Pendulums and the Small-Oscillation Approximation
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You should be comfortable with Newton's second law, basic trigonometry, and the concept of a restoring force. Familiarity with the general solution for simple harmonic motion is assumed.
Why this matters
The pendulum is the historical bridge between discrete mechanics and continuous wave theory.
While the mass on a spring provides the simplest model of oscillation, the pendulum introduces the crucial concept of geometric non-linearity. By examining how the restoring force changes with displacement, we derive the small-oscillation approximation. This approximation is not merely a simplification; it is the foundational step that allows us to linearize complex physical systems, enabling the superposition principle and the eventual derivation of the wave equation in subsequent lectures.
The Geometry of the Pendulum
Consider a point mass suspended from a fixed pivot by a rigid, massless rod of length \(L\). When displaced from its vertical equilibrium position, gravity acts vertically downwards. However, the motion of the mass is constrained to a circular arc. The component of gravity that drives the motion is not the full weight, but the tangential component perpendicular to the rod. As the angle of displacement increases, the effectiveness of gravity in pulling the mass back towards the center decreases. This geometric relationship between the angle and the restoring force is the core physical insight. For very small angles, the arc of the circle becomes indistinguishable from a straight line, and the restoring force becomes directly proportional to the displacement. This linear regime is where the mathematics of simple harmonic motion applies perfectly.
Deriving the Equation of Motion
To formulate the dynamics, we apply Newton's second law in the tangential direction. Let \(\theta\) be the angular displacement from the vertical. The tangential force is the component of weight acting along the arc: \(F_t = -mg \sin\theta\). The negative sign indicates the force opposes the displacement. The tangential acceleration is related to angular acceleration by \(a_t = L\ddot{\theta}\). Thus, Newton's law gives:
Dividing by \(mL\), we obtain the exact equation of motion for a simple pendulum:
This is a non-linear differential equation because of the \(\sin\theta\) term. Unlike the linear spring equation, the frequency of oscillation here depends on the amplitude \(\theta_0\). To recover the simple harmonic motion described in D-057, we must approximate the sine function for small angles.
The Small-Angle Approximation
For small angles measured in radians, the Taylor series expansion of \(\sin\theta\) is \(\theta - \frac{\theta^3}{6} + \dots\). If \(\theta \ll 1\), the higher-order terms are negligible, and we approximate \(\sin\theta \approx \theta\). Substituting this into the equation of motion yields:
This is now identical in form to the standard simple harmonic oscillator equation \(m\ddot{x} = -kx\). By comparing coefficients, we identify the effective stiffness \(k_{\text{eff}}\) and mass \(m_{\text{eff}}\) for the angular coordinate. Here, the "stiffness" is provided by gravity and geometry, \(k_{\text{eff}} = mgL\), and the "mass" is the moment of inertia \(I = mL^2\). The angular frequency \(\omega\) is therefore:
Notice that the mass \(m\) cancels out. This is a profound result: in the small-oscillation limit, the period of a pendulum is independent of the mass of the bob. It depends only on the length of the pendulum and the local gravitational field strength. The period \(T\) is given by \(T = 2\pi\sqrt{L/g}\).
Energy Considerations
We can also analyze the pendulum using energy methods, consistent with D-058. The potential energy \(U\) is measured relative to the lowest point. If the bob rises by a height \(h\), \(U = mgh\). From geometry, \(h = L(1 - \cos\theta)\). For small \(\theta\), \(\cos\theta \approx 1 - \frac{\theta^2}{2}\), so \(h \approx \frac{L\theta^2}{2}\). Thus:
This is a quadratic potential, \(U \propto \theta^2\), which is the hallmark of a harmonic oscillator. The coefficient \(\frac{1}{2} mgL\) plays the role of \(\frac{1}{2}k\). The total energy is conserved, and the exchange between kinetic and potential energy occurs at twice the oscillation frequency, exactly as derived for the linear spring in D-058. The maximum potential energy at amplitude \(\theta_0\) is \(E = \frac{1}{2} mgL \theta_0^2\).
The derivations
Simple Harmonic Motion from a Linear Restoring Force
For a particle of mass \(m\) confined near a stable equilibrium of a smooth one-dimensional potential \(U(x)\), small displacements \(x\) from the minimum obey \(m\ddot{x} = -kx\) with \(k \equiv U''(x_0) > 0\), and the motion is sinusoidal, \(x(t) = A\cos(\omega t + \varphi)\), with a single angular frequency \(\omega = \sqrt{k/m}\) that is independent of amplitude.
Energy Exchange and Conservation in SHM
For a one-dimensional undamped simple harmonic oscillator of mass \(m\) and stiffness \(k\), executing \(x(t)=A\cos(\omega_0 t+\varphi)\) with \(\omega_0=\sqrt{k/m}\), the total mechanical energy \(E=T+U\) is constant in time and equal to \(\tfrac12 kA^2\), while the kinetic part \(T\) and potential part \(U\) each oscillate about the mean value \(\tfrac14 kA^2\) at angular frequency \(2\omega_0\) — twice the oscillation frequency — exactly out of phase with one another.
The small-angle approximation \(\sin\theta \approx \theta\) is valid only when \(\theta\) is measured in radians. If you use degrees, the approximation fails completely because the slope of \(\sin\theta\) at zero is 1 only in the radian measure. Furthermore, the error in the period grows rapidly with amplitude; for \(\theta_0 = 30^\circ\), the period is already about 1.7% longer than the small-angle prediction. C
Worked example
Problem. A simple pendulum has a length \(L = 1.50\ \mathrm{m}\). It is released from rest at an angle \(\theta_0 = 5^\circ\) from the vertical. Calculate the period of oscillation and the maximum speed of the bob. Assume \(g = 9.81\ \mathrm{m/s^2}\).
- Convert the angle to radians for consistency, though it is not needed for the period calculation in the small-angle limit. \(\theta_0 = 5 \times \frac{\pi}{180} \approx 0.0873\ \mathrm{rad}\).
- Calculate the period using \(T = 2\pi\sqrt{L/g}\). \(T = 2\pi\sqrt{1.50/9.81} = 2\pi\sqrt{0.1529} = 2\pi(0.3910) \approx 2.457\ \mathrm{s}\).
- Calculate the angular frequency \(\omega = 2\pi/T \approx 2.557\ \mathrm{rad/s}\).
- The maximum speed occurs at the equilibrium position where potential energy is minimum and kinetic energy is maximum. Using energy conservation: \(\frac{1}{2}mv_{\text{max}}^2 = mgL(1-\cos\theta_0)\). For small angles, \(1-\cos\theta_0 \approx \theta_0^2/2\). Thus \(v_{\text{max}} \approx \sqrt{gL}\theta_0\).
- Compute \(v_{\text{max}} = \sqrt{9.81 \times 1.50} \times 0.0873 = \sqrt{14.715} \times 0.0873 = 3.836 \times 0.0873 \approx 0.335\ \mathrm{m/s}\).
Units check. Period is in seconds. Speed is \(\sqrt{\mathrm{m/s^2} \cdot \mathrm{m}} = \sqrt{\mathrm{m^2/s^2}} = \mathrm{m/s}\). Correct.
Three quick tests
- If the length of a pendulum is quadrupled, what happens to its period?
Answer
It doubles. Since \(T \propto \sqrt{L}\), \(T_{\text{new}} = \sqrt{4}T = 2T\). - Does the mass of the pendulum bob affect the period in the small-angle approximation?
Answer
No. The mass cancels out in the equation of motion \(\ddot{\theta} + (g/L)\theta = 0\). - Why is the small-angle approximation necessary to treat the pendulum as a simple harmonic oscillator?
Answer
Because the exact restoring force is proportional to \(\sin\theta\), which is non-linear. The approximation \(\sin\theta \approx \theta\) linearizes the equation, making it proportional to \(\theta\).
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-057 · D-058 · Threads: force, energy, waves