In a park two swings hang side by side from the same bar, with chains of the same length. A small girl sits on one and her father on the other. Both are pulled back a little and let go together. The father is four times as heavy, yet the two swings move forward and back together, swing after swing. Lengthen one pair of chains, and that swing loses step.
The forces on a bob
A simple pendulum is a small heavy bob hung from a fixed point by a light string that does not stretch. At rest the string hangs straight down, and the bob is at the lowest point, the mean position. Pull the bob aside so that the string makes an angle \(\theta\) with the vertical, and let go. The bob moves along a circular arc of radius \(l\), the length of the string.
Two forces act on the bob: the tension \(T_s\) along the string and the weight \(mg\) straight down. Split the weight into two parts. The part \(mg\cos\theta\) lies along the string, away from the pivot. The part \(mg\sin\theta\) lies along the arc, towards the lowest point. The tension lies along the string, across the arc, so it has no part along the arc. Along the arc, the only force is \(mg\sin\theta\).
That component is the restoring force. Measure the displacement along the arc, \(s = l\theta\), positive to the right. When the bob is to the right, \(mg\sin\theta\) points left, back to the lowest point; when it is to the left, the component points right. So the restoring force is \(F = -mg\sin\theta\), with \(\theta = s/l\). The other part, \(mg\cos\theta\), together with the tension, only keeps the bob on its circle.
The small-angle step
The force \(-mg\sin\theta\) is not quite proportional to the displacement \(s = l\theta\). It grows like \(\sin\theta\), which bends away from \(\theta\) as the angle grows. But for small angles, measured in radians, \(\sin\theta\) is almost equal to \(\theta\) itself. Then the restoring force is \(-mg\theta = -(mg/l)\,s\): a straight line through the origin, as for a spring.
The sim shows how good the small-angle step is. At \(15°\) the restoring force is \(2.59\,\text{N}\), and the straight line gives \(2.62\,\text{N}\), only \(1.2\,\%\) more. At \(30°\) the line is \(4.7\,\%\) too high. At \(60°\) it is \(21\,\%\) too high, and the curve has clearly bent away. For swings of up to about \(15°\) the pendulum is, for every practical purpose, a linear oscillator.
\[ F = -mg\sin\theta \]
Why. step 1, the restoring force along the arc is the component of the weight.
\[ F \approx -mg\,\theta = -\dfrac{mg}{l}\,s \]
Why. for small angles \(\sin\theta \approx \theta\), and the arc length is \(s = l\theta\).
\[ k_{\text{eff}} = \dfrac{mg}{l}, \qquad m_{\text{eff}} = m \]
Why. step 2, the force is linear in \(s\), and it accelerates the bob of mass \(m\) along the arc.
\[ \omega^2 = \dfrac{mg/l}{m} = \dfrac{g}{l}, \qquad T = 2\pi\sqrt{\dfrac{l}{g}} \]
Read it. \(\omega^2 = k_{\text{eff}}/m_{\text{eff}}\); the mass cancels.
The mass cancelled, and the reason is worth a sentence. The restoring force is a part of the weight, so it is proportional to the mass. The inertia that resists it is the same mass. A heavier bob is pulled back harder in exactly the proportion that it is harder to move. So all bobs on strings of one length keep the same time, in the same place.
A simple pendulum swings with a small amplitude. Its bob is replaced by one of twice the mass and the same size, on the same string. What happens to the period?
When a simple pendulum is displaced by a small angle, which force pulls the bob back towards the lowest point?
A simple pendulum has a string \(2.5\,\text{m}\) long. Take \(g = 10\,\text{m/s}^2\). What is its period for small swings, in seconds?
For larger swings the exact period is longer than \(2\pi\sqrt{l/g}\), because \(mg\sin\theta\) falls below the straight line. It is longer by \(0.4\,\%\) at an amplitude of \(15°\), by \(1.7\,\%\) at \(30°\), by \(7.3\,\%\) at \(60°\) and by \(18\,\%\) at \(90°\). The pendulum is isochronous only in the limit of small swings.
A pendulum as a clock and as a gravity meter
A seconds pendulum takes exactly one second to swing from one side to the other, so its period is \(2\,\text{s}\). Its length follows from \(T = 2\pi\sqrt{l/g}\): \(l = gT^2/4\pi^2 = g/\pi^2\). With \(g = 9.8\,\text{m/s}^2\) that is \(0.993\,\text{m}\), very nearly one metre. In 1790 this near-coincidence was even proposed as a definition of the metre.
Turn the formula round and a pendulum becomes a gravity meter: \(g = 4\pi^2 l/T^2\). Measure the length and the period, and \(g\) follows. The same fact makes a pendulum clock sensitive to where it stands. Where \(g\) is smaller, as on a high mountain or near the equator, the period is longer. Each tick takes longer, and the clock runs slow.
A pendulum clock keeps correct time in Chennai. It is carried up to a hill station, where \(g\) is slightly smaller. What does the clock do there?
A child on a park swing makes one complete to-and-fro swing in about \(3\) seconds. Roughly how long are the chains, in metres? Take \(g = 10\,\text{m/s}^2\).
Galileo's first biographer, his student Vincenzo Viviani, tells a story about the cathedral of Pisa. There the young Galileo watched a hanging lamp swing, and timed its swings against the beat of his own pulse. The swings grew smaller, but they seemed to take the same time. Whether or not it happened in the cathedral, Galileo did study pendulums for years. In 1656 Christiaan Huygens used his idea to build the first pendulum clock.
A pendulum in a lift
Hang a pendulum inside a lift. If the lift moves at a steady speed, nothing changes, and \(T = 2\pi\sqrt{l/g}\). But let the lift accelerate upward with \(a\). In the frame of the lift, the chapter Accelerated Objects adds a pseudo force \(ma\) downward on every body. The bob feels \(mg\) and \(ma\) both downward, as if gravity were \(g + a\). That is the effective gravity, \(g_{\text{eff}}\).
Nothing else in the derivation changes, so the period becomes \(2\pi\sqrt{l/g_{\text{eff}}}\). Decide what the three pendulums of the next figure will do before you play it.
Three identical pendulums are released together inside three lifts. One lift is at rest, one accelerates upward and one accelerates downward, both at \(5\,\text{m/s}^2\). Which pendulum swings fastest?
The three lifts soon fall out of step. The pendulum in the lift at rest has \(T = 2\pi\sqrt{1/10} = 1.99\,\text{s}\). Accelerating up, \(g_{\text{eff}} = 15\,\text{m/s}^2\) and \(T = 1.62\,\text{s}\). Accelerating down, \(g_{\text{eff}} = 5\,\text{m/s}^2\) and \(T = 2.81\,\text{s}\). If the cable snapped and the lift fell freely, \(g_{\text{eff}}\) would be zero. The bob would feel no pull back at all, and it would stay wherever it was, never swinging.
A pendulum \(1\,\text{m}\) long hangs from the roof of a car that accelerates forward at \(7.5\,\text{m/s}^2\) on a level road. Take \(g = 10\,\text{m/s}^2\). Find the angle of its mean position and its period for small swings.
In the car's frame the pseudo force \(ma\) points backward, at right angles to \(mg\). The effective gravity is \(\sqrt{10^2 + 7.5^2} = 12.5\,\text{m/s}^2\).
The string hangs along \(\vec g_{\text{eff}}\), tilted backward from the vertical by \(\tan^{-1}(7.5/10) = 36.9°\). That tilted line is the new mean position.
The bob swings about it with \(T = 2\pi\sqrt{1/12.5} = 1.78\,\text{s}\), shorter than the \(1.99\,\text{s}\) of a car at rest.
A pendulum has a period of \(2.0\,\text{s}\) in a lift at rest. The lift then accelerates upward at \(2.5\,\text{m/s}^2\). Take \(g = 10\,\text{m/s}^2\). What is the period now, in seconds?
Back to the two swings in the park. Each rider is a bob on a chain, and the mass cancels from \(T = 2\pi\sqrt{l/g}\). So the father and his daughter keep time together, as long as their chains have the same length and the swings stay small. Only the length and the gravity can change the period. That is why lengthening one pair of chains makes that swing lose step.
- Expecting a heavier bob to swing faster or slower.The restoring force and the inertia both grow with \(m\), so \(T = 2\pi\sqrt{l/g}\) has no mass in it.
- Using \(T = 2\pi\sqrt{l/g}\) for large swings.The formula needs \(\sin\theta \approx \theta\). Above about \(15°\) the true period is noticeably longer.
- Using \(g\) inside an accelerating lift or car.Replace \(g\) by \(g_{\text{eff}} = |\vec g - \vec a|\): \(g + a\) going up, \(g - a\) going down, \(\sqrt{g^2 + a^2}\) sideways.