# 17.6 Simple Pendulum An ideal simple pendulum is a model consisting of a point mass (bob) suspended from a fixed support by a massless, inextensible string. When displaced from its equilibrium position and released, it oscillates in a vertical plane under the influence of gravity. ## Forces on the Pendulum Bob When the pendulum bob of mass $m$ is displaced by an angle $\theta$, two forces act on it: 1. **Tension ($T$):** Exerted by the string, acting along the string towards the support point. 2. **Weight ($mg$):** Acting vertically downward. The weight can be resolved into two perpendicular components: - $mg \cos \theta$: Acts along the string, balancing the tension $T$. - $mg \sin \theta$: Acts tangent to the arc of motion, directed towards the equilibrium position. This is the **restoring force**. The restoring force is given by: $$F = -mg \sin \theta$$ The negative sign indicates that the force is always directed opposite to the displacement, towards the mean position. ## Condition for Simple Harmonic Motion (SHM) For a pendulum to exhibit Simple Harmonic Motion, its acceleration must be directly proportional to its displacement and directed towards the equilibrium position ($a \propto -x$). **Small Angle Approximation:** This condition is met only for small angles of oscillation (typically $\theta < 10^{\circ}$ or 0.2 rad). For such angles, we can approximate $\sin \theta \approx \theta$ (where $\theta$ is in radians). | $\theta$ (degree) | $\theta$ (radian) | $\sin \theta$ | | :---: | :---: | :---: | | 0.00 | 0.0000 | 0.0000 | | 5.00 | 0.0873 | 0.0872 | | 10.00 | 0.1746 | 0.1737 | **Derivation:** 1. Using the approximation, the restoring force becomes $F \approx -mg\theta$. 2. The displacement along the arc is $x = l\theta$, so $\theta = x/l$. 3. Substituting for $\theta$, we get $F = -mg \frac{x}{l}$. 4. From Newton's second law, $F = ma$. 5. Equating the two expressions for force: $ma = -mg \frac{x}{l}$. 6. This simplifies to the acceleration equation: $a = -\frac{g}{l}x$. Since $g$ and $l$ are constants, this shows that $a \propto -x$, which is the defining condition for SHM. ## Time Period of a Simple Pendulum The standard equation for acceleration in SHM is $a = -\omega^2 x$. By comparing this with the pendulum's acceleration equation ($a = -\frac{g}{l}x$), we find the angular frequency $\omega$: $$\omega^2 = \frac{g}{l} \implies \omega = \sqrt{\frac{g}{l}}$$ The time period $T$ is related to $\omega$ by $T = 2\pi/\omega$. Substituting the expression for $\omega$, we get the time period of a simple pendulum: $$T = 2\pi \sqrt{\frac{l}{g}}$$ This shows that the time period depends only on: - The **length of the pendulum ($l$)**. - The **acceleration due to gravity ($g$)**. It is independent of the mass of the bob and the amplitude of oscillation (for small angles). ## Seconds Pendulum A **Seconds Pendulum** is a specific simple pendulum whose time period is exactly 2 seconds. This means it takes 1 second for a single swing (half-vibration) from one extreme to the other. ## Worked Examples **Example 1:** A simple pendulum completes one vibration in one second. What is its length? (Use $g = 9.8 \text{ m/s}^2$) **Solution:** Given $T = 1$ s. Using the formula $T = 2\pi\sqrt{l/g}$: $$1 = 2\pi\sqrt{\frac{l}{9.8}}$$ $$\sqrt{l} = \frac{\sqrt{9.8}}{2\pi}$$ $$l = \left(\frac{\sqrt{9.8}}{2\pi}\right)^2 \approx 0.248 \text{ m} \approx 24.8 \text{ cm}$$ **Example 2:** Find the time period of a 1 m long simple pendulum on the Moon where $g_{moon} = g_{earth}/6$. **Solution:** - On Earth: $T_{Earth} = 2\pi\sqrt{\frac{1}{9.8}} \approx 2.0 \text{ s}$ - On the Moon: $g_{Moon} = \frac{9.8}{6} \approx 1.63 \text{ m/s}^2$ - $T_{Moon} = 2\pi\sqrt{\frac{1}{1.63}} \approx 4.9 \text{ s}$ Alternatively, since $T \propto 1/\sqrt{g}$: $$T_{Moon} = T_{Earth} \sqrt{6} \approx 2.0 \times 2.45 \approx 4.9 \text{ s}$$