C = mg\ell$ , making $T = 2\pi\sqrt{\ell/g}$ .

In this case, we have $K = K_1 + K_2 = \tfrac12 m \ell^2{\dot\theta_1}^2 + \tfrac12 (12m) \ell^2 (\dot\theta_1 + \dot\theta_2)^2, \\U = U_1 + U_2 = -\tfrac12 m g \ell \theta_1^2 - \tfrac12 (12m)\ell(\theta_1^2 + \theta_2^2).$ Since "the large mass hardly moves", we must have $\theta_1 + \theta_2 = 0$ , or $\theta_1 = -\theta_2 = \theta$ , and the equations reduce to $K = \tfrac12 m \ell^2{\dot\theta}^2\ \ \ \ U = -\tfrac12 (25 m) g \ell\theta^2.$ Thus $M = m \ell^2$ and $C = 25 m g \ell$ , making $T' = 2\pi\sqrt{\ell/25 g}$ .

Obviously, $T'/T = \sqrt{1/25} = 1/5 = \boxed{0.20}$ .

I have a simple, nearly no calculation solution: The original has period (root reciprocal of angular acc.) of T. This is with Length, $l$ ; gravity, $g$ ; and mass, $m$ ; and tension $mg$ . With the large weight added the lengths are the same, gravity the same, and the mass is the same, the only things changes is the tension in the strings, namely 12mg and (12+1)mg = 25 mg. So the with the tension x25 the angular acc is x25. So the period is $\boxed{0.2}$ T .

isn't the period supposed to be sqrt(l/g) instead of sqrt(g/l) ?