Suppose we restrict the pendulum’s oscillations to small angles of less than about $10\text{°}$. This restriction allows us to make use of an interesting and important piece of geometry.

FIGURE 15.18 shows an angle $\theta$ and a circular arc of length $\text{s}=r\theta$. A right triangle has been constructed by dropping a perpendicular from the top of the arc to the axis. The height of the triangle is $h=r\text{sin}\theta$. Suppose that the angle $\theta$ is “small,” which, in practice, means $\theta \ll 1$ rad. In that case there is very little difference between h and s. If $h\approx \text{s}$, then $r\text{sin}\theta \approx r\theta$. It follows that

FIGURE 15.18 The geometrical basis of the small-angle approximation.

The result that $\text{sin}\theta \approx \theta$ for small angles is called the small-angle approximation. We can similarly note that $l\approx r$ for small angles. Because $l=r\text{cos}\theta$, it follows that

Finally, we can take the ratio of sine and cosine to find $\text{tan}\theta \approx \text{s}in\theta \approx \theta$. We will have other occasions to use the small-angle approximation throughout the remainder of this text.

How small does $\theta$ have to be to justify using the small-angle approximation? It’s easy to use your calculator to find that the small-angle approximation is good to three significant figures, an error of $\le 0.1\%$, up to angles of $\approx 0.10\text{rad }(\approx 5\text{°})$. In practice, we will use the approximation up to about $10\text{°}$, but for angles any larger it rapidly loses validity and produces unacceptable results.

If we restrict the pendulum swing to angles $\theta <10\text{°}$, we can use $\text{sin}\theta \approx \theta$. In that case, Equation 15.44 for the net force on the mass is

where, in the last step, we used the fact that angle $\theta$ is related to the arc length by $\theta =s\text{/}L$. Then the equation of motion becomes

This is exactly the same as Equation 15.32 for a mass oscillating on a spring. The names are different, with x replaced by s and $k\text{/}m$ by $g\text{/}L$, but that does not make it a different equation.

Because we know the solution to the spring problem, we can immediately write the solution to the pendulum problem just by changing variables and constants:

The angular frequency

is determined by the length of the string. The pendulum is interesting in that the frequency, and hence the period, is independent of the mass. It depends only on the length of the pendulum. The amplitude A and the phase constant ${\varphi }_0$ are determined by the initial conditions, just as they were for an oscillating spring.

EXAMPLE 15.8 The gravimeter

Deposits of minerals and ore can alter the local value of the free-fall acceleration because they tend to be denser than surrounding rocks. Geologists use a gravimeter—an instrument that accurately measures the local free-fall acceleration—to search for ore deposits. One of the simplest gravimeters is a pendulum. To achieve the highest accuracy, a stopwatch is used to time 100 oscillations of a pendulum of different lengths. At one location in the field, a geologist makes the following measurements:

Length (m)	Time (s)
0.500	141.7
1.000	200.6
1.500	245.8
2.000	283.5

What is the local value of $g?$

MODEL Assume the oscillation angle is small, in which case the motion is simple harmonic motion with a period independent of the mass of the pendulum. Because the data are known to four significant figures ($\pm 1\text{mm}$ on the length and $\pm 0.1\text{ s}$ on the timing, both of which are easily achievable), we expect to determine g to four significant figures.

SOLVE From Equation 15.48, using $f=1\text{/}T$, we find

$$T^2={\left(2\pi \sqrt{{\displaystyle \frac{L}{g}}}\right)}^2={\displaystyle \frac{4{\pi }^2}{g}}L$$

That is, the square of a pendulum’s period is proportional to its length. Consequently, a graph of $T^2$ versus $L$ should be a straight line passing through the origin with slope $4{\pi }^2\text{/}g$. We can use the experimentally measured slope to determine $g$. FIGURE 15.19 is a graph of the data, with the period found by dividing the measured time by 100.

FIGURE 15.19 Graph of the square of the pendulum’s period versus its length.

FIGURE 15.19 Full Alternative Text

As expected, the graph is a straight line passing through the origin. The slope of the best-fit line is $4.021{\text{s}}^2\text{/}\text{m}$. Consequently,

$$g={\displaystyle \frac{4{\pi }^2}{\text{slope}}}={\displaystyle \frac{4{\pi }^2}{4.021{\text{s}}^2\text{/}\text{m}}}=9.818\text{m}\text{/}{\text{s}}^2$$

REVIEW The fact that the graph is linear and passes through the origin confirms our model of the situation. Had this not been the case, we would have had to conclude either that our model of the pendulum as a simple, small-angle pendulum was not valid or that our measurements were bad. This is an important reason for having multiple data points rather than using only one length.