Physics · Ch 10 — Oscillations

Oscillations of a simple pendulum in SHM and laws of simple pendulum

10.4.4

Oscillations of a simple pendulum in SHM and laws of simple pendulum

A simple pendulum is a point mass (bob) of mass $m$ hung from a fixed support by a light, inextensible string of length $l$ (measured to the bob's centre of gravity). At equilibrium it hangs vertically; displaced through a small angle and released, it swings to and fro. At any displaced angle $\theta$ , two forces act on the bob: gravity $mg$ (vertically down) and the string tension $T$ (along the string). Resolving gravity along and perpendicular to the string gives a normal component $mg\cos\theta$ (balanced by tension, providing the centripetal force for the bob's circular arc) and a tangential component $mg\sin\theta$ , which always points back towards equilibrium -- this tangential component is the restoring force. Applying Newton's second law along the tangential (arc) direction, with arc-length $s=l\theta$ so $d^2s/dt^2=l\,d^2\theta/dt^2$ , gives $l\dfrac{d^2\theta}{dt^2}=-g\sin\theta$ , i.e. $\dfrac{d^2\theta}{dt^2}=-\dfrac{g}{l}\sin\theta$ -- a non-linear differential equation because of the $\sin\theta$ . Under the small-angle approximation, valid for angular amplitudes up to roughly $10°$ , $\sin\theta\approx\theta$ (radians), which linearises the equation to $\dfrac{d^2\theta}{dt^2}=-\dfrac{g}{l}\theta$ , the standard SHM form. Comparing gives $\omega^2=g/l$ , so $\omega=\sqrt{g/l}$ , $f=\dfrac{1}{2\pi}\sqrt{g/l}$ Hz, and the time period $T=2\pi\sqrt{l/g}$ seconds. The laws of the simple pendulum: (i) Law of length -- for fixed $g$ , $T\propto\sqrt{l}$ ; (ii) Law of acceleration due to gravity -- for fixed $l$ , $T\propto1/\sqrt{g}$ ; the period is also (iii …

Figure 10.21Simple pendulum

What this figure shows. A four-panel figure shows a simple pendulum: panel (a) at its vertical equilibrium position with length l marked from the fixed support O to the bob; panel (b) and (c) showing it displaced through a small angle; and panel (d) a free-body diagram at the displaced position showing the tension T along the string toward the support, the weight mg acting straight down, and this weight resolved into a component mg cos(theta) along the string (balanced by tension, providing centripetal force) and a component mg sin(theta) tangential to the arc of swing, which is the actual re …

Misc Example 10.13Validity of the small-angle approximation

Worked out. The question asks to discuss the small-angle approximation used in the simple pendulum derivation, where sin(theta) is replaced by theta (in radians). A table compares theta in degrees, its value in radians, and sin(theta) side by side, from 0 up to 45 degrees in 5-degree steps: at 5 degrees, theta in radians (0.087) and sin(theta) (0.087) already agree to three decimal places; at 10 degrees they still agree closely (0.174 vs 0.174); by 30 degrees the values (0.524 radians vs sin = 0.500) have started to visibly diverge, and by 45 degrees the gap (0.785 vs 0.707) is substantial. The conclusion is that for angular amplitudes up to roughly 10 degrees, replacing sin(theta) with theta introduces negligible error, which is exactly the regime the whole simple-pendulum period formula T = 2 pi sqrt(l/g) is valid in; beyond that, sin(theta) becomes noticeably smaller than theta an …

Misc Example 10.14Percentage change in period from a 44% length increase

Worked out. The length of a simple pendulum is increased by 44% from its original length, and the task is to find the percentage increase in its time period. Since T is proportional to the square root of length, T_final/T_initial = sqrt(l_final/l_initial) = sqrt(1.44) = 1.2 exactly, because 1.44 is a perfect square (1.2 squared). So the new time period is 1.2 times the original, i.e. T_final = T_initial + 20% of T_initial -- a 44% increase in length produces only a 20% increase in period, a direct consequence of the square-root (rather than line …

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