**Simple Harmonic Motion**

Consider a child sitting on a swing which is in equilibrium position. If a force is applied to pull the swing, it sways forward after releasing it. A returning force will then pull back the swing to its equilibrium position causing a back and forth motion. This type of motion along the same path is called S**imple Harmonic Motion (SHM)**. SHM is the repeating back and forth motion of an object due to a restoring force - a force which is pulling the object back to equilibrium position.

Simple harmonic motion is measured by the amplitude, frequency, and period. **Amplitude **is the maximum distance of the body from the equilibrium position. **Period **is the time needed to repeat one cycle of motion. **Frequency **is the number of cycles per second. A cycle is the complete back and forth motion of the object. To be able to solve these, the minimum as well as the maximum position or displacement, velocity and acceleration of the object must be clearly identified.

Consider an object hanged on a string. If the object is pulled to the right by an applied force, it moves to a maximum displacement - the positive amplitude. After it is released, the object will pass through its equilibrium position continuing to move to the left side of the equilibrium position - the negative amplitude. The continuing back and forth motion of the object following the same path represents simple harmonic motion. The period or one cycle motion of the object can be expressed mathematically as \(T=\frac {1}{f}\) and the the number of cycles per second is expressed as \(f=\frac{1}{T}\).

Simple harmonic motion relative to the uniform circular motion is also decribed as a one-dimensional motion of an object moving in uniform circular motion. The angular displacement of the object is given by \(x=A\cos\theta=A\cos(\omega t)\) where \(A\) is the amplitude, \(\omega\) is the angular frequency and \(t\) is the time.

The relationship between the angular frequency, frequency, and the period can be expressed as \(\omega=2\pi f=\frac{2\pi}{T}\).

**Example 1.**

Anna is riding on a swing. If it takes 15 seconds for the swing to sway back and forth, what is its frequency?

\(f=\frac{1}{T}=\frac{1}{15\;s}=0.07/s\;\text{or}\;0.07\;Hz\)

thus, the swing takes 0.07 cycle per second.

**Example 2.**

If the swing in Example 1 takes 0.1 cycle per second, how long does it take to complete one cycle?

\(T=\frac{1}{f}=\frac{1}{0.1\;s}=10\;s\)

therefore, it takes 10 seconds for the swing to complete one cyle.

**Example 3.**

An object is in simple harmonic motion. What is the angular frequency of the object if takes 70 seconds to complete one cycle of shm? If the object's maximum displacement is 1.0 m, where is its position in 20 seconds?

Given:

\(T=70\;s\\ A=1.0\;m\\ t=20\;s\)

Solution 1.

To solve for angular frequency, we use the equation showing the relationship between angular frequency and period.

\(\omega=\frac{2\pi}{T}=\frac{2\pi}{70\;s}=0.03\pi\;\frac{rad}{s}\)

Solution 2.

Solve for the position of the object at \(20\;s\) using \(x=A\cos(\omega t)\).

\(x=(1.0\;m)\cos(0.03\pi(20\;s))=-0.31\;m\).

Thus, at \(t=20\;s\), the object is found \(0.31\;m\) away from the equilibrium position. The negative sign indicates that the object is in negative side or opposite direction of the equilibrium.