# 15.4 Pendulums

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By the end of this section, you will be able to:
• State the forces that act on a simple pendulum
• Determine the angular frequency, frequency, and period of a simple pendulum in terms of the length of the pendulum and the acceleration due to gravity
• Define the period for a physical pendulum
• Define the period for a torsional pendulum

Pendulums are in common usage. Grandfather clocks use a pendulum to keep time and a pendulum can be used to measure the acceleration due to gravity. For small displacements, a pendulum is a simple harmonic oscillator.

## The simple pendulum

A simple pendulum    is defined to have a point mass, also known as the pendulum bob , which is suspended from a string of length L with negligible mass ( [link] ). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string. The mass of the string is assumed to be negligible as compared to the mass of the bob. A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s , the length of the arc. Also shown are the forces on the bob, which result in a net force of − m g sin θ toward the equilibrium position—that is, a restoring force.

Consider the torque on the pendulum. The force providing the restoring torque is the component of the weight of the pendulum bob that acts along the arc length. The torque is the length of the string L times the component of the net force that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the opposite direction of the angular displacement:

$\begin{array}{ccc}\hfill \tau & =\hfill & \text{−}L\left(mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right);\hfill \\ \hfill I\alpha & =\hfill & \text{−}L\left(mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right);\hfill \\ \hfill I\frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & \text{−}L\left(mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right);\hfill \\ \hfill m{L}^{2}\frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & \text{−}L\left(mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right);\hfill \\ \hfill \frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & -\frac{g}{L}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .\hfill \end{array}$

The solution to this differential equation involves advanced calculus, and is beyond the scope of this text. But note that for small angles (less than 15 degrees), $\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ and $\theta$ differ by less than 1%, so we can use the small angle approximation $\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \approx \theta .$ The angle $\theta$ describes the position of the pendulum. Using the small angle approximation gives an approximate solution for small angles,

$\frac{{d}^{2}\theta }{d{t}^{2}}=-\frac{g}{L}\theta .$

Because this equation has the same form as the equation for SHM, the solution is easy to find. The angular frequency is

$\omega =\sqrt{\frac{g}{L}}$

and the period is

$T=2\pi \sqrt{\frac{L}{g}}.$

The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if $\theta$ is less than about $15\text{°}.$ Even simple pendulum clocks can be finely adjusted and remain accurate.

Note the dependence of T on g . If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity, as in the following example.

## Measuring acceleration due to gravity by the period of a pendulum

What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?

## Strategy

We are asked to find g given the period T and the length L of a pendulum. We can solve $T=2\pi \sqrt{\frac{L}{g}}$ for g , assuming only that the angle of deflection is less than $15\text{°}$ .

## Solution

1. Square $T=2\pi \sqrt{\frac{L}{g}}$ and solve for g :
$g=4{\pi }^{2}\frac{L}{{T}^{2}}.$
2. Substitute known values into the new equation:
$g=4{\pi }^{2}\frac{0.75000\phantom{\rule{0.2em}{0ex}}\text{m}}{{\left(1.7357\phantom{\rule{0.2em}{0ex}}\text{s}\right)}^{2}}.$
3. Calculate to find g :
$g=9.8281{\phantom{\rule{0.2em}{0ex}}\text{m/s}}^{2}.$

## Significance

This method for determining g can be very accurate, which is why length and period are given to five digits in this example. For the precision of the approximation $\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \approx \theta$ to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about $0.5\text{°}$ .

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