## Small angle approximations - an application

• mathematics • teaching • KS5 • PermalinkI thought a bit more about my previous post on small angle approximations, and decided it might be helpful to describe an application of the small angle approximations. While this example contains non-examinable aspects (at least in single maths A-level), the context should be fairly familiar (or can easily be demonstrated), and the mathematics is accessible to single maths students (at least as a demonstration). It also ties together ideas from mechanics and pure maths, so is helpful in this regard.

The question is: what is the period of a pendulum?

We can model the pendulum as a thin rod (inextensible and rigid) of length $L$, freely pivoted about a point $O$, with a single point mass $P$ of mass $m$ on the end of the rod, as shown here (where $T$ is the tension in the rod):

The velocity and acceleration of $P$ are as follows, where $\dot\theta$ means $\dfrac{d\theta}{dt}$ and $\ddot\theta$ means $\dfrac{d^2\theta}{dt^2}$; a derivation of these can be found at the end:

We can now apply Newton’s second law (“$F=ma$”) to the situation: working perpendicular to the rod, this gives $-mg\sin\theta=mL\ddot\theta$ (the minus sign is because the component of the force $mg$ is in the opposite direction to the $L\ddot\theta$ on our diagram). Rearranging this, we get the differential equation:

Unfortunately, this equation is impossible to solve in terms of simple
functions. But if we **assume that the swing of the pendulum is
small**, so that $\theta$ is small, then we can approximate
$\sin\theta$ by $\theta$, and our differential equation becomes

This differential equation (an example of simple harmonic motion) has a solution

(which is easy to check), where $A$ is the amplitude (maximum angle) of the swing. The period of this swing is $2\pi\sqrt{\dfrac{L}{g}}$, which is independent of the amplitude and the mass at the end of the rod! So as long as the swing is relatively small, the period is only dependent upon the length of the pendulum (and the acceleration due to gravity), which is likely to be a surprising result the first time it is met. This would have had great significance for clock-makers in times gone by.

### Deriving the formulae for the velocity and acceleration of $P$

We can work out the velocity and acceleration of $P$ in several different ways. One way is to use coordinates, where $O$ is the origin, and the vertical line is the $y$-axis. Then when $P$ is at an angle of $\theta$, it has a position vector of

A unit vector in the direction of $\overrightarrow{OP}$ is

and a unit vector perpendicular to this in the direction of increasing $\theta$ is

as shown in this diagram:

The velocity of $P$ can be found by differentiating $\mathbf{r}$ with respect to time, giving:

Then the acceleration can be found by differentiating again (using the product rule on both of the components of $\dot{\mathbf{r}}$) to obtain:

These are the components of the velocity and acceleration shown above.

Without as much rigour, one could observe that the distance of $P$ along the circumference of the circle is given by $L\theta$, so it is reasonable to suggest that its speed is $L\dot\theta$ (as $L$ is a constant). Then the acceleration in this direction is plausibly $L\ddot\theta$, while the radial acceleration - which we are not interested in for this application - is a result of the velocity changing direction.