Pendulums

Another way of demonstrating simple harmonic motion is through a pendulum. An object with a mass, \(m\) hanging on a string swaying back and forth repeatedly is a simple example of a pendulum. Applications of pendulum include a swing on the playground and a grandfather's clock.

Consider the mass on the string below.

A force applied to the mass causes it to reach its maximum position from the equilibrium. After the release of the applied force, the mass then turns back to its equilibrium position due to a restoring force caused by gravity. This repeating motion of the mass in the same path is what we call the simple harmonic motion.

SHM in pendulum also demonstrates the relationship between the angular frequency, period and the frequency of the system. The period does not depend on the mass of the object. It depends only on the gravity and the length of the pendulum. An increase in the length of the pendulum increases the period and increasing the field strength or the gravity decreases the period. To show such relationships mathematically, we have the following equations:

For angular frequency: \(\omega=\sqrt {\frac{g}{l}}\), where \(g\) is the acceleration due to gravity and \(l\) is the length of the pendulum.

For period: \(T=\frac{2\pi}{\omega}=2\pi \sqrt{\frac{l}{g}}\).


Example 1.

In the playground, a child is swinging in simple harmonic motion with a period of \(2.5 \;s\). Solve for the length of the swing.

Given: \(T=2.5\;s\)

Solution:

To solve for the period, we take the square of the both sides of the equation \(T=2\pi \sqrt{\frac{l}{g}}\).

\(\begin{pmatrix} T \end{pmatrix}^2=\begin{pmatrix} 2\pi \sqrt{\frac{l}{g}} \end{pmatrix}^2\implies T^2=4\pi^2 \begin{pmatrix} \frac{l}{g} \end{pmatrix}\)

 

thus, the legth of the swing can be solved by the equation \(l=\frac{gT^2}{4\pi^2}\),

\(l=\frac{(9.8\;m/s^2)(2.5\;s)^2}{4\pi^2}=\frac{61.25}{39.48}=1.55\;m\).

The length of the swing is \(1.55\;m\).


Example 2.

A pendulum ride in an amusement park has a height of \(25\;m\). If the ride takes \(0.01 \;\text{cycles/second}\), how long does it take for the ride to complete one cycle of SHM?

Given:

\(l=25\;m\\ f=0.01\;\text{cycles/second}=0.01\;Hz\)


There are two solutions to solve for the period of the ride.

Solution 1.

Using the relationship between the period and the frequency \(T=\frac{1}{f}\).

\(T=\frac{1}{0.01/s}=10\;s\)


Solution 2.

Using the relationship between the length, acceleration due to gravity and the period of the simple harmonic motion \(T=2\pi \sqrt{\frac{l}{g}}\).

\(T=2\pi \sqrt{\frac{25\;m}{9.8\;m/s^2}}=2\pi \sqrt{2.55\;s^2}=10\;s\)

Therefore, the ride's period is \(10\;s\).