For now just consider the magnitude of the torque on the pendulum. However, adding a restriction to the size of the oscillation's amplitude gives a form whose solution can be easily obtained. }\kern0pt is used in the Legendre polynomial solution above. { {{\left( {\frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}} \right)}^2}{k^6} + \ldots } \right. {K\left( k \right) }={ \frac{\pi }{2}\left\{ {1 + {{\left( {\frac{1}{2}} \right)}^2}{k^2} }\right.}+{\left. }\) and \({\left( {2n} \right)!! Again just consider the magnitude of the angular momentum. A pendulum is an object consisting of a mass suspended from a pivot so that it can swing freely. {\left. Pendulum Equation. Dynamics of rotational motion is described by the differential equation\[\varepsilon = \frac{{{d^2}\alpha }}{{d{t^2}}} = \frac{M}{I},\]where \(\varepsilon\) is the angular acceleration, \(M\) is the moment of the force that causes the rotation, \(I\) is the moment of inertia about the axis of rotation.In our case, the torque is determined by the projection of the force of gravity on the tangential direction, that isThe minus sign indicates that at a positive angle of rotation \(\alpha\) (counterclockwise), the torque of the forces causes rotation in the opposite direction.In the case of small oscillations, one can set \(\sin \alpha \approx \alpha.\) As a result, we have a linear differential equationwhere \(\omega = \sqrt {\large\frac{g}{L}\normalsize} \) is the angular frequency of oscillation.The period of small oscillations is described by the well-known formula\[T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{L}{g}} .\]However, with increasing amplitude, the linear equation ceases to be valid.

}\]\[{d\left( {\frac{\alpha }{2}} \right) }={ \frac{{k\cos \theta d\theta }}{{\sqrt {1 – {k^2}\,{{\sin }^2}\theta } }}. Consider Figure 1 on the right, which shows the forces acting on a simple pendulum. If it is assumed that the angle is much less than 1 Therefore, a relatively reasonable approximation for the length and period are, From the last equation and the equation , we have: (9) and (10) The equations and can be compactly written as a single vector equation: (11) The vector equation is a state-space form of the equation of motion .

}\]We denote \(\sin {\large\frac{{{\alpha _0}}}{2}\normalsize} = k\) and introduce the new variable \(\theta\) instead of the angle \(\alpha:\)\[{\sin \frac{\alpha }{2} = \sin \frac{{{\alpha _0}}}{2}\sin \theta }={ k\sin \theta . In this case, the correct description of the oscillating system implies solving the original Suppose that the pendulum is described by the nonlinear second order differential equation\[\frac{{{d^2}\alpha }}{{d{t^2}}} + \frac{g}{L}\sin \alpha = 0.\]We consider the oscillations under the following initial conditions\[{\alpha \left( {t = 0} \right) = {\alpha _0},\;\;\;}\kern-0.3pt{\frac{{d\alpha }}{{dt}}\left( {t = 0} \right) = 0. \]where the double factorials \({\left( {2n – 1} \right)!! When the bob is moved from its rest position and let go, it swings back and forth. This differential equation is like that for the simple harmonic oscillator and has the solution: Index Periodic motion concepts . }\]In the new notation, our equation can be written asNext, we discuss the limits of integration. }\) denote the product, respectively, of odd and even natural numbers.Note that if we restrict ourselves to the zero term of the expansion, assuming that \(K\left( k \right) \approx {\large\frac{\pi }{2}\normalsize},\) we obtain the known formula for the period of small oscillations:\[{{T_0} = 4\sqrt {\frac{L}{g}} K\left( k \right) }\approx {4\sqrt {\frac{L}{g}} \frac{\pi }{2} }={ 2\pi \sqrt {\frac{L}{g}} . so that a pendulum with just the right energy to go vertical will never actually get there. which is the same result as obtained through force analysis. If the initial angle is taken into consideration (for large amplitudes), then the expression for which leads to the following set of coupled differential equations: Its graph is shown below in The function \(K\left( k \right)\) can also be represented as a power series:\[ The equation shown above is the pendulum with no damping (e.g, no resistence by air and any other frictions). We'll assume you're ok with this, but you can opt-out if you wish. The passage of the arc from the lowest point \(\alpha = 0\) to the maximum deviation \(\alpha = {\alpha_0}\) corresponds to a quarter of the oscillation period \(\large\frac{T}{4}\normalsize.\) It follows from the relationship between the angles \(\alpha\) and \(\theta\) that \(\sin \theta = 1\) or \(\theta = {\large\frac{\pi}{2}\normalsize}\) at \(\alpha = {\alpha_0}.\) Therefore, we obtain the following expression for the period of oscillation of the pendulum:The integral on the right cannot be expressed in terms of elementary functions. When a pendulum is displaced sideways from its resting, equilibrium position, it is subject to a restoring force due to gravity that will accelerate it back toward the equilibrium position.

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