The pendulum of a wall clock in a school corridor swings through only a few degrees. Its period is \(2\pi\sqrt{l/g}\), whether it swings through \(3°\) or \(8°\). That formula rests on one step: the chapter Dynamics of SHM replaces \(\sin\theta\) by \(\theta\) itself. Why is that allowed? And how large may the swing be before the step goes wrong?
Three lengths on one circle
Draw a small angle \(\theta\) at the centre of a circle of radius 1. Three lengths belong to it. The height of the tip of the radius above the axis is \(\sin\theta\). The arc from the axis to the tip is \(\theta\) itself, because on a circle of radius 1 the arc equals the angle in radians. The tangent to the circle at the axis, cut off by the extended radius, is \(\tan\theta\).
The height is always a little shorter than the arc, and the arc a little shorter than the tangent: \(\sin\theta < \theta < \tan\theta\). As the angle shrinks, the arc straightens and the three lengths close up on one another. For a small angle, all three may be written as \(\theta\). The cosine is the width, which stays close to 1. The gap \(1 - \cos\theta\) shrinks like the square of the angle, and is close to \(\theta^2/2\).
For an angle \(\theta\) measured in radians and small compared with 1, \(\sin\theta \approx \theta\), \(\tan\theta \approx \theta\) and \(\cos\theta \approx 1 - \theta^2/2\). The approximations hold in radians only.
How the error grows
A more careful calculation, from the chapter Differentiation, gives the next term: \(\sin\theta = \theta - \theta^3/6 + \dots\). So \(\theta\) is larger than \(\sin\theta\) by about \(\theta^2/6\) of its value. The error grows as the square of the angle. Double the angle and the error becomes four times as large.
The numbers are worth remembering. At \(5°\), \(\theta\) is above \(\sin\theta\) by only \(0.13\,\%\). At \(10°\) the gap is \(0.51\,\%\), and at \(15°\) it is \(1.2\,\%\). At \(30°\) it is \(4.7\,\%\), and at \(60°\) it is \(21\,\%\). The tangent parts company faster, as \(\theta^2/3\): \(\tan\theta\) is \(2.3\,\%\) above \(\theta\) at \(15°\). The cosine rule \(1 - \theta^2/2\) is far better, only \(0.02\,\%\) off at \(15°\).
A bob of mass \(1\,\text{kg}\) hangs on a string \(1\,\text{m}\) long. Take \(g = 10\,\text{m/s}^2\). The restoring force along the arc is \(mg\sin\theta\). Compare it with the straight line \(mg\theta\) at \(15°\), \(30°\) and \(60°\). Then find the period for small swings.
\(mg = 10\,\text{N}\). In radians the angles are \(0.2618\), \(0.5236\) and \(1.0472\).
Force \(F = mg\sin\theta\); line \(F_{\text{line}} = mg\theta\); excess \(F_{\text{line}}/F - 1 = \theta/\sin\theta - 1\).
At \(15°\): \(F = 10 \times 0.2588 = 2.59\,\text{N}\) and \(F_{\text{line}} = 2.62\,\text{N}\), \(1.2\,\%\) more. At \(30°\) the line is \(4.7\,\%\) too high, and at \(60°\) it is \(21\,\%\) too high.
Up to about \(15°\) the restoring force is a straight line, \(-(mg/l)s\), within about one per cent. The two steps of SHM then give \(T = 2\pi\sqrt{l/g} = 2\pi\sqrt{0.1} = 1.99\,\text{s}\).
The clock in the corridor now has its answer. A swing of \(8°\) is \(0.140\) rad, and there \(\theta\) exceeds \(\sin\theta\) by only \(0.33\,\%\). So the straight-line force, and with it \(T = 2\pi\sqrt{l/g}\), is as good as exact. A swing of \(30°\) would already be out by nearly five per cent in the force.
A student writes \(\sin 2° \approx 2\). What went wrong?
A slit \(0.2\,\text{mm}\) wide is lit by light of wavelength \(500\,\text{nm}\). The first dark band is at \(\sin\theta = \lambda/a\), on a screen \(2\,\text{m}\) away. With \(\sin\theta \approx \tan\theta \approx \theta\), how far is it from the centre of the screen, in mm?
At \(15°\), \(\theta\) is about \(1.2\,\%\) larger than \(\sin\theta\). About how large is the gap at \(30°\)?
The height of a swinging bob
The cosine approximation answers a height question. A bob on a string of length \(l\) swings out through an angle \(\theta\). Its height above the lowest point is \(h = l - l\cos\theta = l(1 - \cos\theta)\). For a small angle, \(1 - \cos\theta \approx \theta^2/2\), so \(h \approx l\theta^2/2\). The rise grows as the square of the swing, and it is much smaller than the sideways movement \(l\theta\).
A pendulum \(1\,\text{m}\) long swings out to \(0.1\) rad. Use \(\cos\theta \approx 1 - \theta^2/2\) to find how far the bob rises above its lowest point, in mm.
Getting the angle back
Physics often knows a ratio and wants the angle. The inverse sine \(\sin^{-1}x\) is the angle whose sine is \(x\). In the same way, \(\cos^{-1}x\) and \(\tan^{-1}x\) undo the cosine and the tangent. So \(\sin^{-1}0.5 = 30°\), \(\tan^{-1}1 = 45°\) and \(\tan^{-1}(3/4) = 36.9°\). The \(-1\) is not a power; \(\sin^{-1}x\) is not \(1/\sin x\).
One sine belongs to two angles between \(0°\) and \(180°\), because \(\sin(180° - \theta) = \sin\theta\). A calculator gives only one of them, the principal value between \(-90°\) and \(90°\). The other, \(180°\) minus the first, may be the one the physics needs. So after every inverse sine, ask whether the second angle is also possible.
A ball thrown at \(20\,\text{m/s}\) must land \(20\,\text{m}\) away on level ground. Take \(g = 10\,\text{m/s}^2\). At what angles can it be thrown?
The range is \(R = u^2\sin 2\theta/g\), with \(u^2 = 400\) and \(R = 20\,\text{m}\).
\(\sin 2\theta = Rg/u^2 = 20 \times 10/400 = 0.5\).
\(2\theta = \sin^{-1}0.5 = 30°\), or \(2\theta = 180° - 30° = 150°\). So \(\theta = 15°\) or \(\theta = 75°\).
A flat throw at \(15°\) and a lob at \(75°\) both land \(20\,\text{m}\) away. The two angles add to \(90°\), as complementary angles of projection always do.
A ball thrown at \(20\,\text{m/s}\) must land \(20\sqrt3 \approx 34.6\,\text{m}\) away on level ground. Take \(g = 10\,\text{m/s}^2\). Find the larger of the two possible angles of projection, in degrees.
The small-angle approximations fail as the angle grows. Past about \(15°\) the error of \(\sin\theta \approx \theta\) is over one per cent, and the tangent fails sooner, past about \(10°\). For a swing of \(30°\), the true period of a pendulum is \(1.7\,\%\) longer than \(2\pi\sqrt{l/g}\). The approximations hold only in radians; in degrees they are simply wrong. An inverse sine has two answers between \(0°\) and \(180°\), and the calculator shows one.
Dynamics of SHM uses \(\sin\theta \approx \theta\) for the simple pendulum, the physical pendulum, a dipole in a field and a magnet in a field. Each of them becomes an SHM only for small swings. Propagating Waves on a String uses it for the slope of a string. The optics chapters use it for paraxial rays, the apparent depth of a coin in water and the thin prism. They use it again for the fringe width of Young's experiment and the central maximum of a single slit. Elasticity uses it for a shear strain. Circular motion uses it for a small banking angle. Motion in Two Dimensions finds the angle of projection from a range with an inverse sine.
- Using the small-angle approximation with the angle in degrees.Convert first: \(15° = 0.262\) rad. In degrees, \(\sin\theta \approx \theta\) is false.
- Using \(\sin\theta \approx \theta\) for a swing of \(30°\) or more.The error is 4.7 % at \(30°\). Keep the exact sine, or say that the result is approximate.
- Reading \(\sin^{-1}x\) as \(1/\sin x\).\(\sin^{-1}x\) is the angle whose sine is \(x\). The reciprocal \(1/\sin x\) is \(\text{cosec}\,x\).
- Keeping only the calculator's angle.Check \(180°\) minus that angle as well; a range, a refraction or a triangle may need it.